7 concepts18 worked examples32 exercises5 exam-level7 figures
What are you here for?
13Angular momentum, the law that survives when the shape changes, and rotation about an axis that is free to move
A diver leaves the board stretched out and straight, folds into a tight tuck, turns two and a half times, opens out and drops into the water. From the instant her feet lose the board nothing touches her but air. Play the video back frame by frame and time the turns: in the tuck she goes round roughly three times faster than she was on the way up, and she slows again the moment she opens out. Nothing pushed her, and the rate changed twice.
By the end of this section you can predict the factor by which her rate of turning goes up from her shape alone, without knowing her mass, and say exactly where the extra kinetic energy came from and who paid for it.
In 60 seconds
A turning body carries $L = I\omega$ about a stated axis; the honest second law is $\sum\tau = dL/dt$, which allows the moment of inertia to change while $\sum\tau = I\alpha$ does not; with no external torque about an axis, $L$ about it is constant while the kinetic energy generally is not; and once the axis is free to point anywhere, torque and angular momentum become vectors built with the cross product, which is what makes a top go round instead of falling.
Angular momentum of a body about a fixed axis
$$L = I\omega$$
one rigid body, one named axis, and the moment of inertia taken about that same axis
The second law in its general form
$$\sum\tau = \frac{dL}{dt}$$
always; it collapses to $\sum\tau = I\alpha$ only while $I$ is constant
$$\int \tau \, dt = \Delta L$$
the torque varies with time, or you are given a torque and a duration and asked for a change in spin
a top or a wheel whose spin is much faster than the rate at which its axis swings round
Three most common mistakes
Writing an angular momentum without saying which axis or point it is about. The same wheel has three different values of $L$ about three different points, and comparing two instants only means something if both use the same one.
Assuming that conserved angular momentum means conserved kinetic energy. When a skater pulls her arms in, $L$ is unchanged and $K$ goes up, the difference being work she did with her arms.
Using $\sum\tau = I\alpha$ when the moment of inertia is changing. That form came from pulling a constant $I$ out of a derivative, so once the shape changes it is the wrong equation; go back to $\sum\tau = dL/dt$.
This is the last piece of rotation in the course and the natural home of the long question: a collision onto a pivoted body, then an energy line for how far it swings. The syllabus puts 25% on the final and 20% on each midterm, and nothing here claims more. If you drill one thing, drill deciding which law governs which stage, because that is where the marks are lost, not in the arithmetic.
How much time do you have?
10 minutes
You leave able to write the angular momentum of a body on an axle, to recognise the one situation in which it cannot change, and to run the two-line calculation for a shape change. That is the quiz-sized version of most of this page.
In 60 seconds · Formula card · Angular momentum about a fixed axis · When nothing can change it · Mistake ledger
45 minutes
You add the two things that separate a pass from a good mark: the general second law, the only form that survives a changing shape, and the collision-onto-a-pivot setup the long exam question is built from.
In 60 seconds · Conventions used here · Recall first · Angular momentum about a fixed axis · The second law that does not assume a fixed shape · When nothing can change it · Method boxes · Scaffolding comes off · Full exam-style question · Practice B · Check yourself
full read
Everything above plus the vector machinery: the cross product and how to get a direction out of it, the angular momentum of a particle that is not going round anything, what happens when the axis is free to move, and why a top circles instead of falling.
The opening pages · Recall first · Try it yourself first · Notation · All seven concept blocks · Method boxes · Contrast pair · Scaffolding comes off · Full exam-style question · Practice A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
Compute the angular momentum of a rigid body turning about a stated fixed axis, quote it with its axis and its units, and say how the same body can carry two different values about two different axes.
Apply $\sum\tau = dL/dt$ and its integrated form $\int\tau\,dt = \Delta L$ to a body acted on by a torque that varies with time, and explain why the constant-acceleration equations of the previous section are not available there.
Solve a shape-change or a rotational collision problem with $I_1\omega_1 = I_2\omega_2$, then decide separately whether the kinetic energy went up, went down or stayed put, and name who did the work.
Evaluate a cross product both from magnitudes and angles and from components, use the to get the direction of $\vec{\tau}$ and $\vec{L}$, and state the sign convention that makes a plane problem a special case of it.
Calculate the angular momentum of a single particle about a stated origin, including a particle moving in a straight line, and connect its rate of change to the torque of the force acting on it.
Split the angular momentum of a body that travels and spins at once into a centre of mass part and a spin part, and show by example that $\vec{L}$ need not point along $\vec{\omega}$ once the axis is not an axis of symmetry.
Predict the rate and the sense of the steady precession of a fast spinning top or wheel from its weight, its spin and its geometry, and state the condition under which that prediction may be trusted.
Syllabus coverage
Angular Momentum
Angular momentum of a rigid body about a fixed axis; the general form of the second law for rotation and angular impulse; conservation of angular momentum under shape changes and in rotational collisions; the angular momentum of a single particle and of a system of particles
The week line carries no chapter numbers, so none is quoted anywhere here. The scope taken is the standard content of that title in the set textbook.
covered
General Rotation
The vector cross product and the right hand rule; torque and angular momentum as vectors; rotation about an axis that is not fixed and not an axis of symmetry, where the angular momentum is not parallel to the angular velocity; the splitting of the total angular momentum into a centre of mass part and a spin part; steady precession of a fast spinning body
The previous section left two items explicitly to this one: the vector treatment with the right hand rule, and rotation about an axis that moves. Both are taken up here, which is why the vector machinery gets a block of its own.
covered
and the wobble of a real top
The nodding of a top's axis that rides on top of the steady precession
Named in two sentences so that a student who has watched a real top does not think the page is lying, then dropped. Nothing here calculates it.
off_syllabus
Repeating swings of a body about a pivot
How long a pivoted body takes to swing back and forth and whether the motion repeats
Deferred to the next section, where repeating motion is treated. The pivoted rod here is followed once, from just after a collision to the instant it stops rising; nothing asks how long that took.
deferred
Angular momentum in systems where the two bodies are not rigid
Bodies that deform continuously rather than snapping between two shapes
Every shape change here is treated as a jump between one rigid configuration and another, which is all the conservation law needs. Naming the idealisation costs one sentence.
off_syllabus
Recall first
Torque about a stated axis
$\tau = rF\sin\theta = r_{\perp}F$, where $r_{\perp}$ is the perpendicular distance from the axis to the line the force acts along, and the sign is fixed by whichever sense of turning was declared positive.
Everything here says what torques do to angular momentum, so the torque itself has to be computed first.
Moment of inertia and the parallel axis shift
$I = \sum m_i r_i^{2}$ about a stated axis, and $I = I_{\rm cm} + Md^{2}$ for an axis parallel to one through the centre of mass at a distance $d$. Standard values: disc or solid cylinder $\tfrac12 MR^{2}$, hoop $MR^{2}$, solid sphere $\tfrac25 MR^{2}$, rod about its centre $\tfrac{1}{12}ML^{2}$, rod about one end $\tfrac13 ML^{2}$.
The first half of this page is $L = I\omega$, and half the marks usually sit in getting $I$ right about the axis the question names.
The second law for a body on a fixed axle
$\sum\tau = I\alpha$, with all torques taken about the same fixed axis and the body keeping its shape.
This page generalises it, and seeing where the fixed-shape assumption entered is what makes the generalisation necessary rather than decorative.
Rotational kinetic energy and the rolling condition
$K_{\rm rot} = \tfrac12 I\omega^{2}$, a rolling body carries $K = \tfrac12 Mv_{\rm cm}^{2} + \tfrac12 I_{\rm cm}\omega^{2}$, and while it rolls without slipping $v_{\rm cm} = \omega R$.
Several questions here conserve angular momentum in one stage and mechanical energy in the next, and the energy line comes from the previous two sections.
Linear momentum and the perfectly inelastic collision
$\vec{p} = m\vec{v}$, and with no net external force the total is unchanged; when two bodies lock together the momentum survives the collision and the kinetic energy does not.
The rotational collisions here are built on this pattern, and the interleaved questions mix the two so you have to choose which one applies.
Components of a vector and the dot product
A vector in the plane is written $\vec{A} = A_x\hat{\imath} + A_y\hat{\jmath}$, its magnitude is $\sqrt{A_x^{2}+A_y^{2}}$, and $\vec{A}\cdot\vec{B} = AB\cos\theta$ returns a number.
The cross product introduced here is the other way of multiplying two vectors, best held on to by contrast with the one you already own.
Uniform circular motion
A point going round a circle of radius $r$ at speed $v$ completes one circuit in a time $2\pi r/v$, and the equivalent statement for a rate of turning $\Omega$ is that one circuit takes $2\pi/\Omega$.
The axis of a precessing top travels round a circle at a steady rate, and the last block turns that rate into a stopwatch time.
Try it yourself first (3 questions)
1§13.0 — what a spinning skater keeps and what she does not●●○○○
Nobody is expected to get these three right before reading the section; they are here so you find out in ninety seconds which tools from the last two sections you are shaky on. A skater turns on the spot on ice with her arms stretched out, then pulls them in and speeds up visibly. A classmate argues that since the ice is smooth and nothing outside her turns her, her kinetic energy at the end must equal what it was at the start.
Given
smooth ice, so no friction torque about the vertical axis through her
she pulls her arms in using her own muscles
her rate of turning is seen to increase
Find
(a) Decide whether the classmate is right, and say in one sentence what settles it.
Hint 1/4
You are not being asked to compute anything. Ask instead whether anything did work on the skater between the two instants, and remember that a force can do work without exerting any torque about the axis.
Hint 2/4
Work is done whenever a force acts along the displacement of its point of application. Her arms move inwards while she pulls them inwards.
Hint 3/4
The situation again: smooth ice, no outside turning effect, her own muscles pulling her arms from far out to close in, and a visible increase in the rate of turning.
Hint 4/4
The classmate is wrong: her muscles did work on her own arms, and her kinetic energy is larger at the end.
Show solutionSeparate the two claims
$$\text{no external torque} \;\Rightarrow\; \text{something about the turning is fixed}$$
the absence of an outside turning effect is a statement about the turning, and this page is about naming exactly which quantity it fixes
$$\text{no external torque} \;\not\Rightarrow\; \Delta K = 0$$
energy bookkeeping asks who did work, not who exerted a torque, and those are different questions
Find the agent that did the work
$$W_{\rm muscles} = \Delta K > 0$$
the arms move inwards while being pulled inwards, so the muscle force acts along the displacement of its point of application
Answer $$\boxed{\;\text{False: } K \text{ increases, and the skater's own muscles paid for it}\;}$$
Check
Push it to the extreme: if pulling the arms in cost nothing, a skater could pull in and push out repeatedly and spin faster every time, getting energy from nowhere. That it is exhausting is the physical evidence that work is being done.
2§13.0 — the second law for a disc on an axle●●○○○
A uniform disc is mounted on a fixed horizontal axle through its centre and is free to turn. A light cord is wrapped round its rim and pulled steadily with a constant force, tangentially to the rim. The axle is smooth.
cord pulled tangentially with $F = 12.0\ \mathrm{N}$
disc about its own centre: $I = \tfrac12 MR^{2}$
smooth axle, so no friction torque
Find
(a) Find the angular acceleration of the disc.
Hint 1/4
Two quantities are needed before the second law can be used at all: the turning effect of the pull about the axle, and the disc's resistance to being turned about that same axle.
Hint 2/4
$\tau = r_{\perp}F$ with the lever arm equal to the full radius for a tangential pull, $I = \tfrac12 MR^{2}$ for a disc about its centre, and $\sum\tau = I\alpha$.
Hint 3/4
The numbers again: $M = 5.00$ kg, $R = 0.400$ m, $F = 12.0$ N pulled tangentially, smooth axle.
Order of magnitude: a rim point has tangential acceleration $R\alpha = 4.80\ \mathrm{m/s^{2}}$, about half of $g$, which is what a 12 N pull on a 5 kg object should look like. An answer in the hundreds would mean the radius was used the wrong number of times.
3§13.0 — a collision in which the two bodies lock together●●○○○
A lump of putty slides along a smooth horizontal table, hits a wooden block that is standing still, and sticks to it. The two then slide off together. Nothing else touches them horizontally.
Given
putty: $m_1 = 0.600\ \mathrm{kg}$ at $u_1 = 9.00\ \mathrm{m/s}$
block: $m_2 = 2.40\ \mathrm{kg}$, at rest
they stick together on contact
smooth horizontal table
Find
(a) Find the speed of the pair just after the collision.
(b) Find the kinetic energy lost in the collision.
Hint 1/4
Two different quantities are in play and only one of them survives the collision. Deciding which one is the entire question; the arithmetic afterwards is a single division.
Hint 2/4
With no external horizontal force the total momentum is unchanged; when bodies lock together the kinetic energy is not, so it must be computed separately before and after.
Hint 3/4
The data again: 0.600 kg at 9.00 m/s meeting 2.40 kg at rest on a smooth table, sticking together.
Hint 4/4
They move off at 1.80 m/s, and 19.4 J of kinetic energy has gone.
afterwards a single body of the combined mass moves at the common speed
$$\Delta K = 4.86 - 24.3 = -19.4\ \mathrm{J}$$
the loss goes into deforming and warming the putty, which is what sticking together means physically
Answer $$\boxed{\;v = 1.80\ \mathrm{m/s},\qquad \Delta K = -19.4\ \mathrm{J}\;}$$
Check
Independent route to the loss: for a perfectly inelastic collision with one body at rest the fraction of kinetic energy kept is $m_1/(m_1+m_2) = 0.600/3.00 = 0.200$, so 80% is lost, and $0.800\times 24.3 = 19.4$ J. Same number from a formula that never mentions the final speed.
Notation
symbol
reads as
means
watch out
$L$
capital L
angular momentum about a stated axis or point, in kg m squared per second
Never write it without saying about what. The same body has different values about different points, and the symbol carries no memory of which one you meant.
$\vec{L}$
L vector
angular momentum as a vector, pointed by the right hand rule
Its direction lies along the axis, not along the motion of any piece of the body. Nothing physically points that way.
$\Delta L$
delta L
the change in angular momentum between two named instants
It equals the angular impulse. A large torque acting briefly and a small one acting for a long time give the same value.
$\times$
cross
the cross product of two vectors, giving a third vector perpendicular to both
Order matters, and reversing it flips the sign. It is not the dot product from the work section, which gives a number and has no direction.
$\phi$
phi
the angle between the position vector from the origin and the velocity of a particle
Measured tail to tail. A particle heading straight at the origin or away from it has $\phi$ of zero or 180 degrees, and so no angular momentum about it.
$r_{\perp}$
r perpendicular
the perpendicular distance from the chosen origin to the line along which the momentum or the force lies
The same idea as the lever arm for torque, now applied to a velocity. It is not the distance to the particle unless the two happen to be at right angles.
$\Omega$
capital omega
the rate at which the axis of a precessing body swings round, in rad/s
A different quantity from $\omega$, the rate at which the body spins about that axis. Everything on this page assumes the second is much larger than the first.
$\vec{L}_{\rm cm}$
L c m
the angular momentum a body has about its own centre of mass, that is, its spin part
Adding it to the centre of mass part gives the total about an outside point. Using only one of the two is the standard way to lose half the answer.
$I_{\rm cm}$
I c m
the moment of inertia about an axis through the centre of mass
Carried over unchanged from the previous section. Still meaningless without naming the axis direction as well as the point.
