7 concepts18 worked examples32 exercises5 exam-level7 figures
What are you here for?
12Rotational motion: angular kinematics, torque, rotational inertia and rolling
Two cans look identical from the outside and balance each other on a kitchen scale. One of them is packed solid all the way through; in the other the same contents have been pressed against the wall and the middle is empty. Let them go together at the top of a sloping board and one of them reaches the bottom first, by a gap you can see without a stopwatch, and it wins again every single time. Everything you have used so far says they should arrive together: the same drop, the same weight, nothing rubbing.
By the end of this section you can say which can wins before either is released, put a number on the losing margin, and do it without being told either mass or either radius.
In 60 seconds
A turning about a fixed axle is described by one angle for the whole body, so every straight-line result you own comes back with new letters: $\theta$ for position, $\omega$ for velocity, $\alpha$ for acceleration, torque in place of force, and the moment of inertia in place of mass, with $\sum\tau = I\alpha$ as the second law and $\tfrac12 I\omega^{2}$ as the kinetic energy; a body that rolls without slipping does both motions at once, tied together by $v = \omega R$.
Arc, speed and acceleration of a point at radius r
$$s = R\theta,\qquad v = R\omega,\qquad a_{\rm tan} = R\alpha$$
linking what one point on the body does to what the whole body does; every angle in
a wheel, ball or cylinder that is not skidding; the contact point is momentarily at rest and the friction there does no work
Three most common mistakes
Feeding degrees or revolutions into a formula that only accepts radians. Every $s = R\theta$, every $v = R\omega$ and every $W = \tau\theta$ wants radians, and 60 rpm is 6.28 rad/s, not 60.
Using the distance to the point where the force is applied instead of the perpendicular distance to its . A force pointing at the axle has a of zero however far out you apply it.
Treating the moment of inertia as a property of the body. The same rod has $ML^{2}/12$ about its middle and four times as much, $ML^{2}/3$, about its end; quoting $I$ without an axis is quoting nothing.
The syllabus publishes weights and nothing else: two midterms at 20% each, the final at 25%, quizzes 10%, homework 5% and the laboratory 20%. It does not say which paper carries this week, so nothing is claimed here about that. What the table does say is that the laboratory is worth as much as either midterm on its own, and that written papers decide 65% of the course.
How much time do you have?
10 minutes
You leave able to convert an rpm into a usable number, to get a torque right by using the perpendicular distance, and to write the one line that replaces $F = ma$ when the body turns. That is most of a quiz and the first mark of most exam questions.
In 60 seconds · Formula card · Angular position, angular velocity and angular acceleration: one number for the whole body · Torque: the same force does different work depending on where it acts · Newton's second law for a body on a fixed axle · Mistake ledger
45 minutes
You add the two blocks that separate a pass from a good mark: where the moment of inertia comes from and why the axis has to be named with it, and the energy route, which answers in one line what the force route needs three equations to reach.
In 60 seconds · Conventions used here · Angular position, angular velocity and angular acceleration: one number for the whole body · Constant angular acceleration: the equations you already have, wearing new letters · Torque: the same force does different work depending on where it acts · Rotational inertia: not how much mass, but how far out it sits · Newton's second law for a body on a fixed axle · Method boxes · Scaffolding comes off · Practice B · Check yourself
full read
Everything above plus the energy of a turning body, rolling without slipping and the race down the ramp that the opening two cans were about, the full exam-style question with a pulley that has mass, and the interleaved set that makes you choose the tool before you use it.
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
Convert between revolutions, degrees and radians and between rpm and rad/s, and use $s = R\theta$, $v = R\omega$ and $a_{\rm tan} = R\alpha$ to move between what one point does and what the whole body does.
Solve a constant angular acceleration problem by copying the straight-line equations with $\theta$, $\omega$ and $\alpha$ in place of $x$, $v$ and $a$, and report the answer in revolutions when the question asks for turns.
Compute the torque of a force about a stated axis using the perpendicular distance, attach the right sign to it, and add several torques into a net torque.
Calculate a moment of inertia for point masses and for the standard shapes, state the axis it belongs to, and shift it to a parallel axis with $I = I_{\rm cm} + Md^{2}$.
Apply $\sum\tau = I\alpha$ together with $\sum F = ma$ to a cord, a pulley with mass and a hanging body, using the constraint that links the linear and the angular acceleration.
Use $\tfrac12 I\omega^{2}$, $W = \tau\Delta\theta$ and $P = \tau\omega$ to answer a rotation question that never mentions time, and check the result against the force route.
Predict the speed and the ordering of bodies that roll without slipping down a ramp, split the kinetic energy into its two parts, and find the friction a given body needs in order to roll rather than skid.
Syllabus coverage
Rotational Motion
Angular position, velocity and acceleration for a rigid body on a fixed axis; the link between angular and linear quantities; constant angular acceleration; torque; moment of inertia including the parallel axis theorem; rotational dynamics; rotational kinetic energy, work and power; rolling without slipping
The week line carries no chapter numbers, so no chapter number is quoted anywhere in this section. The scope taken is the standard content of that title in the set textbook, spread across the seven blocks.
covered
Angular quantities as vectors and the right hand rule
Giving the angular velocity a direction along the axis rather than a sign
Deferred to the next section, which needs a direction in space rather than a sign because the axis itself is allowed to move there. Every body on this page turns about one fixed axle, and for that a plus or a minus sign carries all the information a direction would.
deferred
Rotation about a moving or changing axis
Bodies whose axis is not fixed, and quantities that survive when the shape changes
Deferred to the next section. Everything here assumes the axis stays put and the body keeps its shape, which is what makes $I$ a constant that can sit outside the derivative.
deferred
Repeating motion of a body that swings back and forth
Period and frequency of a body oscillating about an axis
Deferred to the oscillations section. The rod released from horizontal on this page is followed once, from one stated position to another; how long a full swing takes and whether it repeats belongs there.
deferred
Why a rolling ball eventually stops
from the flattening of the contact
Named in three sentences and no more. A student who is told that static friction does no work is entitled to ask why a marble on a level floor comes to rest, and leaving that unanswered makes the whole block look like a trick. No calculation uses it.
off_syllabus
Recall first
Newton's second law for a body that moves as a whole
$\sum \vec F = m\vec a$, applied to one named body after a free body diagram has been drawn for it. Each body in a connected system gets its own equation.
The pulley problems on this page use it unchanged for the hanging block; only the pulley itself needs the new equation.
Constant acceleration in a straight line
$v = v_0 + at$, $x = v_0 t + \tfrac12 at^{2}$ and $v^{2} = v_0^{2} + 2ax$, valid whenever $a$ is constant.
The rotational equations of this section are these three with the letters changed, so the algebra is already familiar and only the meaning of the symbols is new.
Centripetal and acceleration
A body on a circle of radius $r$ at speed $v$ has an acceleration component $v^{2}/r$ pointing at the centre, and, if the speed is changing, a second component along the motion equal to the rate at which the speed changes.
Both come back rewritten in angular language, $v^{2}/R = \omega^{2}R$ and $a_{\rm tan} = R\alpha$, and the first one explains why a wheel spinning at a steady rate is still accelerating.
Static and kinetic friction
Kinetic friction is $f_k = \mu_k N$ and acts while surfaces slide; static friction takes whatever value equilibrium requires up to a ceiling $f_{s,\max} = \mu_s N$, and acts while they do not slide.
The friction at the contact of a rolling wheel is static, which is the whole reason no energy is lost there, and its ceiling decides whether a body rolls or skids.
Kinetic energy and conservation of mechanical energy
$K = \tfrac12 mv^{2}$, and when only gravity and ideal springs do work, $K_1 + U_1 = K_2 + U_2$ with $U = mgy$ measured from a stated level.
The rolling results come from this line with one extra term added for the spin, and the ramp problems are unreadable without it.
Work done by a constant force and power
$W = Fd\cos\theta$ for a constant force, and average power is the work divided by the time taken, in watts.
The rotational versions $W = \tau\Delta\theta$ and $P = \tau\omega$ are derived from these two in the energy block rather than quoted.
Try it yourself first (3 questions)
1§12.0 — a steady speed is not a zero acceleration●●○○○
Before any of this section's ideas are needed, one thing from earlier has to be firmly in place. A small stone on the end of a string 0.300 m long is being whirled in a horizontal circle at a steady 1.20 m/s.
Given
radius of the circle 0.300 m
speed 1.20 m/s and not changing
the circle is horizontal, so gravity is not the question here
Find
(a) Choose the acceleration of the stone.
Hint 1/4
Ask what acceleration means before reaching for a number: it is the rate of change of the velocity, and velocity has a direction as well as a size.
Hint 2/4
A body on a circle of radius $r$ at speed $v$ has an acceleration of size $v^{2}/r$ directed at the centre, even when the speed never changes.
Hint 3/4
The data again: $v = 1.20$ m/s and $r = 0.300$ m, so $v^{2}/r = 1.44/0.300$.
Hint 4/4
The acceleration is 4.80 m/s$^{2}$, pointing at the centre of the circle at every instant.
Show solution
The starting word has to be velocity, not speed: $\Delta v/\Delta t$ built from the two speeds gives zero, so only the vector definition can answer this at all.
the standard result for circular motion, with the direction towards the centre
Answer $$\boxed{\;a = 4.80\ \mathrm{m/s^{2}}\ \text{towards the centre}\;}$$
Check
Size check: 4.80 m/s$^{2}$ is about half of $g$, so the string tension needed is about half the stone's weight, which is what whirling a light stone slowly feels like.
2§12.0 — the straight line equations you will be copying●○○○○
A car starts from rest and reaches 25.0 m/s in 8.00 s, speeding up at a constant rate. The point of this question is the shape of the working rather than the answer, because in a few pages the same three equations return with different letters.
Given
starts from rest, $v_0 = 0$
final speed 25.0 m/s
time taken 8.00 s
the acceleration is constant
Find
(a) Find the acceleration.
(b) Find the distance covered.
Hint 1/4
Both parts come straight from the constant acceleration equations; the only decision is which equation contains the symbols you already have.
Hint 2/4
$v = v_0+at$ for the first part, and either $x = v_0t+\tfrac12at^{2}$ or $x = \bar v t$ for the second.
Hint 3/4
The data again: $v_0 = 0$, $v = 25.0$ m/s, $t = 8.00$ s.
Hint 4/4
The acceleration is 3.13 m/s$^{2}$ and the distance is 100 m.
Show solution
The distance is taken by the average speed rather than by $x=v_0t+\tfrac12at^{2}$, because both end speeds are already given and that route cannot inherit an error from part (a).
with a constant acceleration the average speed is the average of the two ends, which avoids squaring anything
Answer $$\boxed{\;a = 3.13\ \mathrm{m/s^{2}},\qquad x = 100\ \mathrm{m}\;}$$
Check
Independent check on the distance by the other equation: $x = \tfrac12(3.125)(64.0) = 100$ m.
3§12.0 — a block on a smooth ramp, for comparison later●●○○○
A 2.00 kg block is released from rest at the top of a smooth ramp whose top is 1.50 m above the bottom, and slides down. Keep this answer: a body that rolls down the same ramp instead of sliding will not match it, and the gap is what this section explains.
Given
$m = 2.00\ \mathrm{kg}$
released from rest
vertical drop 1.50 m
the ramp is smooth, so no energy is lost
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the speed of the block at the bottom.
(b) Say whether the answer would change if the mass were doubled.
Hint 1/4
The question mentions no time and no forces, only a height and a speed, which points at one particular tool.
Hint 2/4
With only gravity doing work, $\tfrac12 mv_1^{2}+mgy_1 = \tfrac12 mv_2^{2}+mgy_2$.
Hint 3/4
The data again: released from rest, a drop of 1.50 m, a smooth surface and $g = 9.80$ m/s$^{2}$.
Hint 4/4
The speed at the bottom is 5.42 m/s, and it does not depend on the mass at all.
Show solution
Energy is the cheap route here: the forces route would need the angle of the ramp, which the question never gives, and the energy line never asks for it.
Write the line and delete the zeros
$$mgh = \tfrac12 mv^{2}$$
starts from rest and finishes at the level chosen as zero, so two of the four terms vanish
the mass cancels because it multiplies both surviving terms
Answer $$\boxed{\;v = 5.42\ \mathrm{m/s},\ \text{independent of the mass}\;}$$
Check
Size check: falling freely through 1.50 m would also give 5.42 m/s, and it must, because a smooth ramp only redirects the motion without taking anything from it.
Notation
symbol
reads as
means
watch out
$\theta$
theta
angular position of the body, measured in radians from a stated reference line
In this section $\theta$ is an angle the body has turned through, not the angle between two vectors; the only place the second meaning survives is inside $\tau = rF\sin\theta$.
$\omega$
omega
angular velocity, the rate at which the angle changes, in rad/s
One number for the whole body. A point on the rim and a point halfway in share $\omega$ and disagree about $v$.
$\alpha$
alpha
angular acceleration, the rate at which $\omega$ changes, in rad/s squared
$\alpha$ zero does not mean the acceleration of a rim point is zero; that point still turns, so it still has $\omega^{2}R$ pointing at the axis.
$\tau$
tau
torque about a stated axis, the turning effect of a force, in newton metres
A newton metre is not a joule here even though the units are the same combination; a torque is not an energy and the two never appear on the same side of an equation.
$I$
eye
moment of inertia about a stated axis, in kilogram metres squared
Belongs to a body and an axis together. Changing the axis changes $I$ without touching the body.
$r_{\perp}$
r perpendicular
lever arm: the perpendicular distance from the axis to the line along which the force acts
Not the distance to the point where the force is applied, unless the force happens to be perpendicular to that line.
$R$
capital R
the outer radius of a wheel, cylinder or sphere
Questions like to quote a diameter. Halve it before it enters a formula.
$a_{\rm cm}$
a c m
the acceleration of the centre of mass of a rolling body, along the surface
Tied to $\alpha$ by $a_{\rm cm} = \alpha R$ only while the body rolls without slipping; the moment it skids, the two become independent.
$K_{\rm rot}$
kay rot
the kinetic energy a body has because it is turning, $\tfrac12 I\omega^{2}$
A rolling body has this and $\tfrac12 Mv^{2}$ as well; writing only one of them is the most common way to lose the marks on a ramp question.
$rpm$
turns per minute, the unit machinery is usually rated in
Never substituted directly. Multiply by $2\pi/60$ to get rad/s; 60 rpm is 6.28 rad/s.
Conventions used here
The sense of turning that counts as positive here
Every problem states which way of turning is positive before the first equation, and unless a question says otherwise it is counterclockwise as the figure is drawn. A positive $\alpha$ then means the counterclockwise motion is speeding up or the clockwise motion is slowing down, and a torque that would turn the body counterclockwise enters the sum with a plus sign. The choice is free; keeping it for the whole problem is not.
In a problem with two torques opposing each other, a sign chosen once and then quietly reversed halfway is invisible in the algebra and turns a subtraction into an addition.
Radians, and where degrees and revolutions are allowed
Radians are the only unit that may enter $s = R\theta$, $v = R\omega$, $a_{\rm tan} = R\alpha$, the constant acceleration equations and $W = \tau\theta$. Degrees and revolutions are converted the moment they arrive: one revolution is $2\pi$ rad, and $n$ rpm is $n\,(2\pi/60)$ rad/s. Answers are converted back into revolutions only at the end, when the question asks for turns.
The radian is what makes the arc equal to the radius times the angle, and every one of these formulas is built on that. A degree entering silently multiplies the answer by about 57.
A moment of inertia is never quoted without its axis
Every $I$ in this section is written with the axis it belongs to said out loud, in words: a solid cylinder about its own axis, a rod about one end, a disc about a point on its rim. The tables of standard shapes are tables of body-and-axis pairs, not of bodies.
