Rotational Motion

A ring, a disc and a solid sphere of the same mass and the same radius are released together from the top of an incline. They roll without slipping. Which reaches the bottom first?

Nothing about mass or radius decides it — both cancel. What decides it is shape:

$a=\frac{g\sin \theta }{1+k^2/R^2}$

Sphere, then disc, then ring — and it would be the same order for a marble and a cartwheel.

Mass far from the axis costs you, because moment of inertia weights it by $r^2$. That single fact runs the chapter.

1. The dictionary

Nothing conceptually new happens here. Apply the substitutions and every equation you already know reappears.

Angular kinematics follows the same substitution: $\omega ={\omega }_0+\alpha t$, $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$, ${\omega }^2={\omega }_0^2+2\alpha \theta$.

Linked by the radius: $v=\omega r$, $a_{tan}=\alpha r$, $a_{cen}={\omega }^{2}r$.

The one genuinely new thing. Mass is intrinsic to a body. Moment of inertia is not — it depends on the axis. Same object, different axis, different problem.

Illustration 1

A flywheel spinning at 300 rpm is brought uniformly to rest in 20 s. How many revolutions does it make?

Convert first — rpm is never a usable unit:

${\omega }_0=300\times \frac{2\pi }{60}=10\pi \approx 31.4\text{rad/s}$

$\alpha =\frac{0-31.4}{20}=-1.57{\text{rad/s}}^2$

$\theta =\frac{{\omega }_0^2}{2|\alpha |}=\frac{986}{3.14}=314\text{rad}$

Revolutions $=\frac{314}{2\pi }=50$

Check by the average-speed route: uniform deceleration means $\theta =\bar{\omega }t=15.7(20)=314$ rad. Same answer, no formula needed.

2. Centre of mass

$x_{cm}=\frac{\sum m_i{x}_i}{\sum m_i}$

It moves as if all the mass sat there and all external forces acted there — which is why an irregular body can be treated as a point in a projectile problem.

Internal forces never move it. They cancel in third-law pairs. An exploding shell's centre of mass carries on along the original parabola.

Removed portion trick: treat the missing piece as negative mass. That turns an awkward integration into two point-mass terms.

For symmetric uniform bodies it sits at the geometric centre — which may be outside the material, as at the centre of a ring.

Two particles. With $m_1$ and $m_2$ a distance $d$ apart, the centre of mass lies on the line joining them at

$r_1=\frac{m_{2}d}{m_1+m_2},r_2=\frac{m_{1}d}{m_1+m_2}$

It divides the separation in the inverse ratio of the masses — always nearer the heavier one.

Differentiating the definition gives the two results that make the concept useful:

${\vec{v}}_{cm}=\frac{\sum m_i{\vec{v}}_i}{M},M{\vec{a}}_{cm}={\vec{F}}_{ext}$

The second is Newton's second law for an extended body. It is the reason a spinning, tumbling spanner thrown across a room still has one point tracing a clean parabola.

Illustration 2

Masses of 2 kg and 3 kg sit 1 m apart. Locate the centre of mass. Then the 2 kg mass is moved 40 cm toward the other. How far must the 3 kg mass move to keep the centre of mass fixed?

$r_1=\frac{3(1)}{5}=0.6\text{m}$ from the 2 kg mass — nearer the heavier one, as it must be.

For the centre of mass to stay put, $\sum m_i\Delta x_i=0$:

$2(+0.4)+3(\Delta x_2)=0\Rightarrow \Delta x_2=-0.267\text{m}$

The 3 kg mass must move 26.7 cm in the opposite direction. This is exactly the mechanism by which a person walking forward on a frictionless boat drives the boat backward.

Illustration 3

A square of side $a$ is cut from one corner of a uniform square plate of side $2a$. Locate the centre of mass of what remains.

Put the origin at the plate's centre, so the removed corner square occupies $0\le x\le a$, $0\le y\le a$ and its own centre sits at $\left(\frac{a}{2},\frac{a}{2}\right)$.

Treat the hole as negative mass. Area scales as the side squared, so the removed piece carries mass $M/4$.