$\hat{k}$
k hat
the unit vector along the z axis, out of the page in every figure here
A torque or an angular momentum written as a multiple of it is out of the page when the number is positive and into the page when it is negative.
Conventions used here
Naming the axis or the point before any angular momentum is written
Every $L$ and every $\tau$ here is written with the axis or origin it is taken about said out loud: about the axle, the pivot, the contact point, the centre of mass. When two instants are compared, both use the same one, and if the question does not name it then the first line of the solution does. Angular momentum being conserved is always short for conserved about a particular point.
A rolling wheel carries 1.20 units about its own centre and 3.60 about the ground point under it. Comparing a before-value about one with an after-value about the other compares two different quantities.
Signs in a plane figure, and directions in space
While the axis is fixed and the picture is flat, counterclockwise as drawn is positive and clockwise negative, as in the previous section. When a direction in space is wanted, the same information sits in a vector along the axis, pointed by the right hand rule: curl the fingers the way the body turns and the thumb gives the direction. The two agree, since counterclockwise in a page with $x$ right and $y$ up points out of the page.
Half of this section is written with signs and half with vectors, and a student who does not see that these are one convention will think the page changed its mind.
Radians as the only angular unit allowed inside these formulas
Angular velocities enter every formula here in rad/s and nothing else. Revolutions per minute, revolutions per second and degrees per second are converted the moment they arrive: multiply rpm by $2\pi/60$, and revolutions per second by $2\pi$. Answers are turned back into revolutions only at the very end, and only when the question asks for turns.
A spin rate left in rpm inside $L = I\omega$ makes the answer wrong by a factor of about ten.
The value of g and the rounding carried through every answer here
$g = 9.80\ \mathrm{m/s^{2}}$ throughout, taken positive, with direction carried by the signs in the equations. Final answers are quoted to three significant figures, and intermediate values are kept longer inside the calculation so the rounding does not move the last digit.
Two different values of $g$ in one course make two correct solutions disagree in the third digit, and a student then hunts for a mistake that is not there.
What sticks, locks, light and smooth are allowed to mean here
Sticks and locks together mean the bodies share one angular velocity afterwards, the rotational version of the perfectly inelastic collision. Light means no mass, so no moment of inertia and no kinetic energy. A smooth pivot exerts no friction torque, so it adds nothing to $\sum\tau$, but it usually does exert a large force during a collision, which is why linear momentum is not conserved there and angular momentum about the pivot is.
The pivot is the commonest place to lose marks here: invisible in the picture, no torque about itself, and fatal to conservation of linear momentum. All of those are needed at once.
When the steady precession result may be used
The precession result here assumes $\Omega \ll \omega$ and a steady tilt. Every worked answer using it ends by checking that ratio; if $\Omega$ comes out anywhere near $\omega$, the formula has been used outside its range and the answer is thrown away rather than reported.
Quoting it for a slow top would be quoting a formula outside its conditions, which is the error this course spends the term training out of you.
13.1Angular momentum: what a turning body carries about a named axis
How much turning a body carries about a stated axis: moment of inertia times rate of turning.
The previous section left a rate of turning and a resistance to being turned; multiplying them, as mass times velocity was formed for straight-line motion, gives this quantity.
Solvable with what we have
Find a wheel's angular acceleration from a known pull on its rim.
Get a rolling cylinder's speed at the foot of a ramp from energy.
Add torques about one axle with signs and say which way it turns.
Shift a tabulated moment of inertia to a parallel axis.
Not solvable yet
Say how fast a diver spins after she folds up in mid air.
Handle a lump of clay landing on a spinning turntable.
Give a direction, not a plus or minus sign, once the axis can tip.
Say why a spinning top circles slowly instead of falling over.
Use the tool we have. In the air nothing turns the diver about her own axis, so $\sum\tau = 0$, so $I\alpha = 0$, so $\alpha = 0$: her rate of turning cannot change. On the video it triples, then comes back.
Why it fails
The step from $\sum\tau = 0$ to $\alpha = 0$ quietly divided by $I$, legitimate only while $I$ is fixed. The diver's is not: folding up pulls her mass in towards the axis, to roughly a third of its stretched-out value. So the video is saying that something other than $\omega$ stayed put.
DefinitionDefinition 13.1: angular momentum about a fixed axis
Conditions
One rigid body, or several bodies all turning about the same fixed axis.
The moment of inertia is taken about that same axis, and the answer is quoted with the axis named.
The rate of turning is in rad/s and carries the sign of whichever sense was declared positive.
The units are kg m squared per second; unlike the joule or the newton they have no shorter name.
How much turning a body carries is its unwillingness to be turned multiplied by how fast it is actually turning. Two bodies at the same rate carry different amounts if their mass sits differently, and one body carries different amounts about different axes.
Where the product comes from, piece by piece
Cut the body into small pieces. A piece of mass $m_i$ sitting a distance $r_i$ from the axis moves at $v_i = r_i\omega$, and the amount of turning it carries about the axis is its momentum multiplied by the perpendicular distance from the axis to the line it is moving along, which for circular motion is $r_i$ itself. So the piece contributes $m_i v_i r_i = m_i r_i^{2}\omega$. Every piece shares the same $\omega$, because the body is rigid, so the sum is $\left(\sum m_i r_i^{2}\right)\omega$, and the bracket is the moment of inertia already defined in the previous section.
Two wheels with the same mass, the same radius and the same $\textcolor{#d1690a}{\text{rate of turning}}$. The hoop keeps all of its mass at the rim, so it carries twice the $\textcolor{#d1690a}{\text{angular momentum}}$ of the solid disc, and it will take twice the angular impulse to stop it.
Looks like this, but is not
A body going nowhere cannot be carrying anything, so a wheel spinning in place on its axle has no momentum and no angular momentum.
The first half is true, the second does not follow. The wheel's centre of mass stands still, so its linear momentum is zero. Angular momentum is built differently: two opposite pieces sit on opposite sides of the axis, and their contributions add instead of cancelling. It earns its place by surviving where the old one vanishes.
Angular momentum of a flywheel rated in rpm
A flywheel is a uniform solid disc of mass 12.0 kg and radius 0.400 m turning on a fixed axle through its centre at 300 rpm. Find its angular momentum about the axle.
the definition, with both factors now about the same axis and in the right units
Answer $$\boxed{\;L = 30.2\ \mathrm{kg\,m^{2}/s}\ \text{about the central axle}\;}$$
Check
Independent route, by pieces instead of by the table: cut the disc into four rings of equal mass 3.00 kg, with edges at 0.200, 0.283, 0.346 and 0.400 m. A ring lying between $r_{k-1}$ and $r_k$ has mean square radius $(r_k^{2}+r_{k-1}^{2})/2$, so the four contributions to $L$ are 1.88, 5.65, 9.42 and 13.19 and they add to 30.2. The bracket agrees too: every gram of the disc sits at 0.400 m or less, so $L$ has to stay under the hoop value $MR^{2}\omega = 60.3$, and it does.
One unit conversion, one table lookup, one multiplication. The conversion is where the marks go.
The answer was quoted with the axle named. A bare 30.2 would be unusable in the next line of any problem, since nothing in it says which axis it belongs to.
The same dumbbell about two different axes
Two 0.500 kg balls are fixed to the ends of a light rod 1.20 m long. The assembly turns at 4.00 rad/s. Find its angular momentum first about a vertical axis through the middle of the rod, then about a vertical axis through one of the balls, with the same rate of turning in both cases.
Given
two balls, each $m = 0.500\ \mathrm{kg}$, at the ends of a light rod
rod length 1.20 m, so each ball is 0.600 m from the middle
Check the second value with the parallel axis shift rather than by direct summation: $I_{\rm end} = 0.360 + (1.00)(0.600)^{2} = 0.720$, with the total mass 1.00 kg and the centre of mass at the middle. Same value, from a rule that never looks at the individual balls.
One body, one rate of turning, two answers differing by a factor of two. That is why every angular momentum here is written with its axis attached, and why comparing two instants demands the same axis for both.
Checkpoint
§13.1 — same mass, same radius, same spin●●○○○
A solid disc and a hoop have exactly the same mass and exactly the same radius. Each is mounted on its own identical axle through its centre and spun up until both are turning at the same rate.
Given
equal masses and equal radii
equal rates of turning, both about the central axle
disc about its centre: $I = \tfrac12 MR^{2}$; hoop about its centre: $I = MR^{2}$
Find
(a) Choose the correct comparison of the two angular momenta about their own axles.
Hint 1/4
No numbers are needed. Both share the rate of turning, so the whole comparison is decided by the one factor they do not share.
Hint 2/4
$L = I\omega$, and for the same $M$ and $R$ the hoop's moment of inertia is twice the disc's.
Hint 3/4
The data again: equal $M$, equal $R$, equal $\omega$, with $\tfrac12 MR^{2}$ for the disc and $MR^{2}$ for the hoop.
Hint 4/4
The hoop carries twice as much, because $L$ is proportional to $I$ and its $I$ is twice as large.
Sanity check at the extreme: a body with all its mass at the axis would have $I = 0$ and carry nothing at all however fast it spun. Moving mass outwards can only increase $L$, so the hoop must be the larger, which rules out the two options that say otherwise before any arithmetic.
⚠ Quoting an angular momentum with no axis attached to it
The formula produces a single number and nothing in the symbol reminds you that the number belonged to a particular axis, so the axis gets dropped on the way to the next line.
wrong$$L = 3.60\ \mathrm{kg\,m^{2}/s}$$
right$$L = 3.60\ \mathrm{kg\,m^{2}/s}\ \text{about the contact point}$$
⚠ Putting a rate in rpm straight into the definition
Machines are rated in rpm and the number looks like a perfectly good rate of turning, so it goes in unconverted and the answer comes out about ten times too large.
⚠ Using a tabulated moment of inertia about the wrong axis
The table is indexed by the shape of the body, so it is easy to read off the row for a rod and forget that the row also names an axis that may not be the one in the question.
The total outside turning effect is the speed at which the stored amount of turning is changing. Read backwards, the change between two instants is the whole turning effect accumulated over that stretch of time, so a large torque acting briefly and a small one acting for a long time do the same job. Neither statement mentions the shape of the body, which is what makes this form usable when the shape is what changes.
How the familiar form falls out, and where it stops being available
Differentiate $L = I\omega$ with respect to time. The product rule gives two terms, $dL/dt = I\,d\omega/dt + \omega\,dI/dt$. For a rigid body on a fixed axis $I$ is constant, the second term dies, and what is left is $I\alpha$. So $\sum\tau = I\alpha$ is not a different law; it is this one with a constant moment of inertia substituted in. The moment the shape changes, the second term survives and only the general form may be used.
A $\textcolor{#1f6feb}{\text{torque}}$ that grows in proportion to time, applied to a wheel from rest. The shaded area under the line is the angular impulse, and it is the change in angular momentum. Because the torque is not constant the angular acceleration is not constant either, so the constant-acceleration equations of the previous section are simply unavailable here.
Looks like this, but is not
Since $\sum\tau = I\alpha$ and $\sum\tau = dL/dt$ are both the second law for rotation, either may be used, and the first is easier.
The first is the second with one substitution already made, and the substitution carries a condition. Watch the skater: nothing outside her exerts a torque, so the left side is zero in both versions. The general form says her angular momentum does not change, which is true. The familiar form says her angular acceleration is zero, which is false, because it divided by an $I$ that was in the middle of changing. Whenever a problem says folds, pulls in, lands on or drops onto, the familiar form produces a confident wrong answer rather than an error message.
How long a friction torque takes to stop a flywheel
The flywheel of the previous example, a uniform disc with $I = 0.960\ \mathrm{kg\,m^{2}}$, is turning at 31.4 rad/s when the drive is switched off. A constant friction torque of 0.500 N m at the bearing is the only torque left. Find how long it takes to stop.
Given
$I = 0.960\ \mathrm{kg\,m^{2}}$ about the axle
$\omega_0 = 31.416\ \mathrm{rad/s}$, which is 300 rpm
constant friction torque 0.500 N m, opposing the turning
no other torque after the drive is switched off
Find
the time to come to rest
SolutionWrite the angular impulse the friction has to deliver
the whole of this has to be removed, since the wheel ends at rest and carries nothing then
$$\int\tau\,dt = \tau t = \Delta L = -30.159$$
the torque is constant, so the integral is just the torque times the time, and the sign is negative because the friction opposes the turning
Solve for the time
$$t = \frac{30.159}{0.500} = 60.3\ \mathrm{s}$$
magnitudes are enough once the sign has done its work, and the answer is a positive time
Answer $$\boxed{\;t = 60.3\ \mathrm{s}\;}$$
Check
Independent route through the energy, which never mentions time until the last line. The stored energy is $\tfrac12(0.960)(31.416)^{2} = 473.8$ J, and friction removes it at $\tau$ per radian, so the wheel turns $473.8/0.500 = 947.6$ rad before stopping. The average rate of turning over a uniform slowdown is half the initial one, $15.708$ rad/s, so the time is $947.6/15.708 = 60.3$ s. Same answer, different quantity conserved and different route.
A minute is a long time for a wheel to coast, which is exactly why flywheels are used to store energy: the bearing torque is small compared with what the wheel is carrying.
A wheel driven by a torque that grows with time
A wheel with $I = 0.250\ \mathrm{kg\,m^{2}}$ starts from rest on a smooth axle. A motor applies a torque that grows in proportion to time, $\tau = (0.400\ \mathrm{N\,m/s})\,t$, for 5.00 s. Find the rate of turning at the end of the 5.00 s.
Given
$I = 0.250\ \mathrm{kg\,m^{2}}$, constant, about the axle
starts from rest, $\omega_0 = 0$
$\tau = (0.400)\,t$ in newton metres with $t$ in seconds
smooth axle, so no opposing torque
Find
the rate of turning at $t = 5.00$ s
SolutionNotice which tool is available and which is not
$$\alpha(t) = \frac{\tau(t)}{I} = 1.60\,t \;\; \text{is not constant}$$
the constant-acceleration equations of the previous section have a condition attached and it is not met here, so they are put away rather than adapted
Accumulate the angular impulse
$$\Delta L = \int_0^{5.00} 0.400\,t\,dt = 0.400\cdot\frac{(5.00)^{2}}{2} = 5.00\ \mathrm{kg\,m^{2}/s}$$
the integrated form of the second law is exactly what a time-varying torque calls for
Independent kinematic route: integrate the angular acceleration instead of the torque. $\alpha = 1.60t$ gives $\omega = 0.800t^{2}$, and at $t = 5.00$ that is $0.800(25.0) = 20.0$ rad/s. Note also what the forbidden shortcut would have given: taking the final value $\alpha = 8.00\ \mathrm{rad/s^{2}}$ as though it were constant and writing $\omega = \alpha t$ produces 40.0 rad/s, exactly twice the truth, because it pretends the torque was at its final value the whole time.
One integral, one division. The thinking was all in the first line, where a tool was refused.
The wrong answer is out by a factor of two, and it comes from a formula whose condition was never checked. Reading the condition line under a boxed result is the difference between 20 and 40.
Checkpoint
§13.2 — which form of the second law applies●●●○○
A student on a turntable that spins freely on a smooth bearing holds two heavy weights at arm's length and then pulls them in towards their chest. The turntable is seen to speed up. A tutor asks which equation may legitimately be written down for the interval during which the arms are moving.
Given
smooth bearing, so no external torque about the vertical axis
the student's arms move inwards during the interval
the rate of turning increases during the interval
Find
(a) Choose the statement that is correct for the interval while the arms are moving.