The same rod differs by a factor of four between two perfectly ordinary axes, so an $I$ without an axis is not an incomplete answer but a meaningless one.
The constant g and the rounding used in every answer here
$g = 9.80\ \mathrm{m/s^{2}}$ throughout, positive, with the direction carried by the signs in the equations rather than by the symbol. Answers are given to three significant figures, and intermediate values are kept longer than that inside the calculation so the rounding does not accumulate.
Two different values of $g$ in one course make two correct solutions disagree in the third digit and send a student hunting for a mistake that is not there.
What light, rigid and smooth axle are allowed to mean
A light cord or rod has no mass, so it stores no kinetic energy and needs no torque to turn; a rigid body keeps its shape, so every point keeps its distance from the axis and $I$ stays constant; a smooth axle exerts no friction torque, so the only torques are the ones the question names. A cord that does not slip on a pulley shares the pulley's rim speed, which is what turns $v$ into $\omega R$.
Each of these words removes one term from an equation, and knowing which term it removes is the difference between quoting a formula and using it.
What rolling without slipping is a statement about
Rolling without slipping is a condition on the contact point, not on the surface: the point of the body touching the ground is momentarily at rest with respect to the ground. Its consequences are $v_{\rm cm} = \omega R$ and $a_{\rm cm} = \alpha R$, and the friction acting there is static, so it does no work and no energy is lost to it.
The word friction makes students subtract an energy loss out of habit; here that subtraction is wrong, and knowing why keeps the energy line short.
12.1Angular position, angular velocity and angular acceleration: one number for the whole body
Three quantities for the whole body at once, and the formulas that turn them into what one point does.
A wheel on an axle has no single position, velocity or acceleration: its parts disagree about all three.
Solvable with what we have
Find the speed of a stone after it falls 1.20 m.
Find the acceleration of a block being pulled across a table.
Find the kinetic energy of a 2.0 kg trolley moving at 3.0 m/s.
Find the acceleration of a car rounding a bend at steady speed.
Not solvable yet
State the speed of a wheel spinning on a fixed axle.
Find the kinetic energy of a that is going nowhere.
Say how long a wheel takes to stop under a steady braking effect.
Say why pushing a door near its hinge is harder than pushing it at the handle.
Take the fastest part and call it the speed of the wheel: a 2.0 kg solid disc of radius 0.30 m with its rim at 6.0 m/s would have $\tfrac12(2.0)(6.0)^{2} = 36$ J.
Why it fails
The rim is the only part moving at 6.0 m/s. Halfway out a point does 3.0 m/s, at the axis nothing moves, so the inner mass was credited with a speed it never had. The honest total is 18 J, exactly half. What every part does share is not a speed.
DefinitionDefinition 12.1: the three angular quantities
Conditions
The body is rigid, so the distance from any of its points to the axis never changes
The axis is fixed, so one angle is enough to say where the whole body is
Every angle is in radians; degrees and revolutions are converted before they are used
$r$ in the linking formulas is the distance of that particular point from the axis, not the outer radius of the body
$$\boxed{\;\omega = \frac{d\theta}{dt},\qquad \alpha = \frac{d\omega}{dt},\qquad s = r\theta,\qquad v = r\omega,\qquad a_{\rm tan} = r\alpha,\qquad a_{R} = \omega^{2} r\;}$$
The angular velocity says how fast the body is turning, the angular acceleration how fast that is changing. The next three lines are an exchange rate: multiply an angular quantity by a point's distance from the axis to get the matching linear one. The last line is the old centripetal acceleration rewritten: a point on a steadily turning wheel is still accelerating.
Where v equals r omega comes from
A point at distance $r$ from the axis travels along a circle, and the definition of the radian says that when the body turns through $\theta$ that point covers an arc $s = r\theta$. Divide by the time taken: the left side is the speed of the point, the right is $r$ times the rate the angle changes, so $v = r\omega$. Divide once more for $a_{\rm tan} = r\alpha$. One definition and two divisions: only radians may be substituted.
One wheel, one angular velocity, three different point speeds. The $\textcolor{#1f6feb}{\text{distance from the axis}}$ is the only thing that differs between the three, and the $\textcolor{#d1690a}{\text{speed of a point}}$ is proportional to it.
Looks like this, but is not
Two wheels turning at the same 33 rpm must have their edges moving at the same speed.
They share $\omega$ and nothing else: $\omega = 3.46$ rad/s for both. The record's edge is 0.15 m out and moves at 0.52 m/s; the bicycle rim is 0.335 m out and moves at 1.16 m/s. Reversed, that is the rule: two points of one rigid body always agree about $\omega$ and almost never about $v$.
A grinding wheel rated at 3450 rpm
A bench grinder carries a wheel of radius 9.00 cm and its plate gives the running speed as 3450 rpm. Find the angular velocity in rad/s, the speed of a point on the rim, and the acceleration of that point.
the angular velocity, the rim speed and the rim acceleration
Solution
The rating is converted into rad/s once, on the first line, instead of inside each of the three formulas that follow; converting three times is three chances to drop a factor of $2\pi$.
Independent route to the acceleration: $v^{2}/R = (32.5)^{2}/0.0900 = 1.17\times10^{4}$ m/s$^{2}$, which never uses $\omega$ at all and lands on the same number. Size check: that acceleration is about 1200 times $g$, which is why grinding wheels carry a maximum rated speed and why they are never run above it.
Two unit conversions and two substitutions; the only place a mistake can hide is the first line.
Any machine rating in rpm gets multiplied by $2\pi/60$ on the first line of the solution, before anything else happens. Doing it later means doing it twice.
How far a bicycle goes while the wheel turns 2.50 times a second
A bicycle wheel of radius 0.335 m turns at a steady 2.50 revolutions per second while the bicycle rolls along a road. Find how fast the bicycle is going and how far it travels in one minute.
Given
wheel radius $R = 0.335\ \mathrm{m}$
2.50 rev/s
time interval 60.0 s
the wheel rolls without skidding
Find
the speed of the bicycle and the distance covered in one minute
Solution
The conversion comes first again, and the distance is then found twice, once as an arc and once as speed times time, because a second route costs one line here and catches a dropped $2\pi$.
Independent check by the other route: $s = vt = (5.26)(60.0) = 316$ m, computed without ever mentioning an angle. Size check: 19 km/h on a bicycle is an ordinary cruising speed, and 316 m in a minute is about a lap and a half of a running track.
This is the doing its work quietly. The reason the road distance equals the arc is that the wheel does not skid, and that single sentence is the whole of the last block of this section.
Checkpoint
§12.1 — arc length with the angle in the wrong unit●●○○○
A wheel turns through 30.0 degrees. A ladybird is sitting on it, 0.200 m from the axis, and you want the length of the path it has travelled.
Given
angle turned through: 30.0 degrees
distance of the insect from the axis: 0.200 m
the wheel is rigid and the axis is fixed
Find
(a) Choose the length of the path travelled by the insect.
Hint 1/4
Do not reach for a formula yet. Ask which unit the arc formula was built for, and check whether the number in front of you is in that unit.
Hint 2/4
$s = r\theta$ holds only when $\theta$ is in radians, and 180 degrees is $\pi$ radians.
Hint 3/4
The data again: the angle is 30.0 degrees and the insect sits 0.200 m from the axis. In radians the angle is $30.0\times\pi/180 = 0.524$ rad.
Hint 4/4
The path length is $(0.200)(0.524) = 0.105$ m, about ten centimetres.
Show solution
There is no choice of route in this one, only a choice of units: the degrees are converted before substituting, because $s = r\theta$ is simply false for anything but radians.
$r$ is the distance of that point from the axis, which is what the question gave
Answer $$\boxed{\;s = 0.105\ \mathrm{m}\;}$$
Check
Sanity check without any formula: a full turn would carry the insect around a circle of circumference $2\pi(0.200) = 1.26$ m, and 30 degrees is one twelfth of a turn, so the answer must be $1.26/12 = 0.105$ m.
⚠ Putting degrees into a formula that only accepts radians
The number arrives in degrees, the formula has an angle in it, and nothing on the page objects; degrees are the unit the eye is used to.
12.2Constant angular acceleration: the equations you already have, wearing new letters
When the angular acceleration is constant, the three straight line equations come back with theta, omega and alpha in place of x, v and a.
We have $\omega$ and $\alpha$ defined as a rate and a rate of a rate, which is exactly the relationship $v$ and $a$ have to $x$, so anything proved once for one pair is already proved for the other.
TheoremTheorem 12.2: the constant angular acceleration equations
Conditions
$\alpha$ is constant over the whole interval considered
$\theta$, $\omega$ and $\alpha$ are all measured about the same fixed axis and in the same positive sense
$\theta$ is the angle turned through from the starting position, in radians
The body may pass the same physical orientation many times; $\theta$ keeps counting, it does not reset after each turn
The first says the turning rate climbs at a steady pace, the second says the angle piles up from a starting rate plus a growing correction, and the third is the one to reach for when the question never mentions time. The fourth says that with a constant $\alpha$ the average turning rate is just the average of the first and last values, which is often the fastest line in the whole solution. Nothing here is new physics: the same three equations with the letters changed and radians made compulsory.
Why the same equations come back
The straight line equations were derived from two facts only: that $v$ is the rate of change of $x$ and that $a$, the rate of change of $v$, is constant. Nothing in that derivation used the fact that $x$ was a distance. Since $\omega$ is the rate of change of $\theta$ and $\alpha$ is constant by assumption, the identical derivation runs with the identical result. Replace $x$ by $\theta$, $v$ by $\omega$, $a$ by $\alpha$ and you are finished. This is why the block is short: there is nothing to learn here except a translation table and the discipline of working in radians.
A body whose $\textcolor{#1f6feb}{\text{angular velocity}}$ climbs at a steady rate. The $\textcolor{#d1690a}{\text{slope}}$ of the line is $\alpha$ and the $\textcolor{#8250df}{\text{shaded area}}$ under it is the angle turned through, $96.0$ rad, which is 15.3 revolutions.
Looks like this, but is not
The wheel of a braking car slows from 40 rad/s to rest in 5.0 s, so its angular acceleration is $-8.0$ rad/s$^{2}$ and it turns through $\bar\omega t = 100$ rad.
The arithmetic is right and it rests on one hidden assumption: that the braking is even. Real brakes are not. If $\alpha$ varies, $-8.0$ rad/s$^{2}$ is still the correct average, but $\bar\omega = \tfrac12(\omega_0+\omega)$ is no longer the average turning rate and the 100 rad is wrong. The test is not whether the numbers are clean but whether the question said a constant rate or a steady torque; when it does not, only the definitions and a graph are safe.
straight line
fixed axis rotation
what it says
$x$ (m)
$\theta$ (rad)
where it is
$v$ (m/s)
$\omega$ (rad/s)
how fast that changes
$a$ (m/s$^{2}$)
$\alpha$ (rad/s$^{2}$)
how fast that changes
$v = v_0+at$
$\omega = \omega_0+\alpha t$
rate after a time
$x = v_0t+\tfrac12 at^{2}$
$\theta = \omega_0t+\tfrac12\alpha t^{2}$
how far, given the time
$v^{2}=v_0^{2}+2ax$
$\omega^{2}=\omega_0^{2}+2\alpha\theta$
how far, with no time in the question
Read a row across, not down. The only thing that changes between the two columns is what the letters refer to; the shape of every equation, and the algebra you use to rearrange it, is untouched.
A centrifuge reaching 20000 rpm in half a minute
A centrifuge is switched on and its rotor climbs steadily from rest to 20000 rpm in 30.0 s. Find its angular acceleration and the number of revolutions it makes while getting there.
Given
starts from rest, $\omega_0 = 0$
final speed 20000 rpm
time taken 30.0 s
the climb is steady, so $\alpha$ is constant
Find
the angular acceleration and the number of turns during the spin up
Solution
The average value route is chosen over $\theta = \tfrac12\alpha t^{2}$ because it does not need $\alpha$ at all, so an error in the previous part cannot propagate into this one.
converted back into turns only now, because the question asked for turns
Answer $$\boxed{\;\alpha = 69.8\ \mathrm{rad/s^{2}},\qquad N = 5.00\times10^{3}\ \mathrm{revolutions}\;}$$
Check
Independent check on the revolutions with no radians anywhere: the average rate during the climb is half of 20000 rpm, that is 10000 rev per minute, and the climb lasts half a minute, so 5000 revolutions. Two routes, one number.
When $\alpha$ is constant and you know both end rates and the time, $\bar\omega t$ is almost always the shortest line to the angle.
A wheel that stops in twelve turns
A wheel is turning at 8.00 rad/s when a brake is applied that slows it evenly. It stops after making 12.0 more revolutions. Find the angular acceleration and how long the braking takes.
Given
$\omega_0 = 8.00\ \mathrm{rad/s}$
$\omega = 0$ at the end
12.0 revolutions during the braking
the braking is even, so $\alpha$ is constant
Find
the angular acceleration and the braking time
Solution
Of the four constant acceleration equations, the time free one is chosen because the time is neither given nor wanted in the first part; any other choice carries a second unknown along.
Get the angle into radians
$$\theta = 12.0\times 2\pi = 75.4\ \mathrm{rad}$$
revolutions are data, radians are what the equations accept
two negatives make the time positive, which is a check in itself
Answer $$\boxed{\;\alpha = -0.424\ \mathrm{rad/s^{2}},\qquad t = 18.8\ \mathrm{s}\;}$$
Check
Independent check that uses neither result directly: with a constant $\alpha$ the average rate is 4.00 rad/s, so in 18.85 s the wheel should turn through $(4.00)(18.85) = 75.4$ rad, and that is the 12.0 revolutions we were given. Size check: a wheel losing about half a radian per second each second takes tens of seconds to stop, which is what a freewheeling bicycle wheel does.
Three lines, and the only judgement call was picking the time free equation first.
Checkpoint
§12.2 — a wheel speeding up steadily●●○○○
A wheel is already turning at 3.00 rad/s when a motor starts to speed it up at a constant 2.00 rad/s$^{2}$. You want to know where it is and how fast it is turning 4.00 s later.
Given
$\omega_0 = 3.00\ \mathrm{rad/s}$
$\alpha = 2.00\ \mathrm{rad/s^{2}}$, constant
$t = 4.00\ \mathrm{s}$
Find
(a) Find the angular velocity at the end of the four seconds.
(b) Find the angle turned through, in radians and in revolutions.
Hint 1/4
Nothing new is needed. Ask which of the four equations has exactly the symbols you were given and the symbol you want, and use that one.
Hint 2/4
$\omega = \omega_0+\alpha t$ for the rate, and $\theta = \omega_0 t+\tfrac12\alpha t^{2}$ for the angle.
Hint 3/4
The data again: $\omega_0 = 3.00$ rad/s, $\alpha = 2.00$ rad/s$^{2}$, $t = 4.00$ s.
Hint 4/4
The rate reaches 11.0 rad/s and the wheel has turned through 28.0 rad, which is 4.46 revolutions.
Show solution
Both parts are read straight off the two equations built from $\omega_0$, $\alpha$ and $t$, which are exactly the three quantities given, so nothing has to be rearranged.
The rate
$$\omega = \omega_0+\alpha t = 3.00+(2.00)(4.00) = 11.0\ \mathrm{rad/s}$$
the definition of a constant $\alpha$, used forwards
Independent check on the angle: the average rate over the interval is $\tfrac12(3.00+11.0) = 7.00$ rad/s, and $(7.00)(4.00) = 28.0$ rad. The two routes use different equations and agree.
⚠ Dropping the first term because the motion started earlier
Most textbook problems start from rest, so $\omega_0 t$ has been zero so often that it stops being written.