$x_{cm}=\frac{M(0)-\frac{M}{4}\left(\frac{a}{2}\right)}{M-\frac{M}{4}}=\frac{-Ma/8}{3M/4}=-\frac{a}{6}$

By symmetry $y_{cm}=-\frac{a}{6}$ as well, so the centre of mass sits at $\left(-\frac{a}{6},-\frac{a}{6}\right)$.

It has moved diagonally away from the missing corner — the direction common sense predicts, reached without a single integral.

3. Moment of inertia

$I=\sum m_i{r}_i^2$

$r_i$ is measured perpendicular to the axis, and the squaring means distant mass counts disproportionately.

Parallel axis theorem:

$I=I_{cm}+M{d}^2$

Works only from the centre of mass. To shift between two arbitrary parallel axes, go via the centre of mass in two steps.

Perpendicular axis theorem — planar bodies only:

$I_z=I_x+I_y$

Trap. Applying this to a sphere or a solid cylinder is a standard error. Neither is planar.

Radius of gyration: $I=M{k}^2$ — the distance at which a point mass would match the same $I$.

Since $M{d}^2\ge 0$, $I$ is always smallest about an axis through the centre of mass. That falls straight out of the parallel axis theorem.

Illustration 4

Find the moment of inertia of a uniform ring about (a) a diameter, (b) a tangent lying in its plane, (c) a tangent perpendicular to its plane.

(a) A ring is planar, so the perpendicular axis theorem applies. By symmetry the two in-plane axes are equivalent, $I_x=I_y=I_d$:

$I_z=I_x+I_y\Rightarrow M{R}^2=2I_d\Rightarrow I_d=\frac{1}{2}M{R}^2$

(b) Now shift that diameter out to the rim with the parallel axis theorem, $d=R$:

$I=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

(c) Shift the perpendicular central axis instead:

$I=M{R}^2+M{R}^2=2M{R}^2$

Note the order of operations: perpendicular axis theorem first to get a central value, parallel axis theorem second to move it. Doing it the other way round is invalid, because the perpendicular axis theorem requires all three axes to meet at one point on the plane.

Illustration 5

Four point masses $m$ sit at the corners of a square of side $a$. Find $I$ about (a) an axis through the centre perpendicular to the plane, (b) one side, (c) one diagonal.

Only the perpendicular distance to the axis matters, and masses on the axis contribute nothing.

Check with the perpendicular axis theorem, taking $x$ and $y$ along the two diagonals: $I_x+I_y=m{a}^2+m{a}^2=2m{a}^2$, which matches the perpendicular-plane answer. The two independent routes agree.

4. Torque

$\vec{\tau }=\vec{r}\times \vec{F},|\tau |=rF\sin \theta =Fd$

Only the component of $F$ perpendicular to $\vec{r}$ produces torque. A force pointing straight at the axis produces none.

The fastest route in practice is $\tau =F\times$ (moment arm), where the moment arm $d$ is the perpendicular distance from the axis to the line of action.

${\tau }_{net}=I\alpha$

Valid about a fixed axis, or about the centre of mass even if it is accelerating.

Equilibrium needs both: net force zero and net torque zero. A body can have zero net force and still spin up — that is exactly what a couple does.

Illustration 6

A force $\vec{F}=-2\hat{i}+4\hat{j}$ N acts at the point $\vec{r}=3\hat{i}+2\hat{j}$ m from the axis. Find the torque and the moment arm.

$\vec{\tau }=\vec{r}\times \vec{F}=[(3)(4)-(2)(-2)]\hat{k}=16\hat{k}\text{N m}$

For the moment arm, use $\tau =Fd$ directly:

$|F|=\sqrt{4+16}=4.47\text{N}\Rightarrow d=\frac{16}{4.47}=3.58\text{m}$

Read it back: the point of application is only $|\vec{r}|=\sqrt{13}=3.61$ m from the axis, and the moment arm is 3.58 m — almost the whole of it. The force is nearly perpendicular to $\vec{r}$, so almost none of it is wasted pointing at the axis.