Hint 1/4
Do not ask what happens; ask which of the two forms had a condition attached, and whether that condition holds while the arms are moving.
Hint 2/4
$\sum\tau = dL/dt$ always; $\sum\tau = I\alpha$ only while $I$ is constant, because that is the step in which $I$ was taken out of the derivative.
Hint 3/4
The situation again: no external torque about the axis, arms moving inwards, rate of turning increasing.
Hint 4/4
The general form applies and gives a constant angular momentum; the familiar form is not available because the moment of inertia is changing.
Show solutionTest the condition
$$\frac{dI}{dt} \neq 0 \;\text{while the arms move}$$
the mass is physically getting closer to the axis, and the moment of inertia is defined by exactly those distances
$$\sum\tau = \frac{dL}{dt} = 0 \;\Rightarrow\; L \;\text{constant}$$
the general form carries no condition on the shape, so it survives the interval intact
Test by consequence: the rejected form would give $\alpha = 0$ and therefore no change in the rate of turning, which is directly contradicted by what the turntable is seen to do. A form that predicts the opposite of the observation has failed its own condition check.
⚠ Using the fixed-shape form on a system whose shape is changing
It is the form drilled in the previous section, and nothing in the symbols warns you that its derivation assumed the moment of inertia was a constant.
⚠ Treating a time-varying torque as though it were constant at its final value
The final value is the number written in the question, so it is the one at hand, and the constant-acceleration equations are the reflex from the previous section.
wrong$$\Delta L = \tau(t_2)\,\Delta t = (2.00)(5.00) = 10.0$$
right$$\Delta L = \int_0^{5.00}\!0.400\,t\,dt = 5.00$$
13.3When nothing outside can change it: shape changes and rotational collisions
With no external torque about an axis, the angular momentum about it is fixed even while shape and energy change.
Setting the left-hand side of the general law to zero costs nothing and buys the single most useful statement in this section.
TheoremTheorem 13.3: conservation of angular momentum about an axis
Conditions
The net external torque about the chosen axis is zero; forces may still act, provided their torques about that axis cancel or vanish.
The same axis is used before and after, and it does not move between the two instants.
Nothing is claimed about the kinetic energy, which is a separate question with a separate answer.
Internal forces of any size are allowed, including the impulsive ones in a collision.
If nothing from outside is turning the system about a chosen axis, then whatever the system rearranges itself into, the product of its resistance to turning and its rate of turning comes out the same as before. Make the resistance smaller and the rate goes up in exact proportion. The law says nothing whatever about the energy, and in general the energy does change.
Why the energy is free to move while the angular momentum is not
Write the kinetic energy through the conserved quantity rather than the rate of turning. Since $\omega = L/I$, we get $K = \tfrac12 I\omega^{2} = L^{2}/2I$. With $L$ fixed, halving the moment of inertia doubles both the rate of turning and the energy, and the extra has to come from whoever did the rearranging. Doubling it, as when a lump lands on a turntable, halves the energy, and the loss went into deformation at the contact. Both directions are consistent with a fixed $L$, which is why $L$ tells you nothing about $K$.
The skater seen from above, arms out and then pulled in. The $\textcolor{#d1690a}{\text{angular momentum}}$ row is the only one the law protects, and it is identical in the two columns. The moment of inertia falls by four, the rate of turning rises by four, and so does the kinetic energy, paid for by the $\textcolor{#8250df}{\text{arms}}$.
Looks like this, but is not
Angular momentum is conserved here, so this is a conservation problem and I can equally well write down conservation of energy.
The two laws have different entry conditions. Angular momentum about an axis is protected by the absence of an external torque, which survives collisions, sticking, deformation and muscular effort. Mechanical energy is protected by the absence of losses, which none of those survive. In the skater the energy goes up, in the clay on the turntable it goes down, and in both the angular momentum is untouched. Write the two questions separately: what is the net external torque about my axis, and is anything deforming, sticking or being pulled by a muscle.
A skater pulling her arms in, and where the energy came from
A skater turns on the spot on smooth ice at 1.20 rad/s with her arms outstretched, when her moment of inertia about the vertical axis through her is 4.60 kg m squared. She pulls her arms in, reducing it to 1.15 kg m squared. Find her new rate of turning, and find the change in her kinetic energy.
the rate of turning is squared, so quartering the moment of inertia and quadrupling the rate does not break even
$$\Delta K = 13.2 - 3.31 = +9.94\ \mathrm{J}$$
the increase is work done by her arm muscles, pulling her arms inwards against the outward push they were experiencing
Answer $$\boxed{\;\omega_2 = 4.80\ \mathrm{rad/s},\qquad \Delta K = +9.94\ \mathrm{J}\;}$$
Check
Independent route to the energy, using the conserved quantity instead of the rate of turning: $K = L^{2}/2I$ with $L = 5.52$ fixed gives $K_2/K_1 = I_1/I_2 = 4.00$, so $K_2 = 4.00 \times 3.31 = 13.2$ J. And a plausibility check on the 9.94 J: that is about what it costs to lift a bag of sugar to head height, which is a believable amount of muscular work for one sharp pull.
One conservation line, then two energy lines that the conservation line did not give us for free.
The diver from the opening is this calculation with different numbers. The tuck cuts her moment of inertia to roughly a third, so her rate of turning roughly triples and comes back down when she opens out, and the energy came from the muscles that pulled her in. She never needed anything to push against. The pattern to carry away: get the new rate from the conserved quantity, then treat the energy as a separate question with a named agent who paid.
A lump of clay dropped onto a spinning turntable
A turntable with moment of inertia 2.00 kg m squared about its central axis is turning freely at 3.00 rad/s. A 1.50 kg lump of clay is dropped vertically onto it and sticks at a distance 0.800 m from the axis. Find the new rate of turning and the kinetic energy lost.
writing the energy through the conserved quantity avoids carrying the rounded rate of turning into a square
$$\Delta K = 6.08 - 9.00 = -2.92\ \mathrm{J}$$
the loss goes into the scraping and squashing as the clay is dragged up to speed, which is what sticking means
Answer $$\boxed{\;\omega_2 = 2.03\ \mathrm{rad/s},\qquad \Delta K = -2.92\ \mathrm{J}\;}$$
Check
Independent check on the fraction: with $L$ fixed, $K_2/K_1 = I_1/I_2 = 2.00/2.96 = 0.676$, and $0.676\times 9.00 = 6.08$ J, reached without using $\omega_2$ at all. The sign is also what it must be: this is the rotational twin of two bodies sticking together, and energy always goes down in those.
Same law, opposite sign of energy change from the skater, and the difference is entirely in who did what: a muscle pulling inward adds energy, a collision that ends in sticking removes it.
Checkpoint
§13.3 — what happens to the energy when the shape changes●●●○○
A student sits on a stool that turns freely and holds two dumbbells out at arm's length while turning slowly. They then pull the dumbbells in towards their chest, and the stool speeds up. The bearing is smooth throughout.
Given
smooth bearing, so no external torque about the vertical axis
the dumbbells are pulled in by the student's own arms
the rate of turning is observed to increase
Find
(a) Choose the correct account of the angular momentum and the kinetic energy.
Hint 1/4
Answer the two questions in the right order and separately: first what is protected by the absence of an external torque, then who if anyone did work.
Hint 2/4
No external torque about the axis means $L$ is fixed; the energy at fixed $L$ is $K = L^{2}/2I$, so it rises when $I$ falls.
Hint 3/4
The situation again: smooth bearing, dumbbells pulled inward by the student's arms, rate of turning going up.
Hint 4/4
The angular momentum is unchanged and the kinetic energy increases, the increase being work done by the arms.
Check the bookkeeping closes: the increase must equal the work done by the arms, and pulling a mass inwards while it is being flung outwards is work done against something, so the sign is positive. If the energy had come out lower, the arms would have to be absorbing energy while pulling inwards, which is not what the muscles are doing.
⚠ Assuming that conserved angular momentum means conserved energy
Both are called conservation laws, both are used at two named instants, and in the linear collisions of the previous sections the two were often discussed in the same breath.
right$$I_1\omega_1 = I_2\omega_2,\qquad K = \frac{L^{2}}{2I}\ \text{changes}$$
⚠ Leaving the arriving body out of the final moment of inertia
The final rate of turning is asked about the turntable, so the turntable's own value is the one in mind, and the lump that just landed on it does not feel like part of the rotating body yet.
13.4The cross product, and giving a torque a direction instead of a sign
A way of multiplying two vectors that returns a third, perpendicular to both, whose size measures how far apart their directions are.
Everything so far has been a plus or a minus sign, which works only while the axis is bolted down; the moment the axis is free to tip, a turning effect needs a direction in space.
DefinitionDefinition 13.4: the cross product, and torque as a vector
Conditions
The angle used is the one between the two vectors placed tail to tail, taken between 0 and 180 degrees.
The direction is perpendicular to both, fixed by the right hand rule: fingers curl from the first vector to the second, thumb gives the answer.
Order matters. Swapping the two vectors reverses the result, so this multiplication is not the commutative kind.
Two parallel or antiparallel vectors give zero, which is the vector statement of a force whose line passes through the axis.
The turning effect of a force is the position of its point of application crossed with the force. Its size is the two lengths times the sine of the angle between them, largest at right angles and zero when they lie along one another, and its direction is the axis the force is trying to turn things about. Everything the previous section said with a sign is this statement restricted to one fixed axis.
The component recipe, and why it agrees with the sine rule
In components each output component is built from the two input components that do not share its own label: the $x$ component is $a_yb_z - a_zb_y$, the $y$ component is $a_zb_x - a_xb_z$, and the $z$ component is $a_xb_y - a_yb_x$. The cycle $x \to y \to z \to x$ runs through all three lines, and swapping the two vectors flips every sign. For vectors lying in the page only the last line survives, so the answer points straight out of the page or into it, with size $ab\sin\theta$ and the sign carrying the sense of turning. That is the previous section's convention, derived rather than declared.
Left: a $\textcolor{#8250df}{\text{position vector}}$ of 0.500 m and a $\textcolor{#1f6feb}{\text{force}}$ of 21.6 N with components 12.0 and −18.0 N, whose cross product is a $\textcolor{#d1690a}{\text{torque}}$ of 10.8 N m into the page, drawn as a circle with a cross. Right: the same rule with the vectors along the axes, where the answer comes out of the page.
Looks like this, but is not
Multiplication does not care about order, so $\vec{r}\times\vec{F}$ and $\vec{F}\times\vec{r}$ are the same thing and it does not matter which you write.
For ordinary numbers that is true and here it is false, and not as a technicality: the two answers point in exactly opposite directions, so the wrong one turns the body the wrong way. The right hand rule shows why, since it asks you to curl from the first vector to the second; go the other way and the thumb points backwards. In components every entry changes sign. Build the habit of writing the position first and the force second, every time, and check the direction against the picture.
Torque of a slanted force from components
A force $\vec{F} = (12.0\,\hat{\imath} - 18.0\,\hat{\jmath})$ N is applied at the point whose position vector from the pivot is $\vec{r} = (0.400\,\hat{\imath} + 0.300\,\hat{\jmath})$ m. Find the torque about the pivot, as a vector, and say which way it turns the body.
Given
$\vec{r} = (0.400, 0.300, 0)\ \mathrm{m}$ from the pivot to the point of application
$\vec{F} = (12.0, -18.0, 0)\ \mathrm{N}$
both vectors lie in the plane of the page, with $z$ out of the page
Find
the about the pivot and the sense of turning
SolutionUse the component recipe and let the zeros do the work
Independent route through magnitudes and the angle: $r = 0.500$ m and $F = 21.63$ N, the position 36.87 degrees above the x axis and the force 56.31 degrees below it, so the angle between them is 93.18 degrees with sine 0.9985. Then $rF\sin\theta = 10.8$ N m, the same size from geometry rather than components.
Three lines of the recipe, two of which were zero before they were written.
In any plane problem two of the three lines vanish and the cross product collapses to one subtraction. That line is worth memorising: it appears in every two dimensional torque and angular momentum question here.
Sliding a force along its own line changes nothing
A downward force of 30.0 N acts vertically. Compute its torque about the origin first when it is applied at the point (0.200, 0.400) m and then when it is applied at (0.200, 0.100) m, both distances in metres. Comment on the two answers.
Given
$\vec{F} = (0, -30.0, 0)\ \mathrm{N}$, the same in both cases
first point of application $(0.200, 0.400, 0)\ \mathrm{m}$
second point of application $(0.200, 0.100, 0)\ \mathrm{m}$
both points lie on the same vertical line, $x = 0.200$ m
Find
the two torques about the origin, and what the comparison shows
the y coordinate changed and it multiplied a zero, so it never entered the answer
$$\tau_z^{(1)} = \tau_z^{(2)}$$
the two points lie on the same line of action, and the cross product only ever sees the perpendicular distance to that line
Answer $$\boxed{\;\vec{\tau} = -6.00\,\hat{k}\ \mathrm{N\cdot m}\ \text{for both points}\;}$$
Check
Check with the lever arm picture instead of the components: the line of action is the vertical line $x = 0.200$ m, its perpendicular distance from the origin is 0.200 m whatever the height, and $(0.200)(30.0) = 6.00$ N m. The geometric route never mentions either point of application, which is the whole content of the result.
So a torque question can be answered by drawing the line the force lies along and dropping a perpendicular onto it from the axis. Where the force happens to be applied is then a distraction.
Checkpoint
§13.4 — which way the torque points●●○○○
A body is free to turn about the origin. A force is applied in the positive y direction at a point sitting on the positive x axis, some distance out from the origin. The figure is drawn in the usual way, with x to the right, y upwards and z out of the page.
Given
the point of application lies on the positive x axis
the force points in the positive y direction
axes drawn with x right, y up and z out of the page
Find
(a) Choose the direction of the torque about the origin.
Hint 1/4
You are being asked for a direction, not a size, so no distance and no force value is needed.
Hint 2/4
$\vec{\tau} = \vec{r}\times\vec{F}$, and the result is perpendicular to both, with the right hand curling from the first vector to the second.
Hint 3/4
The setup again: the position along positive x, the force along positive y, and z drawn out of the page.
Hint 4/4
The torque points along positive z, out of the page, which is the counterclockwise sense.
Show solutionWrite only the component that can be non-zero
$$\tau_z = r_xF_y - r_yF_x = r_xF_y > 0$$
the position has no y component and the force has no x component, so one product is all that is left
Answer $$\boxed{\;\vec{\tau}\ \text{along}\ +\hat{k},\ \text{out of the page}\;}$$
Check
Cross-check against physical sense: pushing upwards on a point out to the right of a pivot swings that point anticlockwise, and anticlockwise in a page drawn this way is the positive z direction. The algebra and the picture agree.
⚠ Writing the force first and the position second
The force is the thing that feels active in the problem, so it is the one the hand reaches for first, and the algebra gives no warning.
⚠ Using the cosine of the angle instead of the sine
The dot product from the work section uses the cosine and is drilled first, so the reflex carries over to the other kind of product.
wrong$$|\vec{r}\times\vec{F}| = rF\cos\theta$$
right$$|\vec{r}\times\vec{F}| = rF\sin\theta$$
13.5The angular momentum of a single particle, including one going in a straight line
A single particle has angular momentum about a point whenever its line of travel misses that point.
The vector product just built lets the same quantity be written for something that is not a rigid body at all, which is what a colliding lump of putty is.
DefinitionDefinition 13.5: angular momentum of a particle, and of a system
Conditions
An origin has to be chosen and stated; the same particle has different values about different origins.