12.3Torque: the same force does different work depending on where it acts
The turning effect of a force about an axis, equal to the force times the perpendicular distance from the axis to the line the force acts along.
The angular quantities describe how a body turns; nothing so far says what makes it start, and the honest answer is not the force but where the force is applied.
DefinitionDefinition 12.3: torque about a fixed axis
Conditions
The axis has to be named before a torque means anything; the same force has different torques about different axes
$r$ is the distance from the axis to the point where the force is applied, and $\theta$ the angle between that line and the force
A force whose line passes through the axis has zero torque, whatever its size
The sign is fixed by the sense chosen as positive, and it must be kept for the whole problem
Only the part of the force that acts across the line to the axis turns the body, and it turns it in proportion to how far out it is applied. The two middle forms are the same statement read in the two directions that are useful in practice: either slide the force along its own line until the distance to the axis is perpendicular and multiply by the whole force, or keep the point of application and multiply by the component of the force across the line. Both give the same number; which is less work depends on what the picture gives you.
Why the lever arm and the force component give the same answer
Write $\tau = rF\sin\theta$. The two useful readings come from choosing which pair of factors to group. Grouping as $(r\sin\theta)F$ gives a distance, $r_{\perp} = r\sin\theta$, times the whole force, and that distance is exactly the perpendicular from the axis to the line the force lies along. Grouping as $r(F\sin\theta)$ gives the full distance times $F_{\perp}$, the component of the force perpendicular to $r$. Multiplication does not care how it is bracketed, so a problem can be attacked from whichever end is drawn more clearly.
The same 50 N $\textcolor{#d1690a}{\text{force}}$ on the same spanner, three ways. Only the part of it that acts across the handle turns the bolt, and the $\textcolor{#8250df}{\text{lever arm}}$ is what multiplies it. Pull along the handle and the bolt does not care how hard you pull.
Looks like this, but is not
A newton metre and a joule are the same combination of units, so a torque of 15 N·m is an energy of 15 J.
Same two units, different quantities, because of what the metre is doing in each. In a joule it is measured along the force; in a torque it is measured across it, which is exactly the direction in which a force does no work. You can hold 15 N·m on a bolt all afternoon and transfer nothing, since nothing moves. The two meet only once the body turns, and then the energy is the torque times the angle in radians. Never add a torque to a joule.
A spanner pulled at 60 degrees to its handle
A mechanic pulls with 55.0 N on the end of a spanner 0.280 m long, at an angle of 60.0 degrees to the handle. Find the torque on the bolt, and find how much torque is lost compared with pulling square to the handle.
Given
$F = 55.0\ \mathrm{N}$
$r = 0.280\ \mathrm{m}$ from the bolt to the hand
angle between the handle and the pull: 60.0 degrees
positive sense: the way the pull turns the bolt
Find
the torque, and the comparison with a square pull
Solution
The definition is used with the angle exactly as the question states it, rather than resolving the force into components first; both routes give 13.3 N·m, but the component route adds a step in which the sine can be lost.
a square pull is $\sin 90^{\circ}=1$, the largest the sine can be
$$\frac{13.3}{15.4} = 0.866$$
about 13% of the effort is being spent stretching the spanner rather than turning the bolt
Answer $$\boxed{\;\tau = 13.3\ \mathrm{N\cdot m},\ \text{which is }86.6\%\text{ of the }15.4\ \mathrm{N\cdot m}\text{ available}\;}$$
Check
Independent route, by the lever arm instead of the component: the perpendicular distance from the bolt to the line of the pull is $r_{\perp} = (0.280)\sin 60.0^{\circ} = 0.2425$ m, and $(0.2425)(55.0) = 13.3$ N·m. Different construction, same number.
The sine costs you nothing when it is 1 and everything when it is 0, which is why a mechanic instinctively pulls square and why a long spanner beats a strong arm.
Two cords on a stepped pulley, pulling opposite ways
A stepped pulley turns on a fixed axle. A cord wound on the inner step, of radius 0.300 m, is pulled with 50.0 N so as to turn the pulley clockwise. A second cord on the outer step, of radius 0.500 m, is pulled with 35.0 N the other way. Both cords leave their steps tangentially. Find the net torque and say which way the pulley starts to turn.
both cords tangential, so each is perpendicular to its own radius
positive sense: counterclockwise
Find
the net torque about the axle and the sense of the resulting turn
Solution
A positive sense is declared before any number is written, because the whole question is a subtraction and a subtraction is meaningless until something has been called negative.
Check by asking what would make the pulley balance: the 35.0 N would have to be matched by $ (0.500)(35.0)/0.300 = 58.3$ N on the inner step. The actual 50.0 N is less than that, so the outer cord must win, which is the sign we got.
Two torques, two signs, one addition. The signs did all of the physics.
This is how gears and stepped pulleys buy force: a small pull far out beats a big pull close in, and the exchange rate is exactly the ratio of the two radii.
Checkpoint
§12.3 — which push turns the door hardest●●○○○
A door 0.900 m wide is hinged along one edge, and four people each push on it with exactly 30.0 N. They differ only in where and in which direction they push.
Given
door width 0.900 m, hinge along one edge
every push is 30.0 N
push A: at the outer edge, square to the door
push B: at the outer edge, at 30.0 degrees to the door surface
push C: halfway across, square to the door
push D: at the outer edge, aimed straight at the hinge
Find
(a) Choose the push that produces the largest torque about the hinge.
Hint 1/4
Do not compute four numbers yet. Each push has the same size, so the ranking is decided by the lever arm alone.
Hint 2/4
$\tau = r_{\perp}F$, where $r_{\perp}$ is the perpendicular distance from the hinge to the line the push acts along.
Hint 3/4
Every push is 30.0 N and the door is 0.900 m wide, so the lever arms are 0.900 m, $0.900\sin 30.0^{\circ}$, 0.450 m and zero.
Hint 4/4
Push A wins with 27.0 N·m, because it has the full width as its lever arm and loses nothing to an angle.
Show solution
The force is the same in all four pushes, so it is carried along as a common factor and only the lever arms are compared; four products of one force are easier to rank than four descriptions in words.
halving the distance halves the torque, exactly as the angle did
$$\tau_D = 0$$
its line passes through the axis, so the perpendicular distance is zero
Answer $$\boxed{\;\tau_A = 27.0\ \mathrm{N\cdot m}\ \text{is the largest}\;}$$
Check
Check on the two middle ones without arithmetic: $\sin 30.0^{\circ} = 0.5$ and halving the width is also a factor of one half, so B and C are bound to tie, and they do.
⚠ Using the distance to the hand instead of the perpendicular distance to the line of the force
The picture gives you $r$ for free and the sine has to be remembered, so under time pressure the visible number wins.
12.4Rotational inertia: not how much mass, but how far out it sits
The rotational stand-in for mass, obtained by weighting every scrap of the body by the square of its distance from the axis.
Something has to play the part mass plays, and it cannot be the mass alone: a force far out already turns a body more easily than the same force close in.
DefinitionDefinition 12.4: moment of inertia, and the parallel axis theorem
Conditions
$r_i$ is the perpendicular distance from the chosen axis to the mass $m_i$, not the distance to the centre of the body
The value belongs to a body and an axis together; changing the axis changes the number
For the standard shapes the axis is the one named in the table, usually through the centre of mass
The parallel axis theorem needs the starting axis to pass through the centre of mass; from any other axis it gives the wrong answer
Add up, for every piece of the body, its mass times the square of how far it sits from the axis. Because the distance is squared, mass near the rim counts for far more than mass near the middle: move a lump twice as far out and it becomes four times as hard to spin up. The second formula says that of all parallel axes, the one through the centre of mass is the easiest to turn the body about, and every other one costs you an extra $Md^{2}$, as though the whole mass were sitting at the distance the axis was shifted.
Where the sum of m r squared comes from
Treat the body as a crowd of small masses $m_i$ that all share $\omega$. The piece at distance $r_i$ from the axis moves at $v_i = r_i\omega$, so the kinetic energy of the whole body is the sum of $\tfrac12 m_i r_i^{2}\omega^{2}$. Every term carries the same $\omega^{2}$, so it comes out and what is left is $\sum m_i r_i^{2}$, a number depending on the body and the axis alone. That is the quantity worth naming, and the same combination returns in the next block.
The $\textcolor{#1f6feb}{\text{same two masses}}$ on the same light rod, moved from 0.20 m out to 0.40 m out. Nothing has been added and the moment of inertia is four times larger, because the distance enters squared.
Looks like this, but is not
A 5.0 kg wheel is harder to spin than a 2.0 kg wheel, so the moment of inertia is just another name for the mass.
A 5.0 kg solid disc of radius 0.10 m has $I = 0.025$ kg·m$^{2}$. A 2.0 kg bicycle rim of radius 0.33 m, with almost all of its mass at the rim, has $I = 0.218$ kg·m$^{2}$: nearly nine times as much from a body less than half as heavy. Mass is one factor and geometry is the other, and because geometry is squared it usually wins. The same body also has a different $I$ about every axis, which no property of the mass alone could do.
body and axis
$I$
coefficient $c$
thin hoop or thin walled pipe, about its own axis
$MR^{2}$
1
solid cylinder or disc, about its own axis
$\tfrac12 MR^{2}$
0.5
thin spherical shell, about a diameter
$\tfrac23 MR^{2}$
0.667
solid sphere, about a diameter
$\tfrac25 MR^{2}$
0.4
thin rod of length $L$, about its centre
$\tfrac{1}{12}ML^{2}$
0.0833
thin rod of length $L$, about one end
$\tfrac13 ML^{2}$
0.333
Read the coefficient as an answer to one question: on average, how far out is the mass? The hoop puts every gram at the full radius and scores 1; the solid sphere spreads its mass towards the middle and scores 0.4. The two rod entries are the same rod, and the factor of four between them is the parallel axis theorem in action.
Four bolts on a light square frame, about three axes
Four bolts, each of mass 0.500 kg, are fixed at the corners of a square of side 0.800 m on a frame so light that its own mass can be ignored. Find the moment of inertia about an axis through the centre perpendicular to the square, about an axis through the centre lying in the square and parallel to one side, and about an axis along one side.
Given
four masses, $m = 0.500\ \mathrm{kg}$ each
square of side $0.800\ \mathrm{m}$
the frame itself is light and contributes nothing
treat each bolt as a point mass
Find
the three moments of inertia
Solution
The definition $I = \sum m_i r_i^{2}$ is used directly instead of a table entry, because a square of four bolts is not one of the tabulated shapes; the only real work is getting each perpendicular distance right.
Independent check on the first two by ratio, with no arithmetic: every bolt sits 0.5657 m from the first axis and 0.400 m from the second, so the two values must be in the ratio $(0.5657/0.400)^{2} = 2.00$, and $0.640/0.320$ is exactly 2.00.
Three axes, one body, three different answers, and not one gram of mass moved.
Quoting a moment of inertia without naming the axis is like quoting a speed without saying relative to what. The number is not wrong, it is incomplete.
A rod about its end, from the value about its centre
A uniform rod has mass 1.20 kg and length 0.900 m. Its moment of inertia about a perpendicular axis through its centre is $ML^{2}/12$. Find the moment of inertia about a parallel axis through one end.
Given
$M = 1.20\ \mathrm{kg}$
$L = 0.900\ \mathrm{m}$
$I_{\rm cm} = ML^{2}/12$ about the centre
the new axis is parallel to the old one and passes through an end
Find
the moment of inertia about the end
Solution
The parallel axis theorem is used instead of summing the rod about its end from scratch, which would mean doing the integral again; the theorem turns that integral into one multiplication.
Independent check against the table: the standard entry for a rod about its end is $ML^{2}/3 = (1.20)(0.810)/3 = 0.324$ kg·m$^{2}$, obtained by an integration that never mentions the parallel axis theorem. The two agree exactly, which is the usual way this theorem is checked.
The theorem earns its keep when the table has only the central value and the question asks about a pivot somewhere else, which is most pivoted rod and pivoted disc questions.
Checkpoint
§12.4 — which wheel is harder to spin up●●○○○
Two hoops are lying on a bench. One has mass 2.00 kg and radius 0.100 m; the other has mass 1.00 kg and radius 0.200 m. Both are to be spun about their own axes.
(a) Choose the hoop with the larger moment of inertia.
Hint 1/4
Do not guess from the masses. The two factors in $I$ pull in opposite directions here, so the question is which of them is stronger.
Hint 2/4
For a hoop about its own axis $I = MR^{2}$: the mass enters once, the radius twice.
Hint 3/4
The numbers again: 2.00 kg with 0.100 m, and 1.00 kg with 0.200 m. Halving the mass while doubling the radius is a factor of $\tfrac12$ against a factor of 4.
Hint 4/4
The lighter, wider hoop wins: 0.0400 kg·m$^{2}$ against 0.0200 kg·m$^{2}$, twice as much.
Show solution
Both values are computed rather than argued in words, because the two differences pull opposite ways: one body is heavier, the other is wider, and only the squaring settles which wins.
12.5Newton's second law for a body on a fixed axle
Net torque equals moment of inertia times angular acceleration, used side by side with the ordinary second law whenever a cord links a turning body to a moving one.
We now have the rotational version of force and the rotational version of mass, so the rotational version of $\sum F = ma$ writes itself, and the only new work is proving that it is true.
TheoremTheorem 12.5: rotational form of the second law
Conditions
The axis is fixed and every torque in the sum is taken about that same axis
The body is rigid, so $I$ does not change while it turns
$\alpha$ is in rad/s$^{2}$
Forces acting at the axis contribute nothing, whatever their size, because their lever arm is zero
$$\boxed{\;\sum\tau = I\alpha\;}$$
The net turning effect on the body equals how hard it is to spin, multiplied by how quickly its turning rate is changing. Read it the way you read $\sum F = ma$: the left side is what the outside world does to the body, the right side is the body's response and its reluctance. Everything you learned about the straight line version carries over, including the habit of drawing the diagram before writing the equation and the rule that each body in a linked system gets its own equation.
Adding up the second law over the pieces of the body
Take one piece of mass $m$ at distance $r$ from the axis. It moves on a circle, so $F_{\rm tan} = ma_{\rm tan} = mr\alpha$. Multiply by $r$: the left is the torque on that piece, the right is $mr^{2}\alpha$. Add over every piece. On the right $\alpha$ is shared and comes out, leaving $I\alpha$; on the left the internal forces cancel in pairs, since each pair acts along the line joining them, so only external torques survive.
The standard set-up, split into the two diagrams the solution needs. The $\textcolor{#d1690a}{\text{cord tension}}$ appears in both, once as a force on the block and once as a torque on the pulley, while the $\textcolor{#8250df}{\text{axle force}}$ appears in neither equation because its lever arm is zero.
Looks like this, but is not
The block hangs from the cord, so the tension in the cord is its weight, 19.6 N, and the pulley is turned by a torque of $19.6R$.
True only if nothing accelerates. The block is falling, and its own equation says $mg - T = ma$, so $T = m(g-a)$, less than $mg$ by exactly $ma$. If the tension were the full weight the block would be in equilibrium and could not speed up. The habit comes from the massless pulleys of earlier sections, where the tension really is the same on both sides. Give the pulley mass and that dies: it is a body that must be accelerated, and only an unbalanced pull can do it.
A falling block unwinding a cord from a solid pulley
A cord is wrapped around a solid disc pulley of mass 4.00 kg and radius 0.250 m, free to turn on a smooth fixed axle. A 2.00 kg block hangs from the free end and is released from rest. Find the acceleration of the block, the tension in the cord and the angular acceleration of the pulley.
block: $m = 2.00\ \mathrm{kg}$, released from rest
smooth axle, light cord, no slipping between cord and rim
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the linear acceleration, the tension and the angular acceleration
Solution
The pulley has to get its own torque equation, since a massive pulley is not a point; the two equations are then joined by the rolling of the cord rather than solved apart, because the tension is common to both.