5. Equilibrium of rigid bodies

A rigid body is in equilibrium when both conditions hold:

$\sum \vec{F}=0\text{and}\sum \vec{\tau }=0$

For a point mass the first alone was enough. For an extended body it is not: a couple has zero resultant force and still produces rotation.

Choose the axis to kill an unknown. If the torques are taken about a point where an unknown force acts, that force has zero moment arm and vanishes from the equation. Since $\sum \vec{\tau }=0$ about every point once the body is in equilibrium, you are free to pick the most convenient one — this is the single biggest time-saver in the topic.

Illustration 7

A uniform ladder of mass $m$ rests against a smooth vertical wall, its base on a rough floor at angle $\theta$ to the horizontal. Find the minimum coefficient of friction that stops it slipping.

Three unknowns, three equations.

Vertical: the wall is smooth, so it pushes only horizontally. $N_{floor}=mg$

Horizontal: $f=N_{wall}$

Torques about the base — chosen because both $N_{floor}$ and $f$ act there and so contribute nothing:

$N_{wall}(L\sin \theta )=mg\left(\frac{L}{2}\cos \theta \right)$

$N_{wall}=\frac{1}{2}mg\cot \theta$

Now combine. Since $f\le \mu N_{floor}=\mu mg$:

${\mu }_{min}=\frac{1}{2}\cot \theta$

At $\theta =60°$ this is $\frac{1}{2}(0.577)=0.29$; at $45°$ it rises to $0.5$.

Read it back: the length and the mass both cancel — only the angle matters. A ladder set more steeply needs less friction, which is exactly why you push the base of a slipping ladder inward.

6. Angular momentum

$\vec{L}=\vec{r}\times \vec{p}(\text{particle}),L=I\omega (\text{rigid body, fixed axis})$

${\vec{\tau }}_{net}=\frac{d\vec{L}}{dt}$

$\Rightarrow$ zero external torque means $L$ is conserved. This happens more often than expected, including in cases where linear momentum is not.

Because $L=I\omega$ is fixed, shrinking $I$ must raise $\omega$. A skater pulling their arms in spins faster; a collapsing star becomes a pulsar.

Illustration 8

A particle of mass $m$ moves in a straight line at constant velocity $v$, passing a fixed point $O$ at a perpendicular distance $d$. Find its angular momentum about $O$.

$L=mvr\sin \theta$, and $r\sin \theta$ is precisely the perpendicular distance from $O$ to the line of motion:

$L=mvd\text{— constant.}$

Why it must be constant: the only force is zero, so $\tau =0$, so $dL/dt=0$.

Nothing is rotating, and the angular momentum is still non-zero and conserved. Angular momentum is defined about a point, not about a spin — this is the single most common conceptual gap in the chapter, and it is what makes Kepler's second law fall out in one line.

Illustration 9

A 0.5 kg puck on a frictionless table circles at $2\text{m/s}$ on a string of radius $0.8\text{m}$, the string passing through a hole in the table. The string is pulled from below until the radius is $0.4\text{m}$. Find the new speed and the change in kinetic energy.

The string pulls straight toward the hole, so its torque about the hole is zero and $L=mvr$ is conserved:

$L=0.5(2)(0.8)=0.8{\text{kg m}}^2\text{/s}$

$v^{\prime }=\frac{L}{m{r}^{\prime }}=\frac{0.8}{0.5(0.4)}=4\text{m/s}$

$K_i=\frac{1}{2}(0.5)(4)=1\text{J}$, $K_f=\frac{1}{2}(0.5)(16)=4\text{J}$

Kinetic energy quadrupled, an increase of 3 J.

Where from? The hand. Halving the radius means pulling inward against the required centripetal force over a distance, and that work goes into the motion. Contrast this with the next illustration, where the energy falls — conserving $L$ says nothing about energy in either direction.

Illustration 10

A disc of $I=4{\text{kg m}}^2$ spins at $6\text{rad/s}$. A ring of $I=2{\text{kg m}}^2$ is dropped coaxially onto it. Find the common angular speed and the energy lost.