The angle is the one between the position vector and the velocity, tail to tail.
For several particles the total is the sum of the individual values, all about the same origin.
The rate of change of the total is the net external torque; internal forces between the particles cancel in pairs.
The turning a particle carries about a chosen point is its momentum times how far the line it travels along passes from that point. A particle heading straight at the point, or away from it, carries none, because the miss distance is zero. Everything else carries some, whether or not anything is going round, and the total for a collection is what the outside torques act on.
Why the rigid body result is a special case of this one
Apply the particle definition to one small piece of a rigid body on a fixed axis. The piece moves in a circle of radius $r_i$, so its velocity is perpendicular to its position vector from the axis and the sine is one. Its contribution is $m_i v_i r_i$, and with $v_i = r_i\omega$ that is $m_i r_i^{2}\omega$. Summing gives $I\omega$ again. So there are not two definitions here; there is one, and $L = I\omega$ is what it becomes when everything goes round the same fixed axis.
A 2.00 kg particle crossing the picture at a steady $\textcolor{#d1690a}{5.00\ \mathrm{m/s}}$, drawn at three instants. The $\textcolor{#8250df}{\text{position vector}}$ lengthens and swings round, but the $\textcolor{#8250df}{\text{perpendicular distance}}$ to its line of travel stays at 3.00 m, so the angular momentum about the origin is 30.0 kg m squared per second throughout.
Looks like this, but is not
Angular momentum is the momentum of going round, so a particle travelling in a perfectly straight line has none about any point.
The name is doing the damage. The definition asks for the miss distance of the line of travel from the chosen point, and a straight line misses almost every point in the plane. The particle in the figure carries a fixed 30.0 kg m squared per second about the origin, and zero only about points on its own line. This is not a curiosity: it is how a lump of putty flying at a hinged rod has an angular momentum before it touches anything, and without it no rotational collision could be set up.
x (m)
r (m)
sin of the angle
r sin (m)
L (kg m squared per s)
2.00
3.61
0.832
3.00
30.0
5.00
5.83
0.514
3.00
30.0
8.00
8.54
0.351
3.00
30.0
Two columns change substantially and their product does not move at all. That product is the perpendicular distance from the origin to the line of travel, a property of the line rather than of where the particle has got to along it.
A particle crossing the room in a straight line
A 2.00 kg particle travels in a straight line at a constant 5.00 m/s in the positive x direction, along the line $y = 3.00$ m. Find its angular momentum about the origin, and show that it does not change with time.
Given
$m = 2.00\ \mathrm{kg}$, $v = 5.00\ \mathrm{m/s}$ in the $+x$ direction
the line of travel is $y = 3.00\ \mathrm{m}$
no force acts, so the velocity is constant
origin at the corner of the room, at $(0,0)$
Find
the angular momentum about the origin, and whether it changes
SolutionThe miss distance is the only geometry needed
$$r_{\perp} = 3.00\ \mathrm{m}$$
the line of travel is horizontal at that height, so the perpendicular from the origin onto it is the height itself
no force means no torque about any point, so the general law forbids the value from changing
Answer $$\boxed{\;\vec{L} = -30.0\,\hat{k}\ \mathrm{kg\,m^{2}/s}\ \text{about the origin, constant}\;}$$
Check
Independent check by brute force at two instants. At $x = 2.00$ m the position vector is 3.61 m long and the sine of the angle to the velocity is $3.00/3.61 = 0.832$, giving $mvr\sin\phi = (10.0)(3.61)(0.832) = 30.0$. At $x = 8.00$ m the length is 8.54 m and the sine is 0.351, giving $(10.0)(8.54)(0.351) = 30.0$. Two factors change and their product does not.
The lesson to carry into every collision problem on this page: a body flying in a straight line already has an angular momentum about the pivot it is about to hit, and it equals its momentum times the perpendicular distance from the pivot to its path.
The growing angular momentum of a thrown ball
A 0.400 kg ball is thrown horizontally at 12.0 m/s from a window. Taking the origin at the point of release, find its angular momentum about that point 2.00 s later, and check the answer against the torque that gravity exerted in the meantime. Ignore air resistance.
Given
$m = 0.400\ \mathrm{kg}$, launched horizontally at $12.0\ \mathrm{m/s}$
origin at the release point, $x$ horizontal in the direction of throw, $y$ upwards
$g = 9.80\ \mathrm{m/s^{2}}$, so $a_y = -9.80\ \mathrm{m/s^{2}}$
elapsed time $t = 2.00\ \mathrm{s}$
Find
the angular momentum about the release point at $t = 2.00$ s
negative means into the page, which is the sense in which the ball is swinging round the window as it falls away
Answer $$\boxed{\;\vec{L} = -94.1\,\hat{k}\ \mathrm{kg\,m^{2}/s}\ \text{about the release point}\;}$$
Check
Independent route through the torque, which never uses the position and momentum at the final instant. Gravity acts downwards with magnitude $mg = 3.92$ N, so its torque about the release point is $\tau_z = -mgx = -(3.92)(12.0\,t) = -47.04\,t$. Accumulating that from 0 to 2.00 s gives $-47.04\,(2.00)^{2}/2 = -94.08$, and the ball started with none. Same number by a completely different road.
Four kinematic values, one subtraction, and then a second calculation done purely as a check.
Angular momentum about a point is not a fixed property of a body but a running total a torque keeps adding to. The two routes agreeing is the strongest evidence that neither was done wrong.
Checkpoint
§13.5 — a car driving past a stationary observer●●●○○
A car drives along a perfectly straight road at a constant speed. An observer stands on the pavement, some distance to the side of the road, and considers the car's angular momentum about the spot where they are standing.
Given
the car travels in a straight line at constant speed
the observer stands at a fixed point beside the road, not on it
no forces act along the direction of travel
Find
(a) Choose the correct description of the car's angular momentum about the observer.
Hint 1/4
Two separate things are being asked: whether the value is zero, and whether it changes. Answer them one at a time.
Hint 2/4
$L = mvr_{\perp}$, where $r_{\perp}$ is the perpendicular distance from the chosen point to the line of travel, and $dL/dt$ is the net torque about that point.
Hint 3/4
The situation again: a straight road, a constant speed, and an observer standing to the side of the road rather than on it.
Hint 4/4
It is not zero and it does not change: the miss distance and the momentum are both constant.
Show solutionIs it zero
$$r_{\perp} \neq 0 \;\Rightarrow\; L = mvr_{\perp} \neq 0$$
the observer is beside the road, so their perpendicular distance to the line of travel is not zero
no net force acts on a car at constant velocity, so no torque about any point, so nothing can change the value
Answer $$\boxed{\;L = mvr_{\perp},\ \text{constant and non-zero}\;}$$
Check
Check the limiting case: let the observer step into the middle of the road, so that the miss distance falls to zero. The formula then gives zero, which is right, because the car is coming straight at them. The general answer has to reduce to that special one, and it does.
⚠ Using the distance to the particle instead of the miss distance
The position vector is what is drawn on the figure, so its length is the number in front of you, while the perpendicular distance has to be constructed.
⚠ Forgetting that the answer is tied to the origin that was chosen
For a rigid body on an axle the axis is drawn in the picture and cannot be forgotten; for a free particle the origin is a choice you made, and choices are easy to lose.
wrong$$L = 30.0\ \mathrm{kg\,m^{2}/s}$$
right$$L = 30.0\ \mathrm{kg\,m^{2}/s}\ \text{about the origin at } (0,0)$$
13.6Rotation about an axis that moves, and angular momentum that refuses to line up
A travelling, spinning body carries both kinds of angular momentum, and off a symmetry axis its total misses the spin axis.
Every body so far has been bolted to an axle through a natural axis; taking the bolts out shows what the fixed-axis results were quietly assuming.
TheoremTheorem 13.6: splitting the total, and when the total points along the axis
Conditions
The splitting is exact about any chosen origin, with the first term built from the motion of the centre of mass and the second from the spin about the centre of mass.
For a rigid body turning about a fixed axis, the component of the total along that axis is always the moment of inertia about it times the rate of turning.
The full vector is parallel to the rate of turning only when the axis is an axis of symmetry of the body.
When they are not parallel, keeping the axis fixed requires the bearings to supply a torque even though nothing is speeding up.
The turning a body carries about an outside point is the turning it would have if all its mass sat at the centre of mass and moved with it, plus the turning it has about its own centre. Along the axis it is spun about, the familiar product still gives the right number, but that may be only part of the whole vector, and the rest points sideways and goes round with the body.
Why the two parts do not interfere with each other
Write every position as the centre of mass position plus a displacement measured from it, and every velocity the same way. Substituting into the sum for the total gives four groups of terms. Two are the ones we want. The other two carry a factor $\sum m_i\vec{r}_i\,'$ or $\sum m_i\vec{v}_i\,'$, and both are zero by the definition of the centre of mass, the point about which the mass-weighted displacements cancel. So the cross terms vanish and the total is a clean sum of travel and spin.
Two equal masses on a light rod, clamped at 60 degrees to the $\textcolor{#8250df}{\text{rotation axis}}$ and spun about it. The $\textcolor{#d1690a}{\text{angular momentum}}$ comes out 30 degrees from the axis on the other side, at right angles to the rod, and only 1.80 of its 2.08 units lies along the axis. The rest sweeps round the dashed cone once per turn.
Looks like this, but is not
The angular momentum vector points along the axis the body is spinning about. That is what it means to say it points along the axis of rotation.
It is true for every body in the first half of this section and false in general, and the difference is whether the axis is one the body is symmetric about. The tilted rod in the figure has its angular momentum perpendicular to the rod, 30 degrees from the axis. The mistake is reading a formula derived for a symmetric case as if it were a definition. The consequence is audible: a car wheel with the balance weights knocked off has its angular momentum slightly off the axle, and the bearings supply an alternating torque once per revolution, which is the vibration you feel.
Angular momentum of a rolling cylinder about a point on the ground
A solid cylinder of mass 3.00 kg and radius 0.200 m rolls without slipping along level ground at 4.00 m/s. Find its angular momentum about a fixed point on the ground directly beneath its centre, splitting the answer into a travel part and a spin part.
the centre passes the origin at a perpendicular distance equal to the radius, so the miss distance is R
$$L = 2.40 + 1.20 = 3.60\ \mathrm{kg\,m^{2}/s}$$
both parts turn the same way about the ground point, so they add rather than partly cancel; a wheel rolling right does both clockwise
Answer $$\boxed{\;L = 3.60\ \mathrm{kg\,m^{2}/s}\ \text{about the ground point, of which }2.40\ \text{travel and }1.20\ \text{spin}\;}$$
Check
Independent route using the fact that a rolling body is, at this instant, turning about its contact point. The moment of inertia about that point is $I_{\rm cm}+MR^{2} = 0.0600+0.120 = 0.180\ \mathrm{kg\,m^{2}}$, and $(0.180)(20.0) = 3.60$. One calculation split the motion in two and added; the other treated it as a single rotation about a different point. Same number.
Two terms and one addition, but only after the origin was named in the first line.
The two parts came out two to one, and that ratio is fixed by the shape alone: for a solid cylinder the travel part is always twice the spin part about a ground point.
A tilted dumbbell whose angular momentum leans the wrong way
Two 0.800 kg balls are fixed to the ends of a light rod and the middle of the rod is clamped to a vertical shaft, so that the rod makes 60.0 degrees with the shaft. Each ball is 0.500 m from the clamp. The shaft turns at a steady 6.00 rad/s. Find the component of the angular momentum along the shaft, its total magnitude and its direction, and the torque the bearings must supply.
Given
two balls, each $m = 0.800\ \mathrm{kg}$, each $\ell = 0.500\ \mathrm{m}$ from the clamp
the rod makes 60.0 degrees with the vertical shaft
$\omega = 6.00\ \mathrm{rad/s}$, steady, about the shaft
light rod, so only the two balls carry mass
Find
the component along the shaft, the magnitude and direction of the total, and the bearing torque
SolutionAlong the shaft, where the familiar formula still works
each ball contributes its momentum times the full distance from the clamp, because the position vector is taken from the clamp and is perpendicular to the velocity
Independent geometric check that the vector really is square to the rod. In components with the shaft along z, the angular momentum is $(-1.039,\,0,\,1.800)$ and the rod direction is $(0.866,\,0,\,0.500)$. Their dot product is $-0.900 + 0.900 = 0$ exactly, so the two are perpendicular, which is what the 60 and 30 degrees were saying in words.
The rate of turning never changes and a torque is needed anyway. Zero torque means a steady angular momentum vector, and a steady rate of turning is not the same thing.
Checkpoint
§13.6 — a body spun about a skew axis at a steady rate●●●●○
A rigid body is clamped to a shaft along an axis that is not one of its axes of symmetry, so that its angular momentum does not point along the shaft. It is then driven at a perfectly constant rate of turning by a motor, with the shaft held by two bearings.
Given
the axis is not an axis of symmetry of the body
the rate of turning is constant
the shaft is held in place by bearings
Find
(a) Choose the statement that must be true while the body turns.
Hint 1/4
Ask what the angular momentum vector is doing, not what its size is. A vector can change while its length stays fixed.
Hint 2/4
$\sum\vec{\tau} = d\vec{L}/dt$, and a vector of constant length that is being carried round in a circle is changing all the time.
Hint 3/4
The setup again: a skew axis, a constant rate of turning, and a part of the angular momentum sticking out sideways from the shaft.
Hint 4/4
The sideways part goes round with the body, so its rate of change is not zero and the bearings must be pushing.
the general law reads the rate of change of the vector, not of its length
Answer $$\boxed{\;\text{the bearings must supply a torque of size } \omega L_{\perp}\;}$$
Check
Consistency check with the symmetric case: if the axis were an axis of symmetry, the sideways part would be zero, the required torque would be zero, and the result reduces to the fixed-axis rule that a steady spin needs no torque. A general result that did not reduce correctly would be wrong.
⚠ Treating the fixed-axis formula as a vector statement about any axis
It is written as a product of two quantities and it is true in every example of the first half of the section, so it starts to feel like a definition rather than a special case.
wrong$$\vec{L} = I\vec{\omega}\ \text{for any axis}$$
right$$L_z = I_z\omega\ \text{always};\quad \vec{L}\parallel\vec{\omega}\ \text{only about a symmetry axis}$$
⚠ Using only the spin part for a rolling body about an outside point
The spin is the visible rotation, and the travel part looks like translation, which does not feel like it should count as turning at all.
13.7Why a spinning top circles instead of falling over
A torque at right angles to a large angular momentum turns it sideways, so the axis sweeps a cone instead of tipping.
Everything needed for the last piece is now on the page: a torque is a vector, an angular momentum is a vector, and one is the rate of change of the other.
TheoremTheorem 13.7: steady precession of a fast spinning body
Conditions
The spin about the body's own axis is much faster than the rate at which the axis swings round.
The torque is perpendicular to the angular momentum, which is the case for a weight acting on a horizontal or gently tilted axle.
The angular momentum is taken about the support point, and only the spin part is kept, which is what the fast spin assumption buys.
The tilt of the axis is treated as steady; the small nodding of a real top is not described by this result.
The axis goes round at a rate equal to the turning effect of the weight divided by the spin the body is carrying. Spin it faster and it goes round more slowly, which is backwards from every intuition about fast things; make it heavier or hang it further out and it goes round faster. No term in the result mentions the tilt, which is why a top and a horizontal bicycle wheel obey the same formula.