One equation for each body, and never the same equation twice
Independent check by energy, which uses none of the three answers directly. After the block has fallen 1.00 m it should be doing $v = \sqrt{2(4.90)(1.00)} = 3.13$ m/s. The energy released is $mgh = 19.6$ J, and the energy stored is $\tfrac12(2.00)(9.80)+\tfrac12(0.125)(12.52)^{2} = 9.80+9.80 = 19.6$ J. The two agree exactly.
Two second law equations, one constraint, one substitution. The constraint is the only line that is new.
Notice what $I/R^{2}$ did: it entered the final equation with units of kilograms and behaved like extra mass. That is the general pattern for a pulley with mass, and it is worth recognising rather than rederiving.
Spinning up a playground roundabout against friction
A roundabout is a uniform disc of mass 250 kg and radius 2.00 m turning on a vertical axle. A child pushes tangentially at the rim with a steady 120 N. Friction in the axle opposes the motion with a constant torque of 60.0 N·m. Find the angular acceleration and the time to reach 0.600 revolutions per second from rest.
push $F = 120\ \mathrm{N}$, applied tangentially at the rim
friction torque 60.0 N·m, opposing the turning
starts from rest, target 0.600 rev/s
Find
the angular acceleration and the time to reach the target rate
Solution
The torque route is compulsory here: the question gives a friction torque and asks for $\alpha$, while energy would need the angle turned, which is not given until after $\alpha$ is known.
$\alpha$ is constant because both torques are constant, so the earlier equations are licensed
Answer $$\boxed{\;\alpha = 0.360\ \mathrm{rad/s^{2}},\qquad t = 10.5\ \mathrm{s}\;}$$
Check
Size check against experience: ten seconds of steady pushing to get a heavy roundabout up to one turn every two seconds is about what it feels like, and if the friction is ignored the answer only shortens to 7.85 s, so the friction is a real but not dominant effect. Sign check: the friction torque is smaller than the push, so $\alpha$ came out positive, as it must while the roundabout is speeding up.
Watch what happens if the child stops pushing: the only torque left is the friction, $\alpha$ becomes $-0.120$ rad/s$^{2}$, and the roundabout takes 31.4 s to stop. Same equation, one term deleted.
Checkpoint
§12.5 — doubling the mass of the pulley●●●○○
Take the falling block and pulley of the worked example: a 2.00 kg block on a cord wrapped around a solid disc pulley of radius 0.250 m, released from rest. Now the pulley is replaced by one of the same size but twice the mass, 8.00 kg.
$I = \tfrac12 MR^{2}$ for a solid disc about its own axis
smooth axle, light cord that does not slip
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the new acceleration of the block.
(b) Say whether the tension goes up or down, and by how much.
Hint 1/4
You do not need to start from scratch. Ask what the pulley contributes to the final equation, and what happens to that contribution when its mass doubles.
Hint 2/4
The pulley enters as an extra inertia $I/R^{2}$ added to the mass of the block: $a = mg/(m+I/R^{2})$.
Hint 3/4
The numbers again: $m = 2.00$ kg, $M = 8.00$ kg, $R = 0.250$ m, so $I = \tfrac12(8.00)(0.0625) = 0.250$ kg·m$^{2}$ and $I/R^{2} = 4.00$ kg.
Hint 4/4
The acceleration falls to 3.27 m/s$^{2}$ and the tension rises to 13.1 N.
Show solution
The combined form from the worked example is reused rather than rebuilt from two equations, because only the pulley's mass changed and $I/R^{2}$ is the single number that changed with it.
What the pulley contributes
$$\frac{I}{R^{2}} = \frac{\tfrac12MR^{2}}{R^{2}} = \tfrac12 M = 4.00\ \mathrm{kg}$$
for a solid disc the radius cancels completely, so only half of its mass ever matters
back into the block equation, which is the cheapest route to the tension
Answer $$\boxed{\;a = 3.27\ \mathrm{m/s^{2}},\qquad T = 13.1\ \mathrm{N}\;}$$
Check
Check the limit rather than the arithmetic: as the pulley gets heavier without limit, $a$ should go to zero and $T$ should climb towards the full weight, 19.6 N. Our numbers moved in both of those directions and neither passed its limit.
⚠ Treating the cord tension as the weight of the hanging block
In every earlier pulley problem the pulley was massless, and there the tension really was the same everywhere and the block really did feel a simple weight.
12.6The energy of a turning body, and the work a torque does
A body going nowhere can still carry a great deal of kinetic energy, and a torque acting through an angle is what puts it there.
The force route answers everything, at the price of two or three simultaneous equations; the energy route answered ramp questions in one line before, and it will do the same here once we know what a spinning body's kinetic energy actually is.
TheoremTheorem 12.6: rotational kinetic energy, work and power
Conditions
$I$ is taken about the axis the body actually turns about
$\Delta\theta$ is in radians, always
$W = \tau\Delta\theta$ assumes the torque stays constant over that angle; otherwise it is the area under a torque against angle graph
For a body that turns without going anywhere, this is the whole of its kinetic energy; for one that does both, the two parts add
$$\boxed{\;K_{\rm rot} = \tfrac12 I\omega^{2},\qquad W = \tau\,\Delta\theta,\qquad P = \tau\omega\;}$$
The kinetic energy of a turning body has the same shape as the familiar one, with the moment of inertia standing in for the mass and the angular velocity for the speed. The work a torque does is that torque multiplied by the angle it acts through, exactly as force times distance, and the power it delivers is torque times angular velocity, exactly as force times speed. Every one of the three is the straight line statement with the rotational pair of symbols substituted, and each can be checked by cancelling: torque times angle is newton metre times a pure number, which is a joule.
Both results from the sum over the pieces
For the energy, add $\tfrac12 m_iv_i^{2}$ over the pieces with $v_i = r_i\omega$; the shared $\omega^{2}$ comes out and leaves $\tfrac12 I\omega^{2}$, which is the calculation already done when $I$ was defined. For the work, take a force $F$ applied tangentially at radius $r$ while the body turns through a small angle $d\theta$. Its point of application moves a distance $r\,d\theta$ along the force, so the work done is $F r\,d\theta = \tau\,d\theta$. Divide by the time and the left side becomes power while $d\theta/dt$ becomes $\omega$, giving $P = \tau\omega$.
Three flywheels with the same mass, the same radius and the same turning rate. The $\textcolor{#1f6feb}{\text{energy stored}}$ differs by a factor of two and a half, and the only thing that differs between them is where the mass sits.
Looks like this, but is not
The flywheel is bolted to the floor and its centre is not going anywhere, so it has no kinetic energy; kinetic energy is $\tfrac12 mv^{2}$ and $v$ is zero.
The centre is not moving and almost every other part is. Kinetic energy is a property of the pieces, and $\tfrac12 I\omega^{2}$ is the sum over them. For the flywheel below that is about 50 kJ, what an 80 kg person moving at 35 m/s carries, which is why a failed flywheel does so much damage. The rule: a body has $\tfrac12 Mv_{\rm cm}^{2}$ from its centre moving plus $\tfrac12 I_{\rm cm}\omega^{2}$ from turning, and either may be zero without the other being zero.
How much energy a flywheel holds at 1200 rpm
A flywheel is a solid disc of mass 80.0 kg and radius 0.400 m spinning at 1200 rpm on a smooth axle. Find the kinetic energy stored in it, and find how long it could deliver 500 W if all of that energy could be recovered.
the stored kinetic energy and the running time at 500 W
Solution
The stored energy is asked for directly, so $\tfrac12 I\omega^{2}$ is the whole of the physics; a torque route would need to know how the wheel was spun up, which the question never says.
power is energy per unit time, so the time is the energy divided by the rate
Answer $$\boxed{\;K = 5.05\times10^{4}\ \mathrm{J},\qquad t = 101\ \mathrm{s}\;}$$
Check
Size check by converting the answer into something you can picture: 50.5 kJ would lift the flywheel itself, all 80 kg of it, to a height of $K/Mg = 64.5$ m, or drive an 80 kg cyclist to 35.5 m/s. Both are large, which is the point of a flywheel and the reason they are kept inside steel housings.
The energy goes as $\omega^{2}$, so doubling the running speed quadruples the store. That, and not the mass, is where flywheel design puts its effort.
A motor bringing a wheel up to speed through fifteen turns
A motor applies a constant torque of 3.50 N·m to a wheel of moment of inertia 0.0450 kg·m², initially at rest, and keeps it up for 15.0 revolutions. Find the work done, the angular velocity reached, and the power the motor is delivering at that moment.
Given
$\tau = 3.50\ \mathrm{N\cdot m}$, constant
$I = 0.0450\ \mathrm{kg\cdot m^{2}}$
starts from rest
turns through 15.0 revolutions
no friction torque
Find
the work done, the final angular velocity and the final power
Solution
The work is computed as $\tau\,\Delta\theta$ and then set equal to the stored energy, rather than finding $\alpha$ and running the kinematics; the kinematics route reaches the same $\omega$ three lines later.
Independent check by the force route, which shares no equation with the energy route: $\alpha = \tau/I = 77.8$ rad/s$^{2}$, and $\omega^{2} = 2\alpha\theta = 2(77.78)(94.248) = 1.466\times10^{4}$, giving $\omega = 121$ rad/s. A second check on the power: the run lasts $t = \omega/\alpha = 1.56$ s, so the average power is $330/1.56 = 212$ W, exactly half the final value, as it must be when the power climbs from zero in proportion to $\omega$.
The energy route needed three lines and no simultaneous equations; the force route used to check it needed the same number, so here the two are evenly matched.
A motor rated in watts is telling you $\tau\omega$, not $\tau$. The same motor delivers its rated power as a big torque at low speed or a small torque at high speed, which is what a gearbox is for.
Checkpoint
§12.6 — two flywheels, same mass and radius●●○○○
A workshop has two flywheels of the same mass and the same outer radius, spun up to the same rate on identical axles. One is a solid disc; the other is a hoop, with its material concentrated at the rim.
(a) Choose the statement that correctly compares the energy stored in the two.
Hint 1/4
Nothing needs to be computed. Both have the same $\omega$, so the comparison is entirely a comparison of two moments of inertia.
Hint 2/4
$K = \tfrac12 I\omega^{2}$, so with $\omega$ shared the stored energies are in the same ratio as the values of $I$.
Hint 3/4
The values of $I$ again: $\tfrac12 MR^{2}$ for the disc and $MR^{2}$ for the hoop, with the same $M$ and the same $R$.
Hint 4/4
The hoop stores twice as much as the disc, because its moment of inertia is twice as large.
Show solution
The ratio is formed before a single number is substituted, because the mass, the radius and the rate are all shared and cancel; substituting first means computing two energies only to throw both away.
Check against the figure in this block, which was worked out with real numbers: 101 kJ for the hoop against 50.5 kJ for the disc, a ratio of exactly two.
⚠ Writing the rotational energy with a speed in it
The formula is remembered as a shape, one half something something squared, and the familiar symbol slides into the empty slot.
wrong$$K_{\rm rot} = \tfrac12 Iv^{2}$$
right$$K_{\rm rot} = \tfrac12 I\omega^{2}$$
⚠ Leaving the angle in revolutions inside the work formula
Turns are what the question counts, and $W = \tau\theta$ does not visibly complain about the unit.
If the wheel is not skidding, then in the time it makes one full turn it must move forward exactly one circumference, and that single sentence is what ties the speed to the turning rate. The energy statement says a rolling body carries two kinds of kinetic energy at the same time: one because its centre is going somewhere, and one because it is turning about that centre. Writing only the first is the most expensive mistake in this section: it silently changes every ramp answer.
Why one turn means one circumference, and what that does to the energy
Watch the track a rolling wheel leaves. If it never skids, the piece of rim that touches the ground is laid down onto the ground without sliding, so after one full turn the wheel has advanced by exactly the length of its rim, $2\pi R$. Divide by the time for one turn and you get $v_{\rm cm} = \omega R$; differentiate once more and you get $a_{\rm cm} = \alpha R$. For the energy, substitute $\omega = v/R$ into $\tfrac12 I_{\rm cm}\omega^{2}$ with $I_{\rm cm} = cMR^{2}$: the radius cancels and the rotational part becomes $\tfrac12 cMv^{2}$, so the total is $\tfrac12(1+c)Mv^{2}$: the shape enters only through $c$.
The $\textcolor{#1f6feb}{\text{cylinder}}$ on a slope. The weight and the normal force both point at the centre, so neither has any torque about it, and the only force left to spin the body up is the $\textcolor{#d1690a}{\text{static friction}}$ at the $\textcolor{#8250df}{\text{contact point}}$, which is momentarily at rest and therefore does no work.
Looks like this, but is not
There is friction acting on the rolling cylinder, so some energy must be lost to it, and the energy line needs a $f d$ term like every rough surface problem.
Work is a force times the displacement of the point where it acts, and that point is not moving. In rolling without slipping the piece of rim in contact is momentarily at rest, so it slides nowhere and transfers no energy. It is static friction, and its whole job is to supply a torque. The moment the body skids, the contact really does slide, the friction becomes kinetic, an energy loss term is needed, and $v = \omega R$ is lost as well. Those two arrive and leave together.
body
$c$ in $I=cMR^{2}$
spin's share of the energy
speed at the bottom (m/s)
solid sphere
0.400
28.6%
4.58
solid cylinder or disc
0.500
33.3%
4.43
thin spherical shell
0.667
40.0%
4.20
hoop or thin pipe
1.00
50.0%
3.83
block sliding, no friction
0
0%
5.42
The ordering is fixed by the middle column and by nothing else: the more of the energy a body has to divert into spinning, the less is left to move it along, and the slower it arrives. Neither the mass nor the radius appears anywhere in the table, and that is not an accident of the numbers chosen but a consequence of both cancelling out of $v = \sqrt{2gh/(1+c)}$.
A solid cylinder released at the top of a 1.50 m ramp
A solid cylinder of mass 2.00 kg and radius 0.120 m is released from rest and rolls without slipping down a ramp whose top is 1.50 m above the bottom. Find the speed of its centre at the bottom, its angular velocity there, and how the kinetic energy is divided between moving and turning.
released from rest at a height $h = 1.50\ \mathrm{m}$
rolls without slipping the whole way
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the speed and angular velocity at the bottom, and the split of the energy
Solution
Energy is chosen over torque because only the state at the bottom is wanted: the torque route gives an acceleration first and then still needs a kinematics line, while one energy line joins the two ends directly.
Write the energy line with both kinds of kinetic energy in it
Independent check on the total: the two energies add to 29.4 J and the store released was $Mgh = (2.00)(9.80)(1.50) = 29.4$ J, computed without ever using $v$. Comparison check: a block sliding down the same smooth ramp would arrive at $\sqrt{2gh} = 5.42$ m/s, faster, as it must be, since it has nothing to spin up.
One energy line, one substitution of the rolling condition, and no simultaneous equations at all.
Both the mass and the radius cancelled. That is why a marble and a bowling ball, released together, arrive together, and why the two cans of the opening paragraph do not.
How much friction the cylinder needs in order to roll rather than skid
The same solid cylinder of mass 2.00 kg is on a ramp inclined at 25.0 degrees. Find its acceleration while it rolls without slipping, the friction force required, and the smallest coefficient of static friction that will allow it. Take $\sin 25.0^{\circ} = 0.4226$ and $\cos 25.0^{\circ} = 0.9063$.
the acceleration, the friction force and the least coefficient of static friction
Solution
This one has to go the other way. A friction force is asked for and energy never shows a force, so the second law along the slope and the torque equation about the centre are both written out.
One equation along the slope, one about the centre
$$Mg\sin\theta - f = Ma$$
the ordinary second law along the slope; the normal force is perpendicular to the motion and stays out of it
Independent check on the last answer by a formula with no numbers in it: eliminating $M$ and $g$ gives $\mu_{s,\min} = \tfrac13\tan\theta$ for a solid cylinder, and $\tfrac13\tan 25.0^{\circ} = 0.155$. Consistency check on the acceleration: the energy result of the previous example says $v^{2} = \tfrac43 gh$, and with $h$ the drop over a slope length $L$, $h = L\sin\theta$, this gives $v^{2} = \tfrac43 gL\sin\theta = 2aL$ with $a = \tfrac23 g\sin\theta$. The force route and the energy route agree.
The steeper the slope, the more friction rolling demands, while the ceiling $\mu_sMg\cos\theta$ is falling at the same time. That is why a ball rolls on a gentle slope and skids on a steep icy one, and why the condition is always a comparison, never a single number.
Checkpoint
§12.7 — which body reaches the bottom first●●●○○
Four objects are released from rest at the same instant at the top of the same ramp: a solid sphere, a solid cylinder, a hollow spherical shell and a hoop. All four roll without slipping, and they have different masses and different radii.
Given
all released from rest from the same height
all roll without slipping the whole way
the masses are all different and the radii are all different
$I = cMR^{2}$ with $c$ equal to 0.400, 0.500, 0.667 and 1.00 respectively
Find
(a) Choose the object that arrives at the bottom first.
Hint 1/4
The differing masses and radii are in the question to be dismissed. Work out first whether either of them can affect the answer at all.
Hint 2/4
Energy conservation for a rolling body gives $v = \sqrt{2gh/(1+c)}$, in which neither $M$ nor $R$ appears.
Hint 3/4
The coefficients again: 0.400 for the solid sphere, 0.500 for the cylinder, 0.667 for the shell and 1.00 for the hoop, all from the same height $h$.
Hint 4/4
The solid sphere wins, because it has the smallest $c$ and therefore diverts the smallest share of the energy into spinning.
Show solution
The ranking is done on the general formula rather than on four separate calculations: once $M$ and $R$ have cancelled, one comparison of coefficients answers the question for every ramp at once.
Check the extreme case as a test of the formula: a body with $c = 0$ has all of its mass at the axis, nothing to spin up, and should behave like a sliding block. The formula gives $v = \sqrt{2gh}$, which is exactly the sliding block result.
⚠ Leaving the rotational part out of the energy of a rolling body
The body is visibly moving down the ramp, so $\tfrac12 Mv^{2}$ feels like the whole story, and the spin is easy not to see.
wrong$$Mgh = \tfrac12 Mv^{2}\;\Rightarrow\; v = \sqrt{2gh} = 5.42\ \mathrm{m/s}$$
⚠ Subtracting a friction loss for a body that rolls without slipping
Every earlier rough surface problem had an $f_kd$ term, and the word friction triggers the habit before the type of friction is checked.
wrong$$Mgh - f d = \tfrac34 Mv^{2}$$
right$$Mgh = \tfrac34 Mv^{2}\quad(\text{static friction does no work})$$
A cord, a pulley with mass and a hanging body
Anything in which a turning body is connected by a cord or a belt to something that moves in a straight line, and the question asks for an acceleration, a tension or a time.
Declare the positive sense.
One sentence: down is positive for the block, and the sense in which the cord turns the pulley is positive for the pulley. Choosing them to match each other saves every sign error that follows.
Draw one diagram per body.
Forces on the block; forces and their lever arms on the pulley. Two diagrams, never one combined picture. The axle force belongs on the pulley diagram and will contribute nothing, which is worth seeing rather than remembering.
Write one equation per body.
$\sum F = ma$ for the block, $\sum\tau = I\alpha$ for the pulley. Do not write a third equation; there is no third body.
Add the constraint.
If the cord does not slip, $a = \alpha R$. This is the only line that links the two equations, and forgetting it leaves you with three unknowns and two equations.
Eliminate the tension.
Substituting $\alpha = a/R$ turns the pulley equation into $T = (I/R^{2})a$, which has units of mass times acceleration. Put that into the block equation and the pulley behaves as an extra mass $I/R^{2}$ being dragged along.
Check the limits.
Set $I \to 0$: the answer must collapse to the massless pulley result you already trust. Set $I$ very large: the acceleration must go to zero and the tension to the full weight.
Where it goes wrong
Using $mg$ for the tension. The tension is less than the weight whenever the block accelerates, and the difference is exactly what turns the pulley.
Writing $\sum\tau = Ia$ with the linear acceleration. The units give it away: N·m against kg·m$^{2}$ times m/s$^{2}$.
Taking torques about the wrong point, usually the contact point of the cord rather than the axle. Every torque in one equation must be about the same axis.
Deciding between the torque route and the energy route
At the start of every rotation problem, before any equation is written. The two routes cost very different amounts of work and both are always available.
Read the question for the word time.
If a time, or an angular acceleration, or a force is asked for or given, the torque route is needed: energy has no time in it.
Read it for two positions and two speeds.
If it links a height or an angle to a speed and never mentions time, the energy route is one line and the torque route is three.
Check whether anything dissipates.
Rolling without slipping loses nothing, so energy is conserved. A skidding surface or a friction torque in an axle does lose energy, and then the energy route needs an extra term while the torque route does not.
If both work, use one and check with the other.
This is the cheapest verification available in the whole section, because the two routes share no equation. Every worked example in the last three blocks does exactly this.
Where it goes wrong
Starting the algebra before deciding. Two thirds of the way into a set of simultaneous equations is an expensive place to discover that one energy line would have done it.
Using energy when a force is wanted. Energy will give you a speed and will never tell you what the axle is holding.
Forgetting that the energy route still needs $I$, so the shape of the body and its axis have to be settled first either way.
A block sliding down a smooth 1.50 m ramp
A 2.00 kg block is released from rest and slides down a smooth ramp with a vertical drop of 1.50 m. Find its speed at the bottom.
Given
$m = 2.00\ \mathrm{kg}$
drop $h = 1.50\ \mathrm{m}$
smooth surface, so nothing is lost
the block does not turn
Find
the speed at the bottom
Solution
One energy line settles it. The forces route would need the angle of the ramp and then a kinematics step, and the angle is not given.
One kind of kinetic energy only
$$mgh = \tfrac12 mv^{2}$$
the block slides without turning, so there is nothing to put into a second term
Check: this is the same speed as a free fall through 1.50 m, which it must be, since a smooth ramp changes the direction of the motion and takes nothing from it.
A solid cylinder rolling down the same 1.50 m ramp
A solid cylinder of mass 2.00 kg is released from rest and rolls without slipping down the same ramp, with the same 1.50 m drop. Find the speed of its centre at the bottom.
Given
$M = 2.00\ \mathrm{kg}$
drop $h = 1.50\ \mathrm{m}$
rolls without slipping, so no energy is lost
$I_{\rm cm} = \tfrac12MR^{2}$
Find
the speed of the centre at the bottom
Solution
The same energy line as the sliding block, with one extra term. Substituting the rolling condition into it at once is what keeps the count of unknowns at one.
the radius cancels along with the mass, so neither is needed
Answer $$\boxed{\;v = 4.43\ \mathrm{m/s}\;}$$
Check
Check by share: a third of the released 29.4 J went into spin, leaving 19.6 J to move the centre, and $\sqrt{2(19.6)/2.00} = 4.43$ m/s.
Same ramp, same drop, same mass, same absence of any energy loss, and the rolling body arrives 18% slower, because a third of the energy released went into spinning it up rather than moving it along.
How to tell them apart
Ask one question of the picture: does the body turn as it goes? If it slides, skids or is dragged flat, one kinetic term. If it rolls, two terms, and the second one is $\tfrac12 I_{\rm cm}(v/R)^{2}$. The word smooth does not settle it; a cylinder on a perfectly smooth ramp would slide without turning and arrive at 5.42 m/s like the block.
Scaffolding comes off
The common skeleton
Name the two instants: released from rest at the top, and at the bottom of the ramp.
Choose the zero of height and say where it is; the bottom of the ramp is usually cheapest.
List the forces that do work. Gravity does; a normal force never does; static friction at a rolling contact does not, because the contact point does not move.
Write the energy line with both kinds of kinetic energy in it.
Replace $\omega$ by $v/R$, using the rolling condition, so that one unknown is left.
Solve, then check the answer against the sliding case, which must always be faster.
1 · fully worked
A solid sphere rolling down a 2.00 m drop, fully worked
A solid sphere of mass 1.50 kg and radius 0.0800 m is released from rest and rolls without slipping down a ramp with a vertical drop of 2.00 m. Find the speed of its centre and its angular velocity at the bottom.
the speed of the centre and the angular velocity at the bottom
Solution
Energy is the route because the question links a height to a speed and never mentions time; the torque route would give an acceleration that still has to be pushed through a kinematics equation.
Check against the sliding case: a block on the same smooth ramp would reach $\sqrt{2gh} = 6.26$ m/s, and the rolling sphere must be slower than that but not by much, since a sphere puts only 28.6% of its energy into spin. It came out 15% slower, which fits.
Every question of this type is this same skeleton. What changes between them is one coefficient.
2 · you write the reasoning
Now an easier problem with the reasoning taken out. A hoop is released from rest and rolls without slipping down a ramp with a drop of 1.20 m. Find the speed of its centre at the bottom. The steps are all here and the physics is deliberately lighter than rung 1: the work now is writing the reason for each line before you open the model answers.
State 1: at rest at the top, $y_1 = 1.20$ m. State 2: at the bottom, $y_2 = 0$.
reasoning
Two instants have to exist before an energy line can compare anything, and the zero of height is placed at the bottom so that one of the four terms is zero by construction.
The energy line: $Mgh = \tfrac12 Mv^{2}+\tfrac12 I\omega^{2}$ with $I = MR^{2}$.
reasoning
The licence: the only forces are gravity, the normal force and a static friction at a contact point that is momentarily at rest, so no energy leaves and the line may be written with an equals sign. The hoop's coefficient is 1 because all of its mass sits at the rim.
Substituting $\omega = v/R$ turns the line into $Mgh = Mv^{2}$.
reasoning
The rolling condition is what makes the problem solvable: it removes $\omega$, and in doing so it cancels the radius, so the answer cannot depend on how big the hoop is.
So $v = \sqrt{gh} = \sqrt{(9.80)(1.20)} = 3.43$ m/s.
reasoning
The mass cancels for the same reason it always does here, every surviving term carrying exactly one factor of it, so the 3.43 m/s is the answer for a bicycle rim and for a wedding ring alike.
3 · find the buried error
Harder than rung 2, and this worked solution contains exactly two errors. A solid cylinder of mass 2.00 kg and radius 0.120 m is released from rest and rolls without slipping down a ramp with a vertical drop of 1.50 m. Find the speed of its centre at the bottom. Read the four steps and decide which are wrong.
Step 1. The store released on the way down is $Mgh = (2.00)(9.80)(1.50) = 29.4$ J.
Step 2. At the bottom the body has both kinds of kinetic energy, with $I = MR^{2} = (2.00)(0.120)^{2} = 0.0288$ kg·m².
Step 3. The ramp has to be rough for the cylinder to roll, and that friction removes 9.80 J on the way down, leaving 19.6 J.
The coefficient used is the hoop's. A solid cylinder about its own axis has $I = \tfrac12MR^{2} = 0.0144$ kg·m$^{2}$, half of the value written.
The table has six rows that all look like $cMR^{2}$, and under exam pressure the eye takes the first row it lands on. The wrong value is also perfectly plausible in size, so nothing later in the working looks odd.
No energy is lost. The friction here is static, acting at a contact point that is momentarily at rest, so it does no work at all and the full 29.4 J survives to the bottom.
Every rough surface problem before this one had an $f_kd$ term, and the word friction fires the habit before the type of friction has been checked.
right
The energy line keeps the whole 29.4 J: $Mgh = \tfrac12Mv^{2}+\tfrac12I\omega^{2}$, with nothing subtracted.
4 · the bare problem
§12.7 — a hollow shell down a 0.900 m drop●●●○○
No scaffolding this time. A thin hollow spherical shell rolls without slipping from rest down a ramp with a vertical drop of 0.900 m. For a thin spherical shell about a diameter, $I = \tfrac23 MR^{2}$.
(b) Find what fraction of the kinetic energy at the bottom is rotational.
Hint 1/4
The mass and the radius are not given, and you should expect not to need them; set up the energy line and watch what cancels.
Hint 2/4
$Mgh = \tfrac12 Mv^{2}+\tfrac12 I_{\rm cm}\omega^{2}$ with $\omega = v/R$ and $I_{\rm cm} = \tfrac23 MR^{2}$.
Hint 3/4
The data again: a drop of 0.900 m, a coefficient of $\tfrac23$, and $g = 9.80$ m/s$^{2}$.
Hint 4/4
The speed is 3.25 m/s and two fifths of the kinetic energy, 40.0%, is rotational.
Show solution
The skeleton of the first rung is reused with one coefficient changed, since the shape is the only thing that differs; nothing above the coefficient has to be rebuilt.
Check by placing the answer in the table of this section: a shell sits between a solid cylinder and a hoop, so its speed from a 0.900 m drop must lie between $\sqrt{\tfrac43 gh} = 3.43$ m/s and $\sqrt{gh} = 2.97$ m/s. It does.
Full exam-style question
Full exam question: two blocks, a rough table and a pulley with massexam format
A 2.00 kg block sits on a horizontal table with a coefficient of kinetic friction of 0.200 between block and table. A light cord runs from it, over a pulley at the edge of the table, to a 3.00 kg block hanging freely. The pulley is a uniform disc of mass 1.50 kg and radius 0.100 m turning on a smooth axle, and the cord does not slip on it. The system is released from rest. Work to three significant figures throughout.
Given
block on the table: $m_1 = 2.00\ \mathrm{kg}$, $\mu_k = 0.200$
light cord, no slipping on the pulley, smooth axle
released from rest, $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) the acceleration of the blocks, (b) the tension on the table side, (c) the tension on the hanging side, (d) the speed of the hanging block after it has fallen 0.750 m
Solution
Parts (a) to (c) ask for forces and an acceleration, so the torque route is compulsory; part (d) asks for a speed after a distance with no time mentioned, so it is done in one line and then checked by energy.
Set up: three bodies, three equations, one constraint
Two independent checks. Torque check: the pulley equation was never used to find the tensions, so it can test them, $(T_2-T_1)R = (16.106-12.783)(0.100) = 0.332$ N·m against $I\alpha = (7.50\times10^{-3})(44.313) = 0.332$ N·m. Energy check on part (d), using no acceleration at all: the store released is $m_2gd = 22.05$ J and the friction takes $\mu_km_1gd = 2.94$ J, leaving 19.11 J, while the kinetic energy at that moment is $\tfrac12(m_1+m_2)v^{2}+\tfrac12I(v/R)^{2} = 16.62+2.49 = 19.11$ J.
Three equations, one constraint, one addition to eliminate two unknowns at once. Solving for the tensions individually before finding $a$ would have taken twice as long.
The general shape is worth taking away: a pulley of moment of inertia $I$ and radius $R$ behaves in the final equation exactly like an extra mass $I/R^{2}$, which for a uniform disc is half of the pulley's mass, whatever its radius.