Where the division comes from
In a short time the change in angular momentum is the torque times that time, pointing along the torque. Here the torque is horizontal and at right angles to the angular momentum, so the change is perpendicular to what is already there, and adding a small perpendicular piece to a vector rotates it rather than lengthening it. The tip moves $\tau\,dt$ along a circle of radius $L$, so the angle swept is $d\phi = \tau\,dt/L$ and the rate is $\tau/L$. It feels strange only because intuition expects a downward torque to produce a downward motion.
A spinning wheel held at one end of its axle. Its $\textcolor{#d1690a}{\text{angular momentum}}$ points along the axle, and the $\textcolor{#1f6feb}{\text{weight}}$ at the $\textcolor{#8250df}{\text{distance}}$ D from the support makes a horizontal torque at right angles to it. The change therefore lies sideways, and the axle is carried round the circle shown from above rather than falling.
Looks like this, but is not
The spin holds the wheel up, so a fast enough spin cancels its weight and the support carries nothing.
The weight is still there and the support carries all of it, as scales under the pivot would confirm. What the spin changes is where the weight torque sends the axis. With no spin there is no angular momentum to turn, so the torque produces the falling it looks like it should. With a large angular momentum along the axle, the same torque adds a sideways piece to a large vector and rotates it, so the axis travels horizontally. It is also why a real top topples: friction eats the spin and eventually the assumption fails.
A bicycle wheel precessing on the end of its axle
A bicycle wheel of mass 2.50 kg and radius 0.330 m, with essentially all of its mass at the rim, is spun at 12.0 revolutions per second and its axle is then rested on a support so that the wheel hangs 0.200 m from the support point, with the axle horizontal. Find the rate at which the axle swings round, and how long one complete circuit takes.
Given
wheel treated as a hoop: $I = MR^{2}$, $M = 2.50\ \mathrm{kg}$, $R = 0.330\ \mathrm{m}$
spin 12.0 revolutions per second
distance from the support to the centre of the wheel: $D = 0.200\ \mathrm{m}$
Two checks. The condition: $\Omega/\omega = 0.239/75.4 = 0.0032$, far below one, so the fast spin assumption holds and the answer may be reported. The scale: the far end of the axle, 0.200 m out, travels at $(0.200)(0.239) = 0.048$ m/s, the slow visible crawl of a lecture demonstration.
One conversion, one moment of inertia, one torque, one division.
Read the formula rather than just using it: the spin sits underneath, so spinning the wheel harder makes the precession slower. A top that seems to speed up as it dies is not gaining anything, it is losing spin.
Getting the spin of a top from how fast it goes round
A toy top of mass 0.500 kg has a moment of inertia of $2.00\times 10^{-4}\ \mathrm{kg\,m^{2}}$ about its own axis, and its centre of mass is 0.0300 m from the point it stands on. It is observed to take 3.00 s for its axis to go once round. Find the rate at which the top is spinning.
Given
$M = 0.500\ \mathrm{kg}$, $I = 2.00\times 10^{-4}\ \mathrm{kg\,m^{2}}$ about the top's own axis
centre of mass $D = 0.0300\ \mathrm{m}$ from the contact point
the moment of inertia converts the stored turning back into a rate about the top's own axis
Answer $$\boxed{\;\omega = 351\ \mathrm{rad/s},\ \text{that is } 55.9\ \text{turns per second}\;}$$
Check
The condition check: $\Omega/\omega = 2.09/351 = 0.006$, safely small, so the relation was legitimate. A plausibility check too: 56 turns a second is about 3400 revolutions per minute, the right order for a top launched hard off a string, and fast enough that it looks still rather than visibly spinning.
The same relation read three ways gives three questions: the precession from the spin, the spin from the precession, and the moment of inertia of an awkward body from both. Formulas are worth more run backwards.
Checkpoint
§13.7 — a top slowly losing its spin●●●○○
A top is set spinning on a table and left alone. Friction at the point of contact and against the air slowly bleeds away its spin over half a minute or so, while its weight and the position of its centre of mass stay exactly the same.
Given
the spin rate falls slowly as friction acts
the mass, the moment of inertia and the distance to the centre of mass are unchanged
the axis continues to sweep round steadily while the spin is still fast
Find
(a) Choose what happens to the rate at which the axis sweeps round.
Hint 1/4
Only one quantity in the relation is changing. Find it, find whether it sits on the top or the bottom, and read off the direction.
Hint 2/4
$\Omega = MgD/(I\omega)$, with the spin rate in the denominator.
Hint 3/4
The situation again: the same weight, the same geometry, the same moment of inertia, and a spin rate that is falling.
Hint 4/4
The precession speeds up, and it keeps speeding up until the spin is no longer fast and the relation stops applying.
the numerator is fixed by the weight and the geometry, neither of which friction touches
Answer $$\boxed{\;\Omega\ \text{increases as the spin dies away}\;}$$
Check
Check against the extreme case: a top that is not spinning at all has no angular momentum for the torque to turn, so it simply falls over, which is the limit of an ever faster precession running out of validity. The observed behaviour of a dying top matches that limit exactly.
⚠ Measuring the lever arm to the end of the axle instead of to the centre of mass
The axle is the long visible thing in the picture and its length is the number given, while the centre of mass has to be located.
wrong$$\tau = Mg\,L_{\rm axle}$$
right$$\tau = MgD,\quad D = \text{support to centre of mass}$$
⚠ Reporting the precession rate as a number of turns per second without converting
The answer comes out of a formula in rad/s and the question often asks how long one circuit takes, and the two are easy to confuse when both are small numbers.
wrong$$\text{one circuit in } 1/0.239 = 4.19\ \mathrm{s}$$
right$$\text{one circuit in } 2\pi/0.239 = 26.3\ \mathrm{s}$$
Choosing which law a rotation question is actually giving you
Any question with two named instants and a rotation somewhere in it, before writing a single equation.
Name the axis or the point
Write it down explicitly: about the axle, about the pivot, about the contact point, about the centre of mass. Everything after this line is about that point and nothing else.
Ask what the net external torque about it is
Count only external forces. A weight along the axis, a normal force along the axis and a pivot force at the pivot all have zero torque about it. If the total is zero, the angular momentum about that point is fixed and you have your first equation.
Ask separately whether anything is lost
Look for sticking, gripping, scraping, deforming or a muscle pulling. Any of them means the kinetic energy is not conserved, whatever the angular momentum is doing. None of them, plus a smooth pivot, means the energy line is available.
Check whether the shape changes
If the moment of inertia is different at the two instants, the fixed-shape form of the second law is unavailable and you must use the general form or its conserved consequence.
Write one equation per stage
A collision is one stage and the swing afterwards is another, and they almost never obey the same law. Solve them in order and carry the answer of the first into the second.
Where it goes wrong
Using the same conservation law for both stages of a two-stage problem, the commonest way to lose most of the marks on a long question.
Counting the pivot force as an external torque; it acts at the pivot, so its lever arm there is zero.
Deciding that angular momentum is conserved and concluding that the energy is too.
Setting up a collision onto a body that is hinged or pivoted
Whenever something flies in and hits a rod, a door, a turntable or a wheel that is held at a fixed point.
Take the pivot as the point
This is the whole trick. The pivot exerts a large force during the collision, wrecking conservation of linear momentum, and no torque about itself, leaving conservation of angular momentum about it intact.
Give the incoming body its angular momentum
It is $mvr_{\perp}$, with $r_{\perp}$ the perpendicular distance from the pivot to the line the body is travelling along. Nothing needs to be going round for this to be non-zero.
Build the moment of inertia of what is left afterwards
Add the target's moment of inertia about the pivot to the arriving body's, treating it as a point mass at the distance it stuck at.
Divide
The common rate of turning is the total angular momentum divided by the total moment of inertia, both about the pivot.
Only now start a new stage
If the question asks how far it swings, that is a second stage with its own law: mechanical energy, from the instant just after the collision to the instant it stops rising.
Where it goes wrong
Writing conservation of linear momentum for a collision onto a pivoted body. The pivot supplies whatever impulse it likes.
Using the target's moment of inertia about its own centre instead of about the pivot.
Running an energy line across the collision itself, which throws away the energy that went into the sticking.
Putty hits a rod that is pivoted at one end
A uniform rod of mass 1.50 kg and length 1.00 m hangs at rest from a smooth pivot at its upper end. A 0.400 kg lump of putty flying horizontally at 5.00 m/s strikes it 0.750 m below the pivot and sticks. Find the rate of turning just after the impact, and check what happened to the linear momentum.
Given
rod: $M = 1.50\ \mathrm{kg}$, $L = 1.00\ \mathrm{m}$, pivoted at one end, initially at rest
The extra 0.17 kg m/s came from the pivot, the only thing touching the system from outside. The sign checks independently: for a rod hinged at one end the strike point that leaves the pivot unstressed is two thirds of the way down, at 0.667 m, and hitting below that makes the pivot push forwards. The strike was at 0.750 m, so a forward push is what theory demands and what the numbers gave.
The pivot is invisible in the picture and decisive in the physics: it destroys one conservation law and leaves the other untouched.
The same putty hits the same rod lying free on a smooth table
The same uniform rod, 1.50 kg and 1.00 m long, lies at rest on a smooth horizontal table with nothing holding it. The same 0.400 kg lump of putty slides across the table at 5.00 m/s, strikes the rod at right angles 0.250 m from its centre and sticks. Find the velocity of the centre of mass and the rate of turning afterwards.
Given
rod: $M = 1.50\ \mathrm{kg}$, $L = 1.00\ \mathrm{m}$, free on a smooth table
putty: $m = 0.400\ \mathrm{kg}$ at $v = 5.00\ \mathrm{m/s}$, perpendicular to the rod
it strikes 0.250 m from the rod's centre and sticks
rod about its own centre: $I = \tfrac{1}{12}ML^{2}$
Find
the velocity of the centre of mass and the rate of turning afterwards
SolutionThis time the linear momentum is protected
Energy check, which must show a loss because the putty stuck: before, 5.00 J; after, $\tfrac12(1.90)(1.053)^{2}+\tfrac12(0.1447)(2.727)^{2} = 1.59$ J. A loss of 3.41 J, the right sign, with the two parts of the final energy comparable in size, as they should be for a strike well away from the centre.
With no pivot the body does two things at once, and the answer needs two numbers rather than one.
The same rod and the same lump of putty at the same speed give a single rotation about a fixed point in one case and a translation plus a rotation about a moving centre of mass in the other, and the only difference in the problem is whether there is a pivot.
How to tell them apart
Look for a hinge, a pivot, an axle or a bolt in the picture. If there is one, work about it, use angular momentum, and never write conservation of linear momentum. If there is none, use conservation of linear momentum for the centre of mass and conservation of angular momentum about the centre of mass, and expect two numbers in your answer.
Scaffolding comes off
The common skeleton
Name the axis, and say that both instants are taken about that same axis.
Check the net external torque about it and state that it is zero, with the reason.
Write the moment of inertia at the first instant, adding up every body that is turning.
Write the moment of inertia at the second instant, with whatever has moved or joined.
Set the two products equal and solve for the unknown rate of turning.
If the energy is asked for, compute it separately at each instant, and name who gained or lost it.
1 · fully worked
A child walking to the middle of a spinning roundabout, fully worked
A playground roundabout is a uniform disc of mass 200 kg and radius 2.00 m turning freely at 0.800 rad/s on a smooth central bearing. A 30.0 kg child stands on the rim and then walks in until they are 0.500 m from the centre. Find the new rate of turning and the change in the kinetic energy of the system.
written through the conserved quantity so that the rounded value of the new rate of turning is never squared
$$\Delta K = +45.9\ \mathrm{J}$$
the child did this work walking inwards against the outward push, which is why walking towards the middle of a spinning roundabout is hard
Answer $$\boxed{\;\omega_2 = 1.02\ \mathrm{rad/s},\qquad \Delta K = +45.9\ \mathrm{J}\;}$$
Check
Independent check on the ratio: at fixed angular momentum $K$ goes as $1/I$, so $K_2/K_1 = 520/407.5 = 1.276$ and $166.4\times 1.276 = 212$ J. The speed-up is modest rather than dramatic, which it should be for a 30 kg child on a 200 kg disc.
Every question in this family is these four steps. What changes between them is which body moves and which value you are asked for.
2 · you write the reasoning
Now an easier problem with the reasoning taken out. A turntable of moment of inertia 1.20 kg m squared turns freely at 5.00 rad/s. A metal ring of 0.300 kg m squared about the same axis, not turning, is lowered onto it and the two grip immediately. Find the rate at which they turn afterwards. The steps are all here and the physics is lighter than the last problem: the work is writing the reason for each line before opening the model answers.
Take the axis to be the common one, and note that the net external torque about it is zero.
reasoning
The axis has to be named before either instant can be written, and here the two bodies conveniently share one. Nothing outside exerts a turning effect about it: the bearing is free, and both weights act along the axis, so their lever arms about it are zero.
The angular momentum before is $L = (1.20)(5.00) = 6.00$ kg m squared per second.
reasoning
Only the turntable is turning at the first instant, so the ring contributes nothing and the total is the turntable's product alone. Note that a body at rest still has a moment of inertia; it just has no angular momentum.
Afterwards the two turn as one, with a moment of inertia of $1.20 + 0.300 = 1.50$ kg m squared.
reasoning
The two grip, which is the rotational version of two bodies locking together, so afterwards they share one rate of turning and their moments of inertia about the shared axis simply add.
So the shared rate of turning is $6.00/1.50 = 4.00$ rad/s.
reasoning
The conserved quantity divided by the new moment of inertia gives the new rate. The energy is not conserved here, and if the question had asked, the loss would be found by computing the energy separately at each instant.
3 · find the buried error
Harder than the last one, and this worked solution contains exactly two errors. Disc A, 0.900 kg m squared about its axis, turns at 12.0 rad/s. Disc B, 0.600 kg m squared about the same axis, turns at 4.00 rad/s in the opposite sense. B is lowered onto A and they grip. Find the common rate afterwards and the kinetic energy lost. Decide which two of the four steps are wrong. A step that is only wrong because it faithfully uses a number from an earlier wrong step does not count.
Step 1. The total angular momentum before is the sum of the two discs' contributions, $L = (0.900)(12.0) + (0.600)(4.00) = 13.2$ kg m squared per second.
Step 2. They grip and afterwards turn together, so the moment of inertia is $0.900+0.600 = 1.500$ kg m squared.
Step 3. Dividing the total angular momentum by the new moment of inertia gives $\omega_2 = 13.2/1.500 = 8.80$ rad/s.
Step 4. The energy lost is the final minus the initial, $\tfrac12(1.500)(8.80)^{2} - \tfrac12(0.900)(12.0)^{2} = 58.1 - 64.8 = -6.7$ J, so 6.7 J is lost.
the two buried errors (2)
⚠ step 1
The two discs turn in opposite senses, so their contributions have opposite signs and the second one has to be subtracted: $L = 10.8 - 2.40 = 8.40$ kg m squared per second.
Angular momentum feels like a stock of something, and stocks add. The sign lives in a phrase in the question rather than in a symbol on the page, so it is read and then dropped.
right
Declare a positive sense in the first line and attach a sign to every contribution as it is written, so that $L = (0.900)(12.0) - (0.600)(4.00) = 8.40$, and then $\omega_2 = 8.40/1.500 = 5.60$ rad/s.
⚠ step 4
The initial kinetic energy leaves out disc B, which was turning at 4.00 rad/s and carried $\tfrac12(0.600)(4.00)^{2} = 4.80$ J. The initial total is 69.6 J, not 64.8 J.