Practice
A · concept 4 questions
1§12.1 — what two points on one wheel share●●○○○
A wheel turns on a fixed axle. One ladybird sits on the rim and another sits halfway between the rim and the axle. Somebody claims that the two insects always agree about how fast the wheel is turning but almost never agree about how fast they themselves are moving.
Given
one rigid wheel on a fixed axle
insect 1 at the rim, distance $R$ from the axis
insect 2 at distance $R/2$ from the axis
Find
(a) Decide whether the claim is true or false, and give the reason.
Hint 1/4
Take the two quantities separately. Ask what would have to happen to the wheel for the two insects to disagree about the angular velocity.
Hint 2/4
For a rigid body, $\theta$ is one number for the whole body, so $\omega$ is too; the linear speed of a point is $v = r\omega$ and carries the point's own $r$.
Hint 3/4
The two distances again: $R$ for the first insect and $R/2$ for the second, on the same wheel with the same $\omega$.
Hint 4/4
The claim is true: the same $\omega$ for both, and $v$ in the ratio 2 to 1.
Show solution
The claim hides two different quantities behind one word, so it is tested by computing both for the same wheel: the shared one first, because that is the half of the claim that is true.
each point carries its own distance from the axis into the exchange rate
Answer $$\boxed{\;\text{True: same }\omega,\ v_1 = 2v_2\;}$$
Check
Check at the extreme: a point sitting exactly on the axis has $r = 0$, so $v = 0$, yet it is certainly turning with the wheel. Any description that used a speed would call that point stationary and be wrong.
2§12.2 — zero rate, non zero rate of change●●○○○
A wheel is slowing down, stops for an instant, and then turns the other way, all under one steady influence. Somebody claims that at the instant when the wheel is momentarily at rest, its angular acceleration must also be zero.
Given
the wheel turns one way, stops, then turns the other way
the influence acting on it is steady throughout
we look at the single instant when $\omega = 0$
Find
(a) Decide whether the claim is true or false, and give the reason.
Hint 1/4
Do not answer from the wheel. Ask the same question about a ball thrown straight up, at the top of its flight, where you already know the answer.
Hint 2/4
$\alpha$ is the rate at which $\omega$ changes, not its size, and a quantity can be zero at an instant while changing quickly.
Hint 3/4
At the instant in question, $\omega = 0$ and the wheel is about to turn the other way, so $\omega$ is passing from positive through zero to negative.
Hint 4/4
The claim is false: $\omega$ is zero for one instant and $\alpha$ is not zero at all, which is exactly what makes the wheel reverse.
Show solution
Nothing has to be computed here; the two definitions decide it on their own, and the quickest picture is the graph of $\omega$ against time at the instant it crosses zero.
the second is about how the first is changing, and says nothing about its size
$$\omega = 0\ \text{at one instant},\qquad \alpha \ne 0$$
the standard picture is $\omega$ crossing the axis on a graph against time, where the crossing point has a slope
Answer $$\boxed{\;\text{False}\;}$$
Check
Check by the analogous straight line case, which is already settled: a ball at the top of its flight has $v = 0$ and $a = -9.80$ m/s$^{2}$, and nobody expects it to stay up there.
3§12.4 — what a moment of inertia belongs to●●○○○
A uniform rod is lying on a bench and four students are describing its moment of inertia. Only one of them has said something that is always correct.
Given
one uniform rod of mass $M$ and length $L$
$I = ML^{2}/12$ about a perpendicular axis through its centre
$I = ML^{2}/3$ about a perpendicular axis through one end
Find
(a) Choose the statement that is always correct.
Hint 1/4
Three of the four say something that is true only in a special case or that confuses two different quantities. Test each against the two values given.
Hint 2/4
$I = \sum mr^{2}$ is defined about a chosen axis, so it is a property of a body and an axis together.
Hint 3/4
The two values again: $ML^{2}/12$ about the centre and $ML^{2}/3$ about one end, for the same rod, differing by a factor of four.
Hint 4/4
The correct statement is the one saying that the value changes when the axis changes, even though the body has not.
Show solution
One body with two known values is enough to test all four claims, which is cheaper than arguing each claim in general: any statement that makes $I$ a property of the body alone dies on the first line.
one body, two axes, two answers, so any claim that $I$ is fixed by the body alone fails at once
$$I = \sum m_ir_i^{2}$$
the $r_i$ are measured from the axis, which is where the axis enters the definition
Answer $$\boxed{\;I\ \text{is a property of the body and the axis together}\;}$$
Check
Check with an extreme: spin the rod about its own long axis and almost all of its mass is a fraction of a millimetre from the axis, so $I$ is tiny. Same rod, third axis, third answer.
4§12.7 — the friction under a rolling wheel●●●○○
A cylinder rolls without slipping down a rough ramp, speeding up as it goes. Four students are arguing about what the friction at the contact is doing.
Given
the cylinder rolls without slipping the whole way
the ramp is rough enough for that to happen
the cylinder starts from rest and speeds up
Find
(a) Choose the correct description of the friction force.
Hint 1/4
Two separate questions are hiding here: what kind of friction is it, and does it do any work? Answer them in that order.
Hint 2/4
Work is a force times the displacement of the point where it acts; in rolling without slipping the contact point is momentarily at rest.
Hint 3/4
The situation again: no skidding, so the surfaces do not slide over each other at all, and the contact point has zero velocity at every instant.
Hint 4/4
The friction is static and does no work: it supplies the torque that spins the body up, and takes no energy out.
Show solution
Two separate questions are asked about one force, so they are answered separately, and which kind of friction it is has to be settled first: the work argument only stands while the contact is not sliding.
Which kind of friction
$$\text{no skidding} \Rightarrow \text{surfaces do not slide} \Rightarrow \text{static}$$
kinetic friction is defined by sliding, and there is none here
the point at which the force acts does not move, so nothing is transferred
$$\text{but } \tau = fR \ne 0$$
no work and a real torque are perfectly compatible, and this is the standard example
Answer $$\boxed{\;\text{static friction: supplies torque, does no work}\;}$$
Check
Check by energy bookkeeping: measured speeds at the bottom of a ramp agree with $Mgh = \tfrac12(1+c)Mv^{2}$, with nothing subtracted, and that agreement would fail if the friction were taking joules out.
B · computation 8 questions
1§12.1 — the platter of a hard disc drive●●○○○
The platter of a hard disc drive is 95.0 mm across and spins at 7200 rpm. A single bit of data sits at the outer edge.
Given
platter diameter 95.0 mm, so $R = 0.0475\ \mathrm{m}$
rotation rate 7200 rpm
the rate is steady, so $\alpha = 0$
Find
(a) Find the angular velocity in rad/s.
(b) Find the speed of a point at the outer edge.
(c) Find the acceleration of that point, and compare it with $g$.
Hint 1/4
The rating is in turns per minute and every formula wants radians per second, so decide what the first line of the solution has to be before anything else.
Hint 2/4
$\omega = (\text{rpm})\times 2\pi/60$, then $v = R\omega$ and $a_R = \omega^{2}R$ for a steady rate.
Hint 3/4
The data again: a diameter of 95.0 mm, so a radius of 0.0475 m, at 7200 rpm.
Hint 4/4
The answers are 754 rad/s, 35.8 m/s and $2.70\times10^{4}$ m/s$^{2}$, about 2760 times $g$.
Show solution
The rating is converted once on the first line and the diameter is halved in a line of its own; both are done before any formula, because those two are the errors this question is built to catch.
Independent check on the acceleration by the route that never mentions $\omega$: $v^{2}/R = (35.81)^{2}/0.0475 = 2.70\times10^{4}$ m/s$^{2}$. Size check: 35.8 m/s is 129 km/h at the edge of a disc you can hold in one hand, which is why a dropped drive is a dead drive.
The diameter to radius step is the single most profitable line to double check in this whole block, because it is silent: the answer stays plausible.
2§12.2 — a wheel braked to rest in four seconds●●○○○
A wheel spinning at 12.0 rad/s is braked evenly and comes to rest in 4.00 s.
Given
$\omega_0 = 12.0\ \mathrm{rad/s}$
$\omega = 0$ after $t = 4.00\ \mathrm{s}$
the braking is even, so $\alpha$ is constant
Find
(a) Find the angular acceleration.
(b) Find the number of revolutions made during the braking.
Hint 1/4
Both parts are the constant acceleration equations. Decide which one contains only symbols you already have.
Hint 2/4
$\alpha = (\omega-\omega_0)/t$, and then either $\theta = \bar\omega t$ or $\theta = \omega_0t+\tfrac12\alpha t^{2}$.
Hint 3/4
The data again: from 12.0 rad/s to rest in 4.00 s, with a constant $\alpha$.
Hint 4/4
The angular acceleration is $-3.00$ rad/s$^{2}$ and the wheel makes 3.82 revolutions.
Show solution
Part (b) is taken by the average speed rather than by $\theta = \omega_0 t+\tfrac12\alpha t^{2}$, so that a wrong $\alpha$ in part (a) cannot carry through into the number of turns.
the average value route, which does not need $\alpha$ and so cannot inherit an error from part (a)
$$N = \frac{24.0}{2\pi} = 3.82\ \mathrm{rev}$$
converted at the very end, because turns are what the question asked for
Answer $$\boxed{\;\alpha = -3.00\ \mathrm{rad/s^{2}},\qquad N = 3.82\ \mathrm{rev}\;}$$
Check
Independent check on the angle by the other equation: $\theta = (12.0)(4.00)+\tfrac12(-3.00)(16.0) = 48.0-24.0 = 24.0$ rad. Size check: a wheel doing about two turns a second, stopping in four seconds, ought to manage a few turns, and 3.82 is a few.
3§12.3 — two children on a see-saw●●○○○
A see-saw is a plank pivoted at its middle. A 25.0 kg child sits 2.00 m to the left of the pivot and a 30.0 kg child sits 1.50 m to the right. Take counterclockwise as positive, which for this drawing means the left side going down.
Given
left child: 25.0 kg at 2.00 m from the pivot
right child: 30.0 kg at 1.50 m from the pivot
the plank itself is light and pivoted at its middle
$g = 9.80\ \mathrm{m/s^{2}}$, positive sense counterclockwise
Find
(a) Find the torque each child exerts about the pivot, with its sign.
(b) Find the net torque and say which side goes down.
(c) Find where the 30.0 kg child would have to sit to balance the see-saw.
Hint 1/4
Each child's weight acts straight down at a known distance from the pivot, so each is a lever arm times a force. Get the two signs settled before adding anything.
Hint 2/4
$\tau = r_{\perp}F$ with the weight $mg$ acting vertically, and torques add as signed numbers.
Hint 3/4
The data again: 25.0 kg at 2.00 m on the left, 30.0 kg at 1.50 m on the right, with counterclockwise positive.
Hint 4/4
The torques are $+490$ N·m and $-441$ N·m, the net is $+49.0$ N·m so the left side goes down, and balance needs the heavier child at 1.67 m.
Show solution
A positive sense is fixed before the first torque is written, because parts (b) and (c) are a sum and a cancellation, and neither of them means anything until the signs exist.
Independent check on (c) by ratio, with no arithmetic: the distances must be inversely proportional to the masses, so $d = (2.00)(25.0/30.0) = 1.67$ m. Consistency check on (b): the heavier child was at 1.50 m, closer than the 1.67 m needed, so the lighter child had to win, and the sign agrees.
4§12.4 — three masses on a light rod, two axes●●●○○
Three small masses are fixed on a light rod lying along a line: 2.00 kg at the left end, 3.00 kg at 0.500 m from that end, and 4.00 kg at 1.20 m from that end. The rod's own mass can be ignored.
Given
2.00 kg at position 0
3.00 kg at position 0.500 m
4.00 kg at position 1.20 m
the rod itself is light and contributes nothing
both axes are perpendicular to the rod
Find
(a) Find the moment of inertia about a perpendicular axis through the 2.00 kg mass.
(b) Find it about a perpendicular axis through the 3.00 kg mass.
Hint 1/4
There is no formula to look up here: the definition itself is the method. Decide first, for each axis, how far each of the three masses is from it.
Hint 2/4
$I = \sum m_ir_i^{2}$, where $r_i$ is the distance from the chosen axis to that mass, and a mass sitting on the axis contributes nothing.
Hint 3/4
The positions again: 2.00 kg at 0, 3.00 kg at 0.500 m, 4.00 kg at 1.20 m.
Hint 4/4
The answers are 6.51 kg·m$^{2}$ about the first axis and 2.46 kg·m$^{2}$ about the second.
Show solution
The definition is applied twice from scratch instead of shifting axes with the parallel axis theorem, which would first need the value about the centre of mass, and that is neither given nor asked for.
Axis through the 2.00 kg mass
$$r = 0,\ 0.500,\ 1.20\ \mathrm{m}$$
distances measured from the axis, which is what the definition asks for
Check the direction of the change without arithmetic: the second axis is nearer the heavy 4.00 kg mass and it puts a mass on the axis, so it must give the smaller value, and it does. Order of magnitude: the largest term possible is $4.00(1.20)^{2} = 5.76$, so no answer here can be far from a few kg·m$^{2}$.
Neither answer is the moment of inertia of this rod. Each is the moment of inertia of the rod about one named axis, and the question had to name it before an answer existed.
5§12.4 — a disc pivoted at a point on its rim●●○○○
A uniform disc of mass 1.50 kg and radius 0.200 m is to be pivoted about an axis perpendicular to it through a point on its rim, rather than through its centre. About its centre, $I_{\rm cm} = \tfrac12MR^{2}$.
$I_{\rm cm} = \tfrac12MR^{2}$ about the centre, perpendicular to the disc
the new axis is parallel to that one and passes through a point on the rim
Find
(a) Find the moment of inertia about the rim axis.
(b) Say what fraction of the answer comes from the shift of the axis.
Hint 1/4
The two axes are parallel and one of them passes through the centre of mass, which is exactly the situation one theorem in this section was built for.
Hint 2/4
$I = I_{\rm cm}+Md^{2}$, with $d$ the distance between the two parallel axes.
Hint 3/4
The data again: $M = 1.50$ kg, $R = 0.200$ m, and the shift from the centre to the rim is $d = R = 0.200$ m.
Hint 4/4
The answer is $0.0900$ kg·m$^{2}$, of which two thirds comes from the shift.
Show solution
Here the parallel axis theorem is the cheap route, because the tabulated value about the centre is already known and the shift is exactly one radius; re-summing the disc about its rim would mean an integral.
Check by a bound rather than by repeating the sum: no part of the disc is further than $2R$ from the rim axis, so $I$ cannot exceed $M(2R)^{2} = 0.240$ kg·m$^{2}$, and it must exceed the central value 0.0300. Our 0.0900 sits between them.
6§12.5 — a bucket falling down a well●●●○○
A bucket of mass 3.00 kg hangs from a rope wound around a windlass, a solid cylinder of mass 6.00 kg and radius 0.150 m turning on a smooth axle. The bucket is released from rest.
(c) Find how long the bucket takes to fall 8.00 m.
Hint 1/4
Two bodies are involved and each needs its own equation; then one extra line links them because the rope does not slip.
Hint 2/4
$mg-T = ma$ for the bucket, $TR = I\alpha$ for the windlass, and $a = \alpha R$ to connect them.
Hint 3/4
The data again: $m = 3.00$ kg, $M = 6.00$ kg, $R = 0.150$ m, and $I = \tfrac12(6.00)(0.0225) = 0.0675$ kg·m$^{2}$.
Hint 4/4
The acceleration is 4.90 m/s$^{2}$, the tension is 14.7 N, and the fall takes 1.81 s.