Disc B was subtracted a moment earlier for its angular momentum, so it feels as though it has already been dealt with. Kinetic energy has no sign, though, so a body turning the other way still adds energy rather than removing it.
right
Count every moving body on each side: $K_1 = 64.8 + 4.80 = 69.6$ J, and with the corrected rate $K_2 = \tfrac12(1.500)(5.60)^{2} = 23.5$ J, so 46.1 J is lost.
4 · the bare problem
§13.3 — dumbbells pulled in on a rotating stool●●●○○
A student sits on a stool that turns freely on a smooth bearing and holds a heavy dumbbell in each hand at arm's length. Somebody sets them turning gently, and the student then pulls both dumbbells in towards their chest.
Given
student and stool together, not counting the dumbbells: $I = 3.00\ \mathrm{kg\,m^{2}}$
two dumbbells, each 2.50 kg, starting 0.800 m from the axis
initial rate of turning 1.50 rad/s
the dumbbells end up 0.200 m from the axis
smooth bearing, so no external torque about the vertical axis
Find
(a) Find the rate of turning after the dumbbells have been pulled in.
(b) Find the change in the kinetic energy of the system, and say where it came from or went.
Hint 1/4
Name the axis first, then decide what is protected across the change of shape. The dumbbells are part of the rotating system at both instants, so they belong in both moments of inertia.
Hint 2/4
$\sum\tau_{\rm ext} = 0$ gives $I_1\omega_1 = I_2\omega_2$, with each $I$ the sum over everything turning; the energy is $K = L^{2}/2I$ and is not protected.
Hint 3/4
The data again: 3.00 kg m squared for student and stool, two 2.50 kg dumbbells moving from 0.800 m to 0.200 m from the axis, initial rate 1.50 rad/s.
Hint 4/4
The rate rises to 2.91 rad/s, and the kinetic energy rises by 6.54 J, paid for by the arms.
Show solutionAxis and licence
$$\sum\tau_{\rm ext} = 0\ \text{about the vertical axis}$$
the bearing is smooth and every weight acts parallel to the axis, so nothing outside turns the system
The two moments of inertia and the conserved product
written through the conserved quantity so the rounded new rate is not squared
$$\Delta K = +6.54\ \mathrm{J}$$
supplied by the arms; nothing outside the system did any work, so the source has to be internal
Answer $$\boxed{\;\omega_2 = 2.91\ \mathrm{rad/s},\qquad \Delta K = +6.54\ \mathrm{J}\;}$$
Check
Ratio check that avoids all of the arithmetic above: at fixed angular momentum $K_2/K_1 = I_1/I_2 = 6.20/3.20 = 1.94$, and $6.98\times 1.94 = 13.5$ J. And a scale check: 6.5 J is roughly the work of lifting a 1 kg book to shoulder height, a believable amount for one pull with both arms.
Full exam-style question
Putty strikes a hanging rod, and how far the rod swingsexam format
A uniform rod of mass 2.00 kg and length 1.20 m hangs vertically at rest from a smooth pivot at its upper end. A 0.500 kg lump of putty travelling horizontally at 8.00 m/s strikes it 0.900 m below the pivot and sticks. Find (a) the angular momentum of the putty about the pivot just before the impact, (b) the rate of turning just after it, (c) the kinetic energy lost, and (d) the greatest angle from the vertical the rod reaches.
Given
rod: $M = 2.00\ \mathrm{kg}$, $\ell = 1.20\ \mathrm{m}$, hanging from a smooth pivot at the top, at rest
putty: $m = 0.500\ \mathrm{kg}$, horizontal at $v = 8.00\ \mathrm{m/s}$, sticking at $d = 0.900\ \mathrm{m}$ below the pivot
rod about one end: $I = \tfrac13 M\ell^{2}$; the rod's centre of mass is at its middle
$g = 9.80\ \mathrm{m/s^{2}}$; the pivot is smooth, so it exerts no friction torque
Find
the angular momentum, the rate of turning, the energy lost and the greatest angle
Solution(a) The incoming angular momentum, about the pivot
the pivot exerts a large force but no torque about itself, so angular momentum about the pivot survives and linear momentum does not; writing the latter here is the standard way to lose this part
(c) The energy, counted on both sides of the impact
Two independent checks. A bracketing check on the rate: a massless rod would give $3.60/0.405 = 8.89$ rad/s and an enormously heavy one would give nearly zero, and since the rod carries 0.960 of the 1.365 the answer should sit well below the massless figure; 2.64 is about 30% of it. And a check on the angle from the pendulum side instead of the same energy line: after the impact the rod and putty swing as one physical pendulum, whose small swing rate is $\Omega = \sqrt{16.17/1.365} = 3.44$ rad/s, so a body leaving the bottom at 2.64 rad/s would swing out $2.64/3.44 = 0.766$ rad, that is 43.9 degrees. Real swings run a little wider than the small angle estimate, because the restoring torque falls off as $\sin\theta$, so 45.1 degrees is where it should be: a degree or so above 43.9, not below it and nowhere near 90. It leans on the same two constants, but the machinery is different and it would catch a slip in solving for $\cos\theta$.
Four stages, three different laws, and only one of them is conservation of energy. Choosing the law was worth more marks than any of the arithmetic.
This is the shape of the long question this material generates: a collision governed by angular momentum about the pivot, then a swing governed by energy, the first feeding the second. Practise the join, because that is where the marks concentrate.
Practice
A · concept 4 questions
1§13.1 — is angular momentum a property of the body alone●●○○○
A student is revising and writes on their sheet that once you know a body's mass, its shape and how fast it is turning, its angular momentum is fixed and can be quoted as a single number, in the same way that its mass can.
Given
the body is rigid and is turning at a known rate
its mass and its shape are both known
no axis is mentioned anywhere in the student's note
Find
(a) Decide whether the student's note is right, and say what settles it.
Hint 1/4
Compare the two quantities being equated. A mass is one number for a body; ask whether the definition of the other quantity needs anything besides the body.
Hint 2/4
$L = I\omega$, and $I = \sum m_ir_i^{2}$ counts distances from a particular axis.
Hint 3/4
The claim again: mass, shape and rate of turning are all known, and no axis has been named anywhere.
Hint 4/4
The note is wrong: the same body turning at the same rate carries different angular momenta about different axes.
the rate of turning was identical, so the difference comes entirely from the choice of axis
Answer $$\boxed{\;\text{False: } L \text{ belongs to a body and an axis together}\;}$$
Check
A single counterexample is enough to kill a universal claim, and here it is even stronger than that: the ratio between the two answers can be made anything at all by choosing the axis far enough away, so no single number could ever serve.
2§13.2 — flipping a spinning wheel while standing on a free stool●●●●○
A student stands still on a stool that turns freely on a smooth bearing, holding a spinning bicycle wheel by its axle with its axis pointing straight up. Keeping hold of it, the student turns the wheel over so the axis points straight down, touching nothing else.
Given
smooth bearing, so no external torque about the vertical axis
the wheel is spinning with its angular momentum pointing upwards at the start
the student and the stool are at rest at the start
the wheel is turned over so its angular momentum ends up pointing downwards
Find
(a) Choose what the student and stool do afterwards, and with how much angular momentum compared with the wheel's.
Hint 1/4
Write down the total for the whole system, student and stool and wheel together, at the start and at the end. Only the total is protected.
Hint 2/4
$\sum\tau_{\rm ext} = 0$ about the vertical means the total vertical component of angular momentum is the same before and after.
Hint 3/4
The situation again: wheel's angular momentum up at the start, down at the end, student and stool at rest at the start, smooth bearing throughout.
Hint 4/4
The student ends up turning the way the wheel was first spinning, carrying twice the wheel's angular momentum.
Show solutionTotal before and after
$$L_{\rm total} = +L\ \text{(wheel only, student at rest)}$$
nothing outside can turn the system about the vertical, so this number is the one that must survive
reversing the wheel changes its contribution by two units, and the student has to absorb all of it
Answer $$\boxed{\;\text{student turns the original way with } 2L\;}$$
Check
Check the halfway point: with the axle horizontal the wheel contributes nothing vertically, so the student must be carrying the whole $+L$ and turning at half the final rate. The behaviour partway through is consistent with the endpoints, which a wrong answer would not be.
3§13.4 — a force aimed straight at the pivot●●○○○
During a collision the pivot of a hinged door pushes on the door with an enormous force, far larger than anything else in the problem. A student argues that a force that big cannot be ignored when the torques about the hinge are added up.
Given
the force acts at the hinge itself
its size is very large but finite
torques are being taken about the hinge
Find
(a) Decide whether the force can be left out of the sum of torques about the hinge, and give the reason.
Hint 1/4
The question is about a lever arm, not about a size. Ask where the force acts relative to the point the torques are taken about.
Hint 2/4
$\tau = r_{\perp}F$, and a force whose line of action passes through the chosen point has $r_{\perp} = 0$.
Hint 3/4
The setup again: the force acts at the hinge, torques are taken about that same hinge, and the force is very large.
Hint 4/4
It contributes exactly zero, however large it is, because its lever arm about the hinge is zero.
the lever arm is measured from the point the torques are about, and here that point is where the force acts
$$\sum\tau_{\rm ext} = 0 \;\Rightarrow\; L\ \text{about the hinge is conserved}$$
with the pivot force out of the sum and gravity acting for only a very short time, nothing is left to change L during the impact
Answer $$\boxed{\;\text{True: the pivot force has zero torque about the pivot}\;}$$
Check
Test the same statement about a different point: about a corner of the room the pivot force does have a lever arm and does contribute. The result belongs to the pairing of that force with that point, not to the force, which is why the point has to be chosen deliberately.
4§13.5 — the direction of a straight-line particle's angular momentum●●●○○
In a figure drawn with x to the right, y upwards and z out of the page, a particle travels in the positive x direction along the horizontal line two metres above the origin. Nothing acts on it.
Given
the particle moves in the $+x$ direction
its path is the line $y = +2.00\ \mathrm{m}$
axes drawn with x right, y up, z out of the page
Find
(a) Choose the direction of its angular momentum about the origin.
Hint 1/4
Only a direction is wanted, so no mass and no speed is needed. Work out which single component can be non-zero and then get its sign.
Hint 2/4
$\vec{L} = \vec{r}\times\vec{p}$, and for vectors in the page only $L_z = xp_y - yp_x$ survives.
Hint 3/4
The setup again: motion along $+x$, path at $y = +2.00$ m, and z drawn out of the page.
Hint 4/4
It points into the page, along negative z, because the only surviving term is minus y times the x momentum.
Show solutionKill the terms that cannot survive
$$L_z = xp_y - yp_x,\qquad p_y = 0$$
the velocity is purely horizontal, so the first product is zero before any numbers are put in
$$L_z = -yp_x < 0$$
the path is above the axis and the motion is to the right, so both remaining factors are positive and the minus sign decides it
Answer $$\boxed{\;\vec{L}\ \text{along}\ -\hat{k},\ \text{into the page}\;}$$
Check
Check by watching the line of sight: as the particle crosses from left to right above the origin, an observer at the origin has to turn their head clockwise to follow it, and clockwise in this drawing is the negative z sense. The picture and the algebra agree.
B · computation 8 questions
1§13.1 — angular momentum of a spinning ball●○○○○
A bowling ball is set spinning about a diameter while it is still in the air, before it lands on the lane. It may be treated as a uniform solid sphere.
Independent bracket rather than the same multiplication again: the same 6.00 kg on the same 0.110 m radius, arranged instead as a thin spherical shell, carries a bigger shape factor and would give $L = 0.387$, while pulled in tight to the axis it would give 0. A solid ball, whose mass crowds toward the centre, has to land between those two, and 0.232 does. The usual slip of reaching for $MR^{2}$ gives 0.581, outside the bracket, so the check has teeth.
2§13.2 — a drive torque then a friction torque●●○○○
A wheel on a bearing is started from rest by a motor, run for a while, and then left alone to coast to a stop against friction in the bearing.
Given
$I = 0.450\ \mathrm{kg\,m^{2}}$ about the bearing axis, constant
starts from rest
the motor applies a constant 1.20 N m for 6.00 s and is then switched off
a constant friction torque of 0.150 N m acts at the bearing at all times after that
Find
(a) Find the rate of turning at the moment the motor is switched off.
(b) Find how much longer the wheel takes to come to rest.
Hint 1/4
Two stages, each with its own accumulated turning effect. Do them in order and carry the answer of the first into the second.
Hint 2/4
$\int\tau\,dt = \Delta L$, and with a constant torque the integral is just torque times time.
Hint 3/4
The data again: $I = 0.450$ kg m squared from rest, 1.20 N m for 6.00 s, then 0.150 N m of friction.
Hint 4/4
It reaches 16.0 rad/s and then takes another 48.0 s to stop.
Show solutionStage one, the motor
$$\Delta L = \tau t = (1.20)(6.00) = 7.20\ \mathrm{kg\,m^{2}/s}$$
the torque is constant, so the accumulated turning effect is a product rather than an integral
the moment of inertia is constant here, so angular momentum converts straight back into a rate
Stage two, the friction
$$t = \frac{7.20}{0.150} = 48.0\ \mathrm{s}$$
the whole of the stored angular momentum has to be removed, and the friction removes it at a fixed rate
Answer $$\boxed{\;\omega = 16.0\ \mathrm{rad/s},\qquad t = 48.0\ \mathrm{s}\;}$$
Check
Independent check on (b) through the energy: the wheel stores 57.6 J, friction takes 0.150 J per radian so it turns 384 rad, and at an average 8.00 rad/s that is 48.0 s. Different quantity, same answer.
3§13.3 — one disc dropped onto another●●○○○
Two discs sit on the same vertical shaft in a piece of laboratory equipment. The lower one is spinning freely and the upper one, at rest, is released so that it falls the last millimetre onto it and the two grip immediately.
Given
lower disc: $I_1 = 2.40\ \mathrm{kg\,m^{2}}$ about the shaft, turning at 9.00 rad/s
upper disc: $I_2 = 1.60\ \mathrm{kg\,m^{2}}$ about the same shaft, at rest
they grip on contact and turn together afterwards
the shaft is smooth and vertical
Find
(a) Find the rate at which the pair turns afterwards.
(b) Find the fraction of the kinetic energy that is lost.
Hint 1/4
Decide first what survives the gripping and what does not, then compute the surviving one and use it to answer both parts.
Hint 2/4
$\sum\tau_{\rm ext} = 0$ gives $I_1\omega_1 = (I_1+I_2)\omega_2$; the energy is $K = L^{2}/2I$ and is not protected.
Hint 3/4
The data again: 2.40 kg m squared at 9.00 rad/s meeting 1.60 kg m squared at rest, gripping on a smooth vertical shaft.
Hint 4/4
They settle at 5.40 rad/s, and 40.0% of the kinetic energy has gone.
Show solutionThe conserved quantity
$$L = (2.40)(9.00) = 21.6\ \mathrm{kg\,m^{2}/s}$$
the shaft is smooth and vertical, so no external torque acts about it during the grip
at fixed angular momentum the energy is inversely proportional to the moment of inertia, which avoids computing either energy
$$\text{lost fraction} = 0.400,\qquad \Delta K = -38.9\ \mathrm{J}$$
starting from $K_1 = \tfrac12(2.40)(9.00)^{2} = 97.2$ J
Answer $$\boxed{\;\omega_2 = 5.40\ \mathrm{rad/s},\qquad 40.0\%\ \text{of the energy lost}\;}$$
Check
Independent check from scratch: $K_1 = 97.2$ J and $K_2 = \tfrac12(4.00)(5.40)^{2} = 58.3$ J, a loss of 38.9 J, which is 40.0%. The sign is forced too: bodies that grip lose energy, never gain it.