Show solution
Two bodies, so two equations, joined by the rope through $a = \alpha R$; an energy line would give the speed at the bottom but neither the tension nor the time, and both are asked for.
half the bucket's weight, which is what the equal shares of inertia demand
The fall time
$$d = \tfrac12at^{2} \;\Rightarrow\; t = \sqrt{\frac{2(8.00)}{4.90}} = 1.81\ \mathrm{s}$$
$a$ is constant, so the straight line equation applies to the bucket
Answer $$\boxed{\;a = 4.90\ \mathrm{m/s^{2}},\quad T = 14.7\ \mathrm{N},\quad t = 1.81\ \mathrm{s}\;}$$
Check
Independent check by energy. After 8.00 m the bucket should be doing $v = at = 8.85$ m/s. The store released is $mgd = 235.2$ J; the kinetic energy is $\tfrac12(3.00)(78.4)+\tfrac12(0.0675)(59.03)^{2} = 117.6+117.6 = 235.2$ J, the same number. Size check: 4.90 m/s$^{2}$ is exactly half of $g$, and the bucket must fall more slowly than free fall because it is dragging a windlass with it.
7§12.6 — a torque acting through eight turns●●●○○
A wheel of moment of inertia 2.50 kg·m² about its axle is at rest when a constant torque of 25.0 N·m is applied. The torque is kept up for exactly 8.00 revolutions. There is no friction in the axle.
Given
$I = 2.50\ \mathrm{kg\cdot m^{2}}$
$\tau = 25.0\ \mathrm{N\cdot m}$, constant
starts from rest
the torque acts for 8.00 revolutions
no friction torque
Find
(a) Find the work done by the torque.
(b) Find the angular velocity at the end.
(c) Find the power the torque is delivering at that final instant.
Hint 1/4
The question gives a torque and an angle and asks about energy, so this is the energy route from the start; the only conversion needed is at the very beginning.
Hint 2/4
$W = \tau\Delta\theta$ with the angle in radians, then $W = \tfrac12I\omega^{2}$ since it starts from rest, then $P = \tau\omega$.
Hint 3/4
The data again: $\tau = 25.0$ N·m, $I = 2.50$ kg·m$^{2}$, and $8.00$ revolutions, which is $50.3$ rad.
Hint 4/4
The work is $1.26\times10^{3}$ J, the final rate is 31.7 rad/s, and the power at that moment is 793 W.
Show solution
The work is found first and then handed to the energy statement, because the torque and the angle are both given; going through $\alpha$ and the kinematics arrives at the same $\omega$ with more unknowns on the way.
Independent check by the force route: $\alpha = \tau/I = 10.0$ rad/s$^{2}$ and $\omega^{2} = 2\alpha\theta = 2(10.0)(50.265) = 1005$, giving $\omega = 31.7$ rad/s. Cross check on the power: the run lasts $t = \omega/\alpha = 3.17$ s, so the average power is $1256.6/3.17 = 396$ W, exactly half the final value, as it must be when the power rises linearly from zero.
8§12.7 — a rolling ball running up a slope●●●○○
A solid sphere is rolling without slipping along level ground with its centre moving at 6.00 m/s when it reaches the foot of a slope. It continues to roll without slipping as it climbs.
Given
solid sphere, $I_{\rm cm} = \tfrac25MR^{2}$
speed of the centre at the foot: 6.00 m/s
rolls without slipping throughout, so nothing is lost
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the vertical height it reaches before stopping.
(b) Compare that with the height a block sliding at 6.00 m/s would reach on a smooth slope.
Hint 1/4
The question links a speed to a height and never mentions time or force, which settles the choice of route immediately.
Hint 2/4
$\tfrac12(1+c)Mv^{2} = Mgh$ for a rolling body, with $c = \tfrac25$ for a solid sphere.
Hint 3/4
The data again: $v = 6.00$ m/s at the foot, $c = 0.400$, $g = 9.80$ m/s$^{2}$.
Hint 4/4
The sphere reaches 2.57 m, against 1.84 m for a sliding block, because it arrives with extra energy stored in its spin.
Show solution
Energy again, but run in the unwinding direction: everything the sphere carries at the foot is written down before anything is traded for height, which is how the spin term avoids being quietly dropped.
Check by ratio, with no numbers: the two heights must be in the ratio $(1+c):1 = 1.4$, and $2.571/1.837 = 1.400$. Consistency check: the rolling body must go higher, since it arrives with the same speed of the centre plus a store of spin that also has to be spent.
Going up, the spin is an asset; coming down, it is a tax. Both are the same term with the sign of the trip reversed.
C · exam level 5 questions
1§12.2 — a grinding wheel run up and then left to stop●●●●○
A grinding wheel is a uniform disc of mass 12.0 kg and radius 0.200 m on a smooth axle. Exam format, four parts: a motor drives it from rest with a constant torque of 4.80 N·m up to a working speed of 1500 rpm, and is then switched off, after which a friction torque of 1.20 N·m brings it to rest.
after switch off, a friction torque of 1.20 N·m opposes the motion
Find
(a) Find the angular acceleration during the run up.
(b) Find the time to reach the working speed.
(c) Find the number of revolutions made during the run up.
(d) Find how long the wheel takes to stop after the motor is switched off.
Hint 1/4
Four parts, but only two pieces of physics: one equation turns a torque into an angular acceleration, and the constant acceleration equations do everything else.
Hint 2/4
$\alpha = \sum\tau/I$ with $I = \tfrac12MR^{2}$, then $\omega = \omega_0+\alpha t$ and $\theta = \tfrac12\alpha t^{2}$ from rest.
Hint 3/4
The data again: $M = 12.0$ kg, $R = 0.200$ m, driving torque 4.80 N·m, target 1500 rpm which is 157 rad/s, and a friction torque of 1.20 N·m afterwards.
Hint 4/4
The answers are 20.0 rad/s$^{2}$, 7.85 s, 98.2 revolutions and 31.4 s.
Show solution
Everything here is a torque or a time, so the energy route would answer none of the four parts directly; it is used only as the check below.
the same law with a different net torque; the sign says it opposes the motion
$$t = \frac{0-157.08}{-5.00} = 31.4\ \mathrm{s}$$
four times the run up time, because the friction torque is a quarter of the driving one
Answer $$\boxed{\;\alpha = 20.0\ \mathrm{rad/s^{2}},\ t = 7.85\ \mathrm{s},\ N = 98.2\ \mathrm{rev},\ t_{\rm stop} = 31.4\ \mathrm{s}\;}$$
Check
Independent check on (c) by the time free equation: $\theta = \omega^{2}/2\alpha = (157.08)^{2}/40.0 = 617$ rad, which never uses the 7.85 s. Consistency check on (d): the friction torque is a quarter of the driving torque, so the stopping should take four times as long as the run up, and $4\times7.854 = 31.4$ s.
Notice that the friction torque was acting during the run up too, in reality. The question said the driving torque was 4.80 N·m net of nothing, so it was taken as given; a question that says the motor supplies 4.80 N·m against a friction of 1.20 N·m is a different question with $\alpha = 15.0$ rad/s$^{2}$.
2§12.7 — a race between a sphere and a hoop●●●●○
A solid sphere and a hoop are released from rest at the same instant at the top of a ramp with a vertical drop of 1.20 m. Both roll without slipping. The sphere has twice the mass of the hoop and half its radius. Exam format: three parts, and the last one is the interesting one.
the sphere has twice the mass and half the radius of the hoop
both roll without slipping, $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the speed of each at the bottom.
(b) Find the ratio of the two speeds and say whether the mass or radius mattered.
(c) Find the drop the hoop would need in order to arrive at the sphere's speed.
Hint 1/4
The masses and radii are given in order to be discarded; before using them, find out whether they can survive into the answer at all.
Hint 2/4
$v = \sqrt{2gh/(1+c)}$, with $c = \tfrac25$ for a solid sphere and $c = 1$ for a hoop.
Hint 3/4
The data again: a 1.20 m drop for both, coefficients 0.400 and 1.00, and $g = 9.80$ m/s$^{2}$.
Hint 4/4
The sphere reaches 4.10 m/s and the hoop 3.43 m/s, a ratio of 1.20, and the hoop would need a 1.71 m drop to match the sphere.
Show solution
The general result is derived once and then used three times; substituting numbers into the energy line separately for each body means building the same algebra twice and then inverting it a third time.
(a) The general result, then both bodies
$$Mgh = \tfrac12(1+c)Mv^{2}\;\Rightarrow\; v = \sqrt{\frac{2gh}{1+c}}$$
derived once with the rolling condition substituted, so it can be reused rather than rebuilt
Independent check on (c) by proportion rather than by arithmetic: $v^{2}$ is proportional to $h/(1+c)$, so matching speeds needs heights in the ratio of the coefficients, $2.00/1.400 = 1.43$, and $1.43\times1.20 = 1.71$ m. Sanity check on (a): both are slower than the $\sqrt{2gh} = 4.85$ m/s of a sliding block, as they must be.
This is the answer to the two cans of the opening paragraph, in numbers. The packed can rolls like a solid cylinder and the hollow one like a hoop, and the table of this section has already raced that pair: 4.43 m/s against 3.83 m/s down the same ramp, so the solid one arrives about 15% faster, whatever they weigh and whatever their size.
3§12.5 — a rod released from horizontal●●●●●
A uniform rod of mass 1.20 kg and length 0.900 m is hinged at one end and held out horizontally, then released. Exam format: four parts. About a perpendicular axis through one end, $I = \tfrac13ML^{2}$.
hinged at one end, released from rest in the horizontal position
$I = \tfrac13ML^{2}$ about the hinge
the weight acts at the centre of the rod, $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the torque about the hinge at the instant of release.
(b) Find the angular acceleration at that instant.
(c) Find the acceleration of the free end at that instant and compare it with $g$.
(d) Find the angular velocity when the rod reaches the vertical.
Hint 1/4
Parts (a) to (c) are about one instant and involve a force, so they belong to the torque route. Part (d) links two positions to a speed and mentions no time, which points elsewhere.
Hint 2/4
$\tau = Mg(L/2)$ at release, $\alpha = \tau/I$, $a_{\rm tan} = \alpha L$ at the end, and for (d) $Mg(L/2) = \tfrac12I\omega^{2}$.
Hint 3/4
The data again: $M = 1.20$ kg, $L = 0.900$ m, $I = \tfrac13(1.20)(0.810) = 0.324$ kg·m$^{2}$, and the centre of the rod falls $L/2$ on the way to vertical.
Hint 4/4
The torque is 5.29 N·m, $\alpha$ is 16.3 rad/s$^{2}$, the free end starts with 14.7 m/s$^{2}$ which is larger than $g$, and the rod reaches the vertical at 5.72 rad/s.
Show solution
Part (d) is done by energy because $\alpha$ changes continuously as the rod swings, so no constant acceleration equation applies; energy does not care how $\alpha$ varied along the way.
Independent check on (b) and (d) with the mass and length eliminated: $\alpha = 3g/2L = 29.4/1.80 = 16.3$ rad/s$^{2}$ and $\omega = \sqrt{3g/L} = \sqrt{32.67} = 5.72$ rad/s, neither of which contains $M$. Reality check on (c): put a coin on the far end of a ruler held horizontally and let go, and the ruler outruns the coin, which is this result rather than a trick.
The tip beating $g$ is not a violation of anything: the rod is rigid, so the hinge and the near half of the rod are pulling the tip down faster than gravity alone could.
4§12.6 — a winch lifting a load, steadily and then faster●●●●●
A motor drives a winch drum, a uniform cylinder of mass 30.0 kg and radius 0.250 m, with a cable wound on it lifting a 60.0 kg load. Exam format, four parts. First the load is being raised at a constant 0.500 m/s.
second phase: the motor supplies a constant torque of 200 N·m
light cable, no slipping, smooth axle, $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the cable tension while the load rises at constant speed.
(b) Find the torque the motor must supply during that phase.
(c) Find the power it is delivering, and check it a second way.
(d) Find the acceleration of the load at the instant the motor torque is raised to 200 N·m.
Hint 1/4
Constant speed is the phrase that does the work in the first three parts: it fixes both the acceleration and the angular acceleration at zero.
Hint 2/4
At constant speed $T = mg$ and $\tau = TR$, with $P = \tau\omega = Fv$; when it accelerates, $\tau_{\rm motor}-TR = I\alpha$ together with $T-mg = ma$ and $a = \alpha R$.
Hint 3/4
The data again: $M = 30.0$ kg and $R = 0.250$ m so $I = 0.938$ kg·m$^{2}$, load 60.0 kg, lifting speed 0.500 m/s, later torque 200 N·m.
Hint 4/4
The tension is 588 N, the torque 147 N·m, the power 294 W, and the acceleration in the last part is 2.83 m/s$^{2}$.
Show solution
Parts (a) to (c) are settled by the constant speed condition alone, which makes every acceleration zero; the coupled pair of equations is not written until part (d), where something finally accelerates.
(a) and (b) Constant speed means nothing is accelerating
$$T-mg = ma = 0 \;\Rightarrow\; T = mg = (60.0)(9.80) = 588\ \mathrm{N}$$
constant speed makes the load's equation trivial and fixes the tension exactly
with $I = \tfrac12(30.0)(0.0625) = 0.938$ kg·m$^{2}$ and $\alpha = a/R$
$$200-147-15.0a = 3.75a \;\Rightarrow\; a = \frac{53.0}{18.75} = 2.83\ \mathrm{m/s^{2}}$$
the surplus torque above the 147 N·m of the steady phase is what does the accelerating
Answer $$\boxed{\;T = 588\ \mathrm{N},\ \tau = 147\ \mathrm{N\cdot m},\ P = 294\ \mathrm{W},\ a = 2.83\ \mathrm{m/s^{2}}\;}$$
Check
Independent check on (d) by putting the answer back into the untouched torque equation: $T = 60.0(9.80+2.827) = 758$ N, so $TR = 189$ N·m, and $I\alpha = (0.938)(11.31) = 10.6$ N·m, giving a total of 200 N·m, which is the motor torque we were given. Size check: 294 W to lift 60 kg at half a metre per second is about the output of a fit cyclist, which is plausible for a small winch.
Constant speed is worth spotting early: it kills two terms and turns a four unknown problem into two one line answers.
5§12.2 — five readings from a slowing wheel●●●●○
A wheel on a smooth axle is slowing under a constant braking torque, and its angular velocity is measured every two seconds. Exam format, four parts, and everything you need is in the five readings.
Given
reading at $t = 0$: $\omega = 24.0\ \mathrm{rad/s}$
reading at $t = 2.00\ \mathrm{s}$: $\omega = 19.0\ \mathrm{rad/s}$
reading at $t = 4.00\ \mathrm{s}$: $\omega = 14.0\ \mathrm{rad/s}$
reading at $t = 6.00\ \mathrm{s}$: $\omega = 9.00\ \mathrm{rad/s}$
reading at $t = 8.00\ \mathrm{s}$: $\omega = 4.00\ \mathrm{rad/s}$
the wheel's moment of inertia is 3.00 kg·m$^{2}$
Find
(a) Show from the readings that the angular acceleration is constant, and find it.
(b) Find the angle turned through between $t = 0$ and $t = 8.00$ s, in revolutions.
(c) Find the instant at which the wheel stops.
(d) Find the braking torque.
Hint 1/4
Do not fit a curve. Look at what happens to $\omega$ between one reading and the next, and ask whether that change is the same every time.
Hint 2/4
A constant $\alpha$ shows up as equal changes of $\omega$ in equal times; then $\theta = \bar\omega t$, and $\tau = I\alpha$.
Hint 3/4
The readings again, in rad/s: 24.0, 19.0, 14.0, 9.00 and 4.00, taken 2.00 s apart, with $I = 3.00$ kg·m$^{2}$.