4§13.4 — a torque from two awkward vectors●●●○○
A bracket is bolted to a wall at the origin. A cable pulls on a point of the bracket with a force whose components are given, and the point itself sits at a stated position relative to the bolt.
Given
$\vec{r} = (0.250, -0.400, 0)\ \mathrm{m}$ from the bolt to the point where the cable is attached
$\vec{F} = (-30.0, 15.0, 0)\ \mathrm{N}$
axes drawn with x right, y up and z out of the page
Find
(a) Find the torque about the bolt, as a vector, and say which way it would turn the bracket.
Hint 1/4
Both vectors lie in the page, so the answer can only point along one axis. Find out which, and then get its sign right.
Hint 2/4
$\tau_z = r_xF_y - r_yF_x$ is the only surviving line of the cross product when both vectors lie in the page.
Hint 3/4
The data again: $\vec{r} = (0.250, -0.400, 0)$ m and $\vec{F} = (-30.0, 15.0, 0)$ N.
Hint 4/4
The torque is 8.25 N m into the page, that is, clockwise.
Independent route through magnitudes: $r = 0.472$ m, $F = 33.5$ N, angle 148.6 degrees with sine 0.521, so $rF\sin\theta = 8.25$ N m. The size agrees, computed without a single component.
5§13.5 — angular momentum of a particle from its position and velocity●●●○○
A tracking system reports the position and velocity of a single object at one instant, both measured from the same fixed origin, and asks for its angular momentum about that origin.
Given
$m = 3.00\ \mathrm{kg}$
$\vec{r} = (2.00, -1.00, 0)\ \mathrm{m}$ from the origin
$\vec{v} = (4.00, 5.00, 0)\ \mathrm{m/s}$
axes drawn with x right, y up and z out of the page
Find
(a) Find the angular momentum about the origin, as a vector.
Hint 1/4
Turn the velocity into a momentum before doing anything else, so the mass does not get forgotten halfway through.
Hint 2/4
$\vec{L} = \vec{r}\times\vec{p}$, and for vectors in the page only $L_z = xp_y - yp_x$ survives.
Hint 3/4
The data again: 3.00 kg at $(2.00, -1.00, 0)$ m moving at $(4.00, 5.00, 0)$ m/s.
Hint 4/4
The angular momentum is 42.0 kg m squared per second out of the page.
Independent route: $r = 2.24$ m, $p = 19.2$ kg m/s, angle 77.9 degrees with sine 0.978, giving 42.0. Same number without touching a component.
6§13.6 — a rolling ball taken about a point on the floor●●●○○
A solid ball rolls without slipping in a straight line across a level floor. A fixed point is marked on the floor directly beneath the ball's centre at the instant of interest.
the centre moves along a line whose perpendicular distance from the floor point is the radius
$$L = 0.900+0.360 = 1.26\ \mathrm{kg\,m^{2}/s}$$
both parts turn the same way about that point, so they add
Answer $$\boxed{\;L = 1.26\ \mathrm{kg\,m^{2}/s},\ \text{of which } 0.900\ \text{travel and } 0.360\ \text{spin}\;}$$
Check
Second method as promised: a rolling body is momentarily turning about its contact point, where $I = 0.0180+0.0450 = 0.0630$ kg m squared, giving $(0.0630)(20.0) = 1.26$. Different pictures, exact agreement.
7§13.6 — a rod clamped at an angle to its shaft●●●●○
In a balancing rig, a light rod carrying a mass at each end is clamped at its middle to a vertical shaft, but the clamp has been set at an angle rather than square, so the rod is skew to the shaft.
Given
two masses, each $m = 1.20\ \mathrm{kg}$, each $\ell = 0.400\ \mathrm{m}$ from the clamp
the rod makes 40.0 degrees with the shaft
$\omega = 8.00\ \mathrm{rad/s}$ about the shaft
the rod itself is light and contributes nothing
Find
(a) Find the component of the angular momentum along the shaft.
(b) Find the magnitude of the total angular momentum and the angle it makes with the shaft.
Hint 1/4
Two different distances are needed and mixing them up is the whole difficulty: one is the distance from the shaft, the other the distance from the clamp.
Hint 2/4
$L_z = I_z\omega$ with $I_z = 2m(\ell\sin\theta)^{2}$; the total has magnitude $2m\omega\ell^{2}\sin\theta$.
Hint 3/4
The data again: two 1.20 kg masses, each 0.400 m from the clamp, rod at 40.0 degrees to the shaft, turning at 8.00 rad/s.
Hint 4/4
The axial component is 1.27 and the total is 1.97 kg m squared per second, leaning 50.0 degrees from the shaft.
Show solutionAlong the shaft, where the familiar product still works
Independent geometric check: 40.0 degrees for the rod and 50.0 for the angular momentum add to exactly 90, so the vector is perpendicular to the rod, which is what the theory demands for two equal masses on a light rod. Had they not added to a right angle, one calculation would be wrong.
8§13.7 — the precession rate of a laboratory ●●●○○
A demonstration gyroscope is spun up and its axle is then placed on a support at one end, with the axle horizontal, and released.
Given
$I = 0.0140\ \mathrm{kg\,m^{2}}$ about the gyroscope's own axis
spinning at $\omega = 240\ \mathrm{rad/s}$
$M = 0.850\ \mathrm{kg}$, centre of mass 0.0650 m from the support point
$g = 9.80\ \mathrm{m/s^{2}}$, axle horizontal
Find
(a) Find the rate at which the axle sweeps round.
(b) Find the time for one complete circuit, and check that the result may be trusted.
Hint 1/4
Two quantities feed the answer: the turning effect of the weight about the support, and the amount of turning stored in the spin.
Hint 2/4
$\Omega = \tau/(I\omega)$ with $\tau = MgD$, and one circuit takes $2\pi/\Omega$.
Hint 3/4
The data again: $I = 0.0140$ kg m squared at 240 rad/s, mass 0.850 kg with its centre 0.0650 m from the support.
Hint 4/4
It sweeps round at 0.161 rad/s, taking 39.0 s per circuit, and the fast spin condition is comfortably met.
both quantities are taken about the same support point, which is what the relation requires
$$t = \frac{2\pi}{0.1611} = 39.0\ \mathrm{s}$$
converting a rate of going round into a time for one circuit, exactly as for any uniform circular motion
Answer $$\boxed{\;\Omega = 0.161\ \mathrm{rad/s},\qquad t = 39.0\ \mathrm{s}\;}$$
Check
Two checks. The condition: $\Omega/\omega = 0.00067$, so the fast spin assumption is safe. The scale: the end of the axle creeps round at $(0.0650)(0.161) = 0.010$ m/s, about what a demonstration gyroscope visibly does.
C · exam level 5 questions
1§13.2 — a motor whose torque dies away and reverses●●●●○
A grinding wheel is driven from rest by a motor whose torque falls steadily and has reversed by the time it is switched off. The wheel then coasts to a stop against a constant friction torque at the bearing.
Given
$I = 0.850\ \mathrm{kg\,m^{2}}$ about the bearing axis, constant, and the wheel starts from rest
the motor supplies $\tau = 3.00 - 0.750\,t$ in newton metres, with $t$ in seconds, for $0 \le t \le 5.00\ \mathrm{s}$
at $t = 5.00\ \mathrm{s}$ the motor is switched off
a constant friction torque of 0.200 N m then acts at the bearing
Find
(a) Find the instant at which the wheel is turning fastest, and the rate it reaches then.
(b) Find the rate of turning at $t = 5.00$ s, when the motor is switched off.
(c) Find how much longer the wheel takes to come to rest after that.
Hint 1/4
The rate of turning is largest when it stops rising, and it stops rising when the thing driving it changes sign. Find that instant before computing anything.
Hint 2/4
$\int\tau\,dt = \Delta L$, and the rate of turning peaks when $\tau = 0$; afterwards $\int\tau\,dt$ takes angular momentum away.
Hint 3/4
The data again: $I = 0.850$ kg m squared from rest, $\tau = 3.00 - 0.750t$ N m up to $t = 5.00$ s, then 0.200 N m of friction.
Hint 4/4
The peak is 7.06 rad/s at $t = 4.00$ s, the rate at 5.00 s is 6.62 rad/s, and the wheel then takes another 28.1 s to stop.
Show solution(a) Where the rate peaks
$$\tau = 0 \;\Rightarrow\; t = \frac{3.00}{0.750} = 4.00\ \mathrm{s}$$
the rate of turning stops rising exactly when the torque changes sign, not when it is largest
Independent check on (c) through the energy: the wheel holds 18.6 J, friction takes 0.200 J per radian so it turns 93.1 rad, and at an average 3.31 rad/s that is 28.1 s. And on (a): using the constant-acceleration equations with the initial torque would give $\omega = 17.6$ rad/s, nearly three times too large.
2§13.3 — a ball thrown at the rim of a mounted disc●●●●○
A uniform disc is mounted on a fixed frictionless axle through its centre and is at rest. A small ball of clay is thrown at it so that it arrives moving horizontally along the tangent to the rim, and sticks where it lands.
Given
disc: $M = 4.00\ \mathrm{kg}$, $R = 0.350\ \mathrm{m}$, $I = \tfrac12 MR^{2}$, on a fixed frictionless axle, at rest
clay: $m = 0.200\ \mathrm{kg}$ at $v = 12.0\ \mathrm{m/s}$, tangential at the point of contact
the clay sticks to the rim on arrival
Find
(a) Find the angular momentum of the clay about the axle just before it lands.
(b) Find the rate at which the disc and clay turn just after.
(c) Find the kinetic energy lost.
(d) Say what the answer to (b) would have been if the clay had instead been thrown straight at the axle, and why.
Hint 1/4
The axle is what holds the disc in place, so it is also the point everything should be taken about. Work out what the incoming clay carries about it.
Hint 2/4
$L = mvr_{\perp}$ for the clay, then $L = I_{\rm total}\omega$ afterwards; the axle exerts no torque about itself, so $L$ about the axle survives the impact.
Hint 3/4
The data again: a 4.00 kg disc of radius 0.350 m at rest, struck tangentially at the rim by 0.200 kg of clay at 12.0 m/s that sticks.
Hint 4/4
The clay brings 0.840 kg m squared per second, the pair ends up at 3.12 rad/s, 13.1 J is lost, and a shot aimed at the axle would produce no turning at all.
Independent check on the energy fraction: the fraction kept is the arriving body's moment of inertia over the total, $0.0245/0.2695 = 0.0909$, and $0.0909 \times 14.4 = 1.31$ J. That route never mentions the final rate of turning, and it is the exact rotational twin of the mass ratio in a linear inelastic collision.
3§13.3 — a falling rod that picks up a lump on the way●●●●○
A uniform rod is held horizontal, hinged at one end to a smooth pivot, and released. As it swings down through the vertical it strikes a small lump of putty resting against a light stop, and the putty sticks to the rod.
Given
rod: $M = 1.20\ \mathrm{kg}$, $\ell = 0.900\ \mathrm{m}$, hinged at one end, $I = \tfrac13 M\ell^{2}$
released from rest in the horizontal position
putty: $m = 0.300\ \mathrm{kg}$, at rest, at 0.600 m from the pivot
$g = 9.80\ \mathrm{m/s^{2}}$, smooth pivot
Find
(a) Find the rate of turning of the rod just before it reaches the putty.
(b) Find the rate of turning just after the putty has stuck.
(c) Find the kinetic energy lost in the sticking.
Hint 1/4
Two stages with two different laws. Decide which law governs the swing and which governs the sticking before writing anything.
Hint 2/4
Swing: mechanical energy, with the rod's centre of mass falling half the length. Sticking: angular momentum about the pivot, since the pivot exerts no torque about itself.
Hint 3/4
The data again: a 1.20 kg rod 0.900 m long hinged at one end and released from horizontal, meeting 0.300 kg of putty at rest 0.600 m from the pivot.
Hint 4/4
It arrives at 5.72 rad/s, leaves at 4.29 rad/s, and 1.32 J is lost in the sticking.
Independent check: at fixed angular momentum the energy keeps the fraction $0.324/0.432 = 0.750$, so a quarter of 5.292 J goes, which is 1.32 J. The rate drops by the same factor, $5.72 \times 0.750 = 4.29$.
4§13.6 — a cylinder rolling down a slope, taken about the slope●●●●●
A solid cylinder is released from rest on a slope and rolls without slipping down it. All the angular momentum in this question is taken about a point fixed on the surface of the slope, the point the cylinder happens to be touching at the start.
slope at 20.0 degrees to the horizontal, rolling without slipping, released from rest
$g = 9.80\ \mathrm{m/s^{2}}$
the origin is a fixed point on the slope surface
Find
(a) By taking torques about that fixed point, find the acceleration of the cylinder along the slope.
(b) After it has travelled 2.00 m along the slope, find its angular momentum about the same fixed point.
Hint 1/4
Choosing the point on the slope is what makes part (a) short: two of the three forces then have no lever arm at all.
Hint 2/4
About the contact point the only torque is from the component of the weight along the slope, acting at the centre, and $I_{\rm contact} = I_{\rm cm}+MR^{2}$; then $\tau = I\alpha$ and $a = \alpha R$.
Hint 3/4
The data again: a 5.00 kg solid cylinder of radius 0.250 m released from rest on a slope of 20.0 degrees and rolling without slipping.
Hint 4/4
The acceleration is 2.23 m/s squared, and after 2.00 m the angular momentum about the slope point is 5.61 kg m squared per second.
Show solution(a) Torques about the point of contact
the centre travels parallel to the slope at a perpendicular distance R from it, so the travel part uses the radius whatever point on the slope was chosen
Answer $$\boxed{\;a = 2.23\ \mathrm{m/s^{2}},\qquad L = 5.61\ \mathrm{kg\,m^{2}/s}\;}$$
Check
Three checks. The acceleration against $a = g\sin\theta/(1+I_{\rm cm}/MR^{2}) = 2.23$ m/s squared. The speed against energy: the drop is 0.684 m and $Mgh = \tfrac34 Mv^{2}$ gives 2.99 m/s. The angular momentum against the single-rotation picture: $I_{\rm contact}\omega = (0.4688)(11.96) = 5.61$.
5§13.7 — a precessing wheel, spun fast and then slowly●●●●○
A wheel with all of its mass at the rim is mounted at the middle of a light axle. One end of the axle is placed on a pivot and released, with the axle horizontal, so that the wheel is a quarter of a metre from the pivot.
Given
wheel: $M = 3.20\ \mathrm{kg}$, $R = 0.280\ \mathrm{m}$, all mass at the rim, so $I = MR^{2}$
the centre of the wheel is $D = 0.250\ \mathrm{m}$ from the pivot, axle horizontal
first case: the wheel spins at 30.0 rad/s
second case: the wheel has slowed to 10.0 rad/s; $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the angular momentum of the wheel and the torque of its weight about the pivot.
(b) Find the precession rate and the time for one circuit while it spins at 30.0 rad/s.
(c) Repeat for the wheel spinning at 10.0 rad/s, and say whether the answer should be trusted.
Hint 1/4
Parts (b) and (c) are the same calculation with one number changed, so set it up once. Part (c) also asks a second question, about whether the setup still applies.
Hint 2/4
$\Omega = MgD/(I\omega)$, one circuit takes $2\pi/\Omega$, and the relation assumes $\Omega \ll \omega$.
Hint 3/4
The data again: a 3.20 kg rim wheel of radius 0.280 m, centre 0.250 m from the pivot, spinning first at 30.0 rad/s and then at 10.0 rad/s.
Hint 4/4
At 30.0 rad/s it goes round every 6.03 s; at 10.0 rad/s the formula gives 2.01 s but the fast spin condition has failed, so that figure is not reliable.
Check the trend against the physical picture: a dying top swings round faster and faster and then wobbles and falls, which is the sequence here, a rising precession rate followed by the breakdown of the description. A falling rate would contradict what everybody has watched happen.
D · interleaved 4 questions
1§13.0 — two carts and a spring at the end of the track●●●○○
A cart runs along a smooth level track, hits a heavier cart standing still, and the two lock together. They then run on and squash a spring fixed at the end of the track.
Given
moving cart: 1.20 kg at 4.00 m/s
stationary cart: 2.80 kg
they lock together on contact
the track is smooth and level, and the spring has stiffness 250 N/m
Find
(a) Find the speed of the pair just after they lock together.
(b) Find the kinetic energy lost in the locking.
(c) Find how far the spring is squashed at the moment the carts are momentarily at rest.
Hint 1/4
Read the question for what survives each stage. The two stages obey different rules and picking the wrong one for the first stage is the whole trap.
Hint 2/4
In a collision where the bodies lock, total momentum is unchanged and kinetic energy is not; afterwards, on a smooth track with a spring, mechanical energy is unchanged and $\tfrac12 kx^{2}$ is the store.
Hint 3/4
The data again: 1.20 kg at 4.00 m/s meeting 2.80 kg at rest, locking together, then meeting a spring of stiffness 250 N/m.
Hint 4/4
They move off at 1.20 m/s, 6.72 J is lost in the locking, and the spring is squashed by 0.152 m.
about fifteen centimetres, a reasonable squash for a soft laboratory spring
Answer $$\boxed{\;v = 1.20\ \mathrm{m/s},\quad \Delta K = -6.72\ \mathrm{J},\quad x = 0.152\ \mathrm{m}\;}$$
Check
Independent check on the loss: the fraction kept is $m_1/(m_1+m_2) = 0.300$, so 70.0% goes and $0.700\times 9.60 = 6.72$ J. On the compression too: the spring force at maximum squash is 38 N, a firm but ordinary push.
2§13.0 — a hoop and a disc released together on a ramp●●●●○
A hoop and a solid disc have the same mass and the same radius. Both are released from rest at the top of the same ramp and roll down it without slipping, through a vertical drop of 1.50 m.
Given
both bodies: $M = 2.00\ \mathrm{kg}$, $R = 0.200\ \mathrm{m}$
Independent check by splitting the hoop's answer: $\omega = 19.2$ rad/s, spin part 1.53, travel part 1.53, total 3.07. The two being equal is what a hoop must give, since its moment of inertia about the centre matches the travel term's factor.
3§13.0 — a comet at its nearest and its farthest●●●●○
A comet travels on a long closed path round the Sun. At its nearest approach and at its farthest point its velocity is at right angles to the line joining it to the Sun. Nothing acts on it except the Sun's pull.
Given
nearest approach: $8.00\times 10^{10}\ \mathrm{m}$ from the Sun, moving at $5.40\times 10^{4}\ \mathrm{m/s}$
farthest point: $5.20\times 10^{12}\ \mathrm{m}$ from the Sun
at both points the velocity is perpendicular to the line to the Sun
the only force on the comet is the Sun's pull, directed straight at the Sun
Find
(a) Find the comet's speed at its farthest point.
(b) State what property of the Sun's pull makes the method work.
Hint 1/4
The comet's mass is not given and does not need to be, which is a strong hint about which relation is wanted. Ask what stays the same between the two points.
Hint 2/4
A force whose line of action passes through a point has zero torque about that point, so $L = mvr_{\perp}$ about the Sun is unchanged, and at both named points $r_{\perp} = r$.
Hint 3/4
The data again: $8.00\times 10^{10}$ m at $5.40\times 10^{4}$ m/s, and $5.20\times 10^{12}$ m at the far end, with the velocity perpendicular to the radius at both.
Hint 4/4
The far speed is 831 m/s, and the method works because the Sun's pull always points straight at the Sun.
the line of action passes through the point the torques are taken about, so the lever arm is zero however strong the pull is
Answer $$\boxed{\;v_2 = 831\ \mathrm{m/s}\;}$$
Check
Check the ratio directly: $r_2/r_1 = 65.0$ and $5.40\times 10^{4}/65.0 = 831$ m/s. A scale check too: under a kilometre a second at the far end against fifty at the near end is why comets spend almost all their time far away and sweep past the Sun in weeks.
4§13.0 — a puck on a cord pulled through a hole in the table●●●●○
A puck slides on a smooth horizontal table, tied to a light cord that passes down through a small hole in the middle of the table. Somebody below pulls steadily on the cord, and the puck spirals inwards while continuing to circle the hole.
Given
puck: $m = 0.250\ \mathrm{kg}$, on a smooth horizontal table
at first it circles at radius 0.600 m with speed 2.00 m/s
the cord is pulled until the radius is 0.300 m
the cord is light and the table is smooth
Find
(a) Find the puck's speed at the smaller radius.
(b) Find the tension in the cord at the smaller radius.
(c) Find the work done by whoever pulled the cord.
Hint 1/4
Ask which way the cord's pull points relative to the hole, and what that means for turning effects about the hole. Then decide whether the energy is protected as well.
Hint 2/4
The tension is aimed straight at the hole, so $\tau = 0$ about it and $mvr$ is unchanged; the tension does do work, so the energy is not; and while circling, $T = mv^{2}/r$.
Hint 3/4
The data again: a 0.250 kg puck circling at 0.600 m and 2.00 m/s on a smooth table, pulled in to 0.300 m by a cord through a hole.
Hint 4/4
The speed doubles to 4.00 m/s, the tension is 13.3 N, and the person did 1.50 J of work.
Show solution(a) What the pull cannot change
$$\vec{T}\ \text{points at the hole} \;\Rightarrow\; \tau = 0\ \text{about the hole}$$
the line of action passes through the point, so the lever arm is zero and the angular momentum there is fixed
nothing else does work on the puck, so the whole change in its kinetic energy came from the person pulling
Answer $$\boxed{\;v_2 = 4.00\ \mathrm{m/s},\quad T = 13.3\ \mathrm{N},\quad W = 1.50\ \mathrm{J}\;}$$
Check
Independent check by scaling: at fixed angular momentum $v \propto 1/r$, so $T \propto 1/r^{3}$ and halving the radius multiplies the tension by eight. The starting tension is 1.67 N, and $8 \times 1.67 = 13.3$ N. The work is positive too, as it must be when the person pulls inwards while the puck moves inwards.
Mistake ledger (19 entries)
⚠ Quoting an angular momentum with no axis attached to it
The formula produces a single number and nothing in the symbol reminds you that the number belonged to a particular axis, so the axis gets dropped on the way to the next line.
wrong$$L = 3.60\ \mathrm{kg\,m^{2}/s}$$
right$$L = 3.60\ \mathrm{kg\,m^{2}/s}\ \text{about the contact point}$$
⚠ Putting a rate in rpm straight into the definition
Machines are rated in rpm and the number looks like a perfectly good rate of turning, so it goes in unconverted and the answer comes out about ten times too large.
⚠ Using a tabulated moment of inertia about the wrong axis
The table is indexed by the shape of the body, so it is easy to read off the row for a rod and forget that the row also names an axis that may not be the one in the question.
⚠ Treating a time-varying torque as though it were constant at its final value
The final value is the number written in the question, so it is the one at hand, and the constant-acceleration equations are the reflex from the previous section.
wrong$$\Delta L = \tau(t_2)\,\Delta t = (2.00)(5.00) = 10.0$$
right$$\Delta L = \int_0^{5.00}\!0.400\,t\,dt = 5.00$$
⚠ Assuming that conserved angular momentum means conserved energy
Both are called conservation laws, both are used at two named instants, and in the linear collisions of the previous sections the two were often discussed in the same breath.
right$$I_1\omega_1 = I_2\omega_2,\qquad K = \frac{L^{2}}{2I}\ \text{changes}$$
⚠ Leaving the arriving body out of the final moment of inertia
The final rate of turning is asked about the turntable, so the turntable's own value is the one in mind, and the lump that just landed on it does not feel like part of the rotating body yet.
⚠ Using the cosine of the angle instead of the sine
The dot product from the work section uses the cosine and is drilled first, so the reflex carries over to the other kind of product.
wrong$$|\vec{r}\times\vec{F}| = rF\cos\theta$$
right$$|\vec{r}\times\vec{F}| = rF\sin\theta$$
⚠ Using the distance to the particle instead of the miss distance
The position vector is what is drawn on the figure, so its length is the number in front of you, while the perpendicular distance has to be constructed.
⚠ Forgetting that the answer is tied to the origin that was chosen
For a rigid body on an axle the axis is drawn in the picture and cannot be forgotten; for a free particle the origin is a choice you made, and choices are easy to lose.
wrong$$L = 30.0\ \mathrm{kg\,m^{2}/s}$$
right$$L = 30.0\ \mathrm{kg\,m^{2}/s}\ \text{about the origin at } (0,0)$$
⚠ Treating the fixed-axis formula as a vector statement about any axis
It is written as a product of two quantities and it is true in every example of the first half of the section, so it starts to feel like a definition rather than a special case.
wrong$$\vec{L} = I\vec{\omega}\ \text{for any axis}$$
right$$L_z = I_z\omega\ \text{always};\quad \vec{L}\parallel\vec{\omega}\ \text{only about a symmetry axis}$$
⚠ Using only the spin part for a rolling body about an outside point
The spin is the visible rotation, and the travel part looks like translation, which does not feel like it should count as turning at all.
⚠ Measuring the lever arm to the end of the axle instead of to the centre of mass
The axle is the long visible thing in the picture and its length is the number given, while the centre of mass has to be located.
wrong$$\tau = Mg\,L_{\rm axle}$$
right$$\tau = MgD,\quad D = \text{support to centre of mass}$$
⚠ Reporting the precession rate as a number of turns per second without converting
The answer comes out of a formula in rad/s and the question often asks how long one circuit takes, and the two are easy to confuse when both are small numbers.
wrong$$\text{one circuit in } 1/0.239 = 4.19\ \mathrm{s}$$
right$$\text{one circuit in } 2\pi/0.239 = 26.3\ \mathrm{s}$$
⚠ Writing conservation of linear momentum for a collision onto a pivoted body
It is the reflex from the linear momentum section, and the picture gives no hint that the pivot is delivering a large impulse of its own.
spin much faster than the precession; torque perpendicular to the angular momentum; steady tilt
Check yourself
Close the page and write from memory: angular momentum for a body on an axle and for a particle; the general second law and the one thing it can do that the familiar form cannot; the condition for conservation and the separate question about the energy; the two ways of computing a cross product; the two parts of a rolling wheel's angular momentum; and the sweeping rate of a top. Then open the formula card and mark what you missed.
Compute the angular momentum of a body on an axle, quote it with its axis, and explain why the same body has two values about two axes?
c-angular-momentum
Handle a torque that varies with time by accumulating it, and say why the constant-acceleration equations are unavailable there?
c-tau-dl-dt
Solve a shape-change or a gripping problem and then decide separately whether the energy went up or down, naming who paid?
c-conservation
Get a torque from components and from magnitudes and an angle, and get its direction right without guessing?
c-cross-product
Give the angular momentum of a particle moving in a straight line about a point beside its path, and use it to set up a collision onto a pivot?
c-particle-L
Split a rolling body's angular momentum into two parts about a ground point, and check the answer by the contact point route?
c-general-rotation
Find the sweeping rate of a top from its weight, geometry and spin, and check that the fast spin condition really holds?
c-precession
Glossary (16 terms)
angular momentumaçısal momentum
What a turning body carries about a stated axis: the moment of inertia about that axis times the rate of turning, or for a particle its momentum times the perpendicular distance from the point to its line of travel. Meaningless until the axis is named.
angular impulseaçısal impuls
The torque accumulated over an interval of time, equal to the change in angular momentum it produces. A large torque acting briefly and a small torque acting for a long time deliver the same one.
cross productvektörel çarpım
A multiplication of two vectors returning a third, perpendicular to both, of size equal to their lengths times the sine of the angle between them. Reversing the order reverses the answer.
right hand rulesağ el kuralı
The convention fixing the direction of a cross product and of an angular velocity: curl the right hand from the first vector to the second, and the thumb points along the answer.
conservation of angular momentumaçısal momentumun korunumu
With no net external torque about a chosen axis, the angular momentum about that axis is the same at every instant, whatever the system does to its shape. It says nothing about the kinetic energy.
symmetry axissimetri ekseni
An axis about which the body looks the same from all sides, so that turning about it gives an angular momentum pointing straight along it. About any other axis it leans away and the mounting has to push it round.
rotational collision
Bodies joining or gripping and afterwards sharing one rate of turning. Angular momentum about the axis crosses it unchanged and kinetic energy does not, so it is the rotational counterpart of a perfectly inelastic collision.
The part taken about the body's own centre of mass, as against the part it has because that centre travels. A rolling wheel carries both, and about a ground point they add.
ani dönme ekseni
The line a body may be regarded as turning about at one instant. For rolling without slipping it passes through the contact point, which is why the moment of inertia there gives the whole angular momentum in one product.
merkezi kuvvet
A force whose line always passes through one fixed point, such as the Sun's pull on a comet. Its torque about that point is zero however strong it is.
precessionpresesyon
The slow sweeping of a spinning body's axis round a cone, caused by a torque at right angles to its angular momentum. Its rate is the torque divided by the angular momentum, so a faster spin makes it slower.
gyroscopejiroskop
A wheel mounted so that it spins rapidly about its own axis while that axis is free to turn. Its resistance to being moved and its sideways response to a torque are what make it useful.
nutationnutasyon
The small nodding of a real top's axis that rides on the steady sweeping motion. Named here only so the gap between the idealised description and a real top is not hidden.
açısal hız vektörü
The rate of turning written as a vector along the axis by the right hand rule. It carries the same information as the plus or minus sign, and becomes necessary when the axis can move.
torque vectortork vektörü
The turning effect written as the position of the point of application crossed with the force. It points along the axis the force is trying to turn the body about, perpendicular to both inputs.
The point at which a blow on a pivoted body leaves the pivot free of sudden sideways force. Struck nearer in, the pivot is pushed backwards; struck further out, forwards.
What comes next
§14 · Oscillations
Everything here was followed once, from one named instant to another: the putty struck the rod, the rod swung up, and we stopped watching. The next section takes the same pivoted rod and the same spring and asks what happens when the body comes back through where it started and does it all again. The new question is not what the state is, but how long the repetition takes.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook. Its treatment of angular momentum and general rotation covers this ground in this order, and its end of chapter problems are the right next step after the practice set here.
Course syllabus: the week line and the assessment table The scope comes from the week line, which reads Angular Momentum and General Rotation and quotes no chapter numbers, so none is quoted here. The weightings on the summary card come from the assessment table.
SI units: the kilogram metre squared per second and the newton metre Angular momentum has no named unit of its own, so an answer must always be written out in full. The newton metre of torque is kept distinct from the joule, though the two are the same combination of base units.