Hint 4/4
The wheel loses 5.00 rad/s every 2.00 s, so $\alpha = -2.50$ rad/s$^{2}$; it turns 17.8 revolutions in the eight seconds, stops at 9.60 s, and the braking torque is 7.50 N·m.
Show solution
The constancy of $\alpha$ is tested against the five readings before any constant acceleration equation is allowed in, because every later part, and the extrapolation in (c) most of all, rests on that test.
(a) Test for constancy before assuming it
$$24.0\to19.0\to14.0\to9.00\to4.00$$
four intervals, each losing exactly 5.00 rad/s, which is what constant means here
Independent check on (b) by a different equation: $\theta = \omega_0t+\tfrac12\alpha t^{2} = 192-80.0 = 112$ rad. Check on (c) by the time free equation: the total angle to a stop is $\omega_0^{2}/2|\alpha| = 576/5.00 = 115$ rad, and from 112 rad at 8.00 s the remaining 3.2 rad at an average of 2.0 rad/s takes about 1.6 s, landing at 9.6 s.
A table of readings is doing two jobs at once: it supplies the numbers and it justifies using the constant acceleration equations at all. Show the constancy before you use them, because that is where the marks for part (a) live.
D · interleaved 4 questions
1§12.0 — two trolleys on a level track●●●○○
A 2.00 kg trolley moving at 3.00 m/s along a level track runs into a 4.00 kg trolley standing still, and the two lock together and move off as one. They then run up a smooth slope at the end of the track.
Given
moving trolley: 2.00 kg at 3.00 m/s
stationary trolley: 4.00 kg
they lock together on contact
the track is level and smooth, and so is the slope
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the speed of the pair just after they lock together.
(b) Find how much kinetic energy was lost in the process.
(c) Find the vertical height the pair rises to on the slope.
Hint 1/4
Read the question for what is preserved and what is not. Two different quantities are involved in the two halves of it, and choosing the wrong one for the first half is the whole trap.
Hint 2/4
In a collision in which the bodies lock together, the total momentum is unchanged and the kinetic energy is not; afterwards, on a smooth slope, the mechanical energy is unchanged.
Hint 3/4
The data again: 2.00 kg at 3.00 m/s meeting 4.00 kg at rest, then a smooth slope, with $g = 9.80$ m/s$^{2}$.
Hint 4/4
The pair moves off at 1.00 m/s, 6.00 J of kinetic energy is lost in the collision, and they rise 0.0510 m.
Show solution
Nothing here rotates. The trolleys have wheels in real life and the question says nothing about them, so they are treated as light, and the whole problem is momentum followed by energy.
(a) The collision: use the quantity that survives it
three times the mass at a third of the speed, and the speed is the one that is squared
$$\Delta K = -6.00\ \mathrm{J}$$
two thirds of it gone, into deformation and sound at the coupling
(c) The climb, where energy is conserved again
$$\tfrac12(6.00)(1.00)^{2} = (6.00)gh \;\Rightarrow\; h = \frac{v^{2}}{2g} = 0.0510\ \mathrm{m}$$
the slope is smooth and nothing turns, so one kinetic term and one store
Answer $$\boxed{\;v = 1.00\ \mathrm{m/s},\quad \Delta K = -6.00\ \mathrm{J},\quad h = 0.0510\ \mathrm{m}\;}$$
Check
Check on the height by size: at 1.00 m/s a body has very little energy, and 5 cm is about the height it would reach if thrown straight up at that speed, which is the same calculation. Check on (b): the fraction lost in a collision of this kind is $m_2/(m_1+m_2) = 2/3$, and 6.00 J out of 9.00 J is two thirds.
The habit worth taking from this: decide, for each stage separately, which quantity is conserved. A single problem can conserve momentum in one stage and energy in the next, and neither in a third.
2§12.0 — a coin on a turntable●●●○○
A coin is lying on a turntable 0.120 m from the centre. The turntable is switched on and settles at a steady 45.0 rpm. The coefficient of static friction between coin and turntable is 0.150.
Given
distance of the coin from the centre: 0.120 m
turntable rate: 45.0 rpm, steady
coefficient of static friction 0.150
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the acceleration the coin would need in order to stay put.
(b) Decide whether it stays on or slides off.
(c) Find the largest distance from the centre at which a coin would stay put.
Hint 1/4
Two separate ideas meet here: one gives the acceleration a body needs in order to go round, the other gives the largest force the surface can supply. Compare them.
Hint 2/4
$a_R = \omega^{2}r$ for the acceleration needed, and the friction available has a ceiling of $\mu_smg$, so the coin stays if $\omega^{2}r \le \mu_sg$.
Hint 3/4
The data again: $r = 0.120$ m, 45.0 rpm which is 4.71 rad/s, and $\mu_s = 0.150$ with $g = 9.80$ m/s$^{2}$.
Hint 4/4
The coin needs 2.66 m/s$^{2}$ but friction can only supply 1.47 m/s$^{2}$, so it slides off; a coin closer than 0.0662 m would stay.
Show solution
This looks like a rotation problem and the only rotational thing in it is converting the rpm into $\omega$; the physics that decides the answer is the friction ceiling.
Check on (c) by ratio: the required acceleration is proportional to $r$, and the coin at 0.120 m needed 1.81 times what was available, so the limiting radius must be $0.120/1.81 = 0.0662$ m. Reality check: coins do fly off record players, and they always go from the outside in.
A body on a turntable has no choice about its acceleration: it is fixed by $\omega^{2}r$. The only question is whether something can supply it.
3§12.0 — a ball that rolls off a table●●●●○
A solid sphere is released from rest and rolls without slipping down a ramp with a vertical drop of 0.800 m, arriving on a horizontal table top. It rolls along the table and leaves the edge, which is 0.900 m above the floor.
Given
solid sphere, $I_{\rm cm} = \tfrac25MR^{2}$, rolling without slipping
vertical drop of the ramp: 0.800 m, released from rest
height of the table top above the floor: 0.900 m
the table top is horizontal and the sphere leaves the edge moving horizontally
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the speed of the centre as it leaves the edge.
(b) Find the time it spends in the air.
(c) Find how far from the foot of the table it lands.
Hint 1/4
This is two problems joined end to end, and they use different tools. Deal with the ramp completely before thinking about the fall.
Hint 2/4
On the ramp, $Mgh = \tfrac12(1+c)Mv^{2}$ with $c = \tfrac25$; in the air, the centre of mass is a projectile with $y = \tfrac12gt^{2}$ and $x = vt$.
Hint 3/4
The data again: a 0.800 m drop on the ramp, a table 0.900 m high, $c = 0.400$ and $g = 9.80$ m/s$^{2}$.
Hint 4/4
It leaves the edge at 3.35 m/s, is in the air for 0.429 s, and lands 1.43 m from the foot of the table.
Show solution
The spin still exists during the flight and it is not asked about; the path of the centre follows from the forces on the body, and in the air the only force is gravity acting through the centre.
the vertical motion starts from rest and does not care about the horizontal one
(c) Put them together
$$x = vt = (3.347)(0.4286) = 1.43\ \mathrm{m}$$
no horizontal force acts in the air, so the horizontal speed is unchanged from the moment it left
Answer $$\boxed{\;v = 3.35\ \mathrm{m/s},\quad t = 0.429\ \mathrm{s},\quad x = 1.43\ \mathrm{m}\;}$$
Check
Check on (a) by bounds: the answer must be less than $\sqrt{2gh} = 3.96$ m/s and more than the 2.80 m/s a hoop would manage, and 3.35 m/s sits between them. Check on (c) by size: half a second of flight at about three metres per second has to give something near a metre and a half.
The rolling is entirely confined to part (a). Once the sphere is airborne it is an ordinary projectile, and mixing the two stages is what makes this question long rather than hard.
4§12.0 — a ball on a string in a vertical circle●●●●○
A 0.150 kg ball is swinging on the end of a light string 0.800 m long, in a vertical circle. At the lowest point of its path its speed is 5.00 m/s. Air resistance is negligible.
Given
ball mass 0.150 kg
string length 0.800 m, light and inextensible
speed at the lowest point: 5.00 m/s
air resistance negligible, $g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the tension in the string at the lowest point.
(b) Find the speed of the ball when the string is horizontal.
(c) Find the tension in the string at that horizontal position.
Hint 1/4
Nothing here is rigid and nothing is rolling, so ask instead which two ideas from earlier sections handle a body on a circular path at a known speed.
Hint 2/4
Along the radius, the net force towards the centre equals $mv^{2}/r$; between two positions with only gravity doing work, $\tfrac12mv_1^{2}+mgy_1 = \tfrac12mv_2^{2}+mgy_2$.
Hint 3/4
The data again: $m = 0.150$ kg, $r = 0.800$ m, 5.00 m/s at the bottom, and the horizontal position is 0.800 m higher.
Hint 4/4
The tension is 6.16 N at the bottom, the speed at the horizontal position is 3.05 m/s, and the tension there is 1.75 N.
Show solution
No moment of inertia appears anywhere: the ball is a point on a light string, so this is circular motion and energy, not rotation of a rigid body. Recognising that is most of the question.
(a) The lowest point
$$T-mg = \frac{mv^{2}}{r}$$
at the lowest point the centre of the circle is straight up, so the weight opposes the tension
Check on (a) by size: the tension came out at about four times the weight, 1.47 N, and a ball whirled fast on a short string should indeed pull much harder than it weighs. Check on (c) against (a): both speed and the helpful direction of gravity have gone, so the tension had to drop sharply, and it fell by a factor of three and a half.
The radial equation has to be rewritten at every position, because the angle between the weight and the radius changes. The energy line does not, which is why it is the one used to travel between positions.
Mistake ledger (17 entries)
⚠ Putting degrees into a formula that only accepts radians
The radian is what makes the arc equal to the radius times the angle; a degree entering silently multiplies the answer by about 57.
⚠ Subtracting a friction loss for a body that rolls without slipping
Every earlier rough surface problem carried an $f_kd$ term, and the word friction fires the habit before the kind of friction is checked.
wrong$$Mgh - f d = \tfrac34 Mv^{2}$$
right$$Mgh = \tfrac34 Mv^{2}\quad(\text{static friction does no work})$$
⚠ Quoting a moment of inertia with no axis attached
The same body has a different value about every axis, so a bare $I$ is not an incomplete answer but a meaningless one, and in a marked solution it usually loses the method mark as well.
Both are a newton times a metre, but in a joule the metre is measured along the force and in a torque it is measured across it. A torque becomes an energy only after it is multiplied by an angle in radians.
⚠ Forgetting the constraint that links a cord to a pulley
The two second law equations are the visible part of the method and the constraint is the invisible one, so it is the line most often left out; without it the system cannot be solved and the usual response is to invent a value for the tension.
rolling without slipping, with $I_{\rm cm} = cMR^{2}$
Speed at the bottom of a ramp for a rolling body
$$v = \sqrt{\frac{2gh}{1+c}}$$
released from rest, rolling without slipping, no other losses; $h$ is the vertical drop
Check yourself
Close the page and write from memory: the three formulas that turn an angular quantity into what one point of the body is doing; the two ways of computing a torque and the distance each of them uses; what a moment of inertia belongs to besides the body; the equation that replaces $F = ma$ and the extra line needed when a cord runs over the pulley; the kinetic energy of a body that is turning; and the two consequences of a wheel not skidding. Then open the formula card and mark what you missed, not what you got.
Turn an rpm rating into rad/s, and get the speed and the acceleration of a point at a stated distance from the axis?
c-angular-quantities
Solve a constant angular acceleration problem and give the answer in revolutions when the question asks for turns?
c-constant-alpha
Compute a torque when the force is not perpendicular, attach a sign to it, and add several torques correctly?
c-torque
Compute a moment of inertia for point masses about two different axes, and shift a tabulated value to a parallel axis?
c-moment-of-inertia
Set up a block, cord and massive pulley problem with one equation per body and the constraint that links them?
c-rotational-dynamics
Get an angular velocity from a torque acting through an angle, and a power from a torque and a rate?
c-rotational-energy
Find the speed at the bottom of a ramp for a rolling body, say why the mass and radius drop out, and rank shapes without computing anything?
c-rolling
Glossary (17 terms)
rigid bodykatı cisim
A body whose parts keep their distances from one another, so that every point stays at a fixed distance from the axis and one angle describes the whole thing.
açısal yer değiştirme
The angle a body has turned through, measured in radians from a stated starting orientation and counted continuously rather than reset at each full turn.
radianradyan
The angle that cuts an arc equal in length to the radius, so that a full turn is $2\pi$ of them. It is the only angular unit the formulas of this section accept.
angular velocityaçısal hız
The rate at which the angular displacement changes, in rad/s. It is one number for the whole rigid body, shared by every point of it.
angular accelerationaçısal ivme
The rate at which the angular velocity changes, in rad/s squared. It can be large at an instant when the angular velocity itself is zero.
revolutions per minutedakikadaki devir sayısı
The unit machinery is rated in, written rpm. It becomes a usable angular velocity only after multiplying by $2\pi/60$.
torquetork
The turning effect of a force about a stated axis, equal to the force times the perpendicular distance from the axis to the line the force acts along, in newton metres.
lever armkuvvet kolu
The perpendicular distance from the axis to the line of action of a force. It is what multiplies the force to give the torque, and it is zero for a force aimed at the axis.
line of actionetki çizgisi
The infinite straight line along which a force acts. A force may be slid anywhere along it without changing its torque about any axis.
moment of inertiaeylemsizlik momenti
The rotational counterpart of mass, equal to the sum of each piece of mass times the square of its distance from the axis, in kg·m squared. It belongs to a body and an axis together.
parallel axis theoremparalel eksen teoremi
The rule that the moment of inertia about any axis equals the value about the parallel axis through the centre of mass plus the total mass times the square of the distance between the two axes.
rotational kinetic energydönme kinetik enerjisi
The energy a body has because it is turning, equal to half the moment of inertia times the square of the angular velocity, in joules. A body whose centre never moves can still carry a great deal of it.
flywheelvolan
A heavy wheel used to store energy in its rotation, usually with its mass concentrated near the rim so that the moment of inertia is as large as the material allows.
rolling without slippingkaymadan yuvarlanma
Motion in which the point of a body touching the surface is momentarily at rest relative to it, so that the speed of the centre and the angular velocity are locked together by $v = \omega R$.
rolling conditionyuvarlanma koşulu
The pair of statements $v_{\rm cm} = \omega R$ and $a_{\rm cm} = \alpha R$ that follow from not skidding, and that reduce a rolling problem to a single unknown.
rolling resistanceyuvarlanma direnci
The small loss that stops a real rolling body, caused by the wheel and the surface flattening slightly at the contact. It is named here only to explain why a marble stops, and it is not calculated anywhere in this section.
tangentialteğetsel
Along the direction of motion of a point on a circle, at right angles to the line joining it to the axis. A tangential force is the one that changes the speed of that point.
What comes next
§13 · Angular Momentum; General Rotation
Every body on this page turned about an axle bolted in place, and a sign was enough to say which way. The next section takes the axle away: the axis is allowed to point anywhere and to move, which forces the turning to be described by a direction in space rather than a plus or a minus, and it brings with it a quantity that survives even when the shape of the body changes.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook. Its chapter on rotational motion covers this ground in this order, and its end of chapter problems are the right next step once the practice set here feels comfortable.
Course syllabus: the week line and the assessment table The scope comes from the week line, which reads Rotational Motion and quotes no chapter numbers, so no chapter number is quoted here either. The weightings on the summary card come from the assessment table and nothing beyond them is claimed.
SI units: the radian, the newton metre and the watt The radian is a ratio of two lengths and therefore carries no dimension, which is why $v = r\omega$ comes out in m/s with no conversion factor. Torque is quoted in N·m and never in joules, even though the two are the same combination of base units.