No external torque about the axis, so $L$ is conserved:

$4(6)=(4+2)\omega \Rightarrow \omega =4\text{rad/s}$

$K_i=\frac{1}{2}(4)(36)=72\text{J}$, $K_f=\frac{1}{2}(6)(16)=48\text{J}$

24 J lost — to friction between the surfaces as they came to a common speed. This is the rotational twin of a perfectly inelastic collision, where momentum survives and kinetic energy does not.

7. Rolling motion

Rolling without slipping means the contact point is instantaneously at rest:

$v_{cm}=\omega R$

$K=\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}{I}_{cm}{\omega }^2=\frac{1}{2}M{v}_{cm}^2\left(1+\frac{k^2}{R^2}\right)$

The shape factor $k^2/R^2$ is the only thing distinguishing one rolling body from another: 1 for a ring, $\frac{1}{2}$ for a disc, $\frac{2}{5}$ for a solid sphere.

Down an incline:

$a=\frac{g\sin \theta }{1+k^2/R^2},{\mu }_{min}=\frac{\tan \theta }{1+R^2/k^2}$

Both mass and radius cancel from $a$ — only shape and angle survive.

Friction is essential for rolling but does no work, because the contact point is instantaneously at rest, so nothing slides and nothing is dissipated. Below ${\mu }_{min}$ the body slips, rolling fails, and friction becomes kinetic and does dissipate.

The instantaneous axis trick. Since the contact point is at rest, treat the whole body as purely rotating about it. Then every point's speed is just $\omega \times$(its distance from contact):

This is why the top of a rolling wheel blurs in a photograph while the bottom stays sharp — the top really is moving twice as fast as the car.

Illustration 11

A solid sphere rolls without slipping down a $30°$ incline from a height of $1.4\text{m}$, $g=10$. Find its speed at the bottom.

Energy conservation with the shape factor $k^2/R^2=\frac{2}{5}$:

$mgh=\frac{1}{2}m{v}^2\left(1+\frac{2}{5}\right)=\frac{7}{10}m{v}^2$

$v^2=\frac{10gh}{7}=\frac{10(10)(1.4)}{7}=20\Rightarrow v\approx 4.5\text{m/s}$

Compare: sliding frictionlessly would give $\sqrt{2gh}=5.3\text{m/s}$. The rolling body is slower because some of the energy went into spin, not translation — and the mass never entered.

Illustration 12

Find the minimum coefficient of friction needed for (a) a disc and (b) a ring to roll without slipping down a $30°$ incline.

${\mu }_{min}=\frac{\tan \theta }{1+R^2/k^2}$, with $\tan 30°=0.577$.

The ring needs more friction. Its mass sits entirely at the rim, so it demands more torque to spin up at the required rate, and only friction can supply it.

Consistency check: the ring is also the slowest down the incline, from the table in the opening. More friction required, less acceleration achieved — both follow from the same large $k^2/R^2$.

Illustration 13

A solid cylinder rolling at $3\text{m/s}$ runs up a $30°$ incline. How far along the incline does it travel? $g=10$.

Rolling friction does no work, so mechanical energy is conserved. For a solid cylinder $k^2/R^2=\frac{1}{2}$:

$\frac{1}{2}m{v}^2\left(1+\frac{1}{2}\right)=mgh\Rightarrow h=\frac{3v^2}{4g}=\frac{3(9)}{40}=0.675\text{m}$

Distance along the incline $=\frac{h}{\sin 30°}=1.35\text{m}$

Compare: a body that slid up without friction would rise only $v^2/2g=0.45$ m. The rolling cylinder goes higher, because it arrives carrying rotational kinetic energy as well, and all of it converts to height.

Summary

• Rotational mechanics is translational mechanics with the dictionary applied. Only $I$ is genuinely new.
• $I$ depends on the axis, not just the body. Change the axis and it is a different problem.
• $I=\sum m_i{r}_i^2$ with $r$ perpendicular to the axis — distant mass counts as the square.
• Two-particle centre of mass divides the separation in the inverse ratio of the masses. Removed material is negative mass.
• $M{\vec{a}}_{cm}={\vec{F}}_{ext}$ — internal forces never move the centre of mass.
• Parallel axis $I=I_{cm}+M{d}^2$, from the centre of mass only. Perpendicular axis $I_z=I_x+I_y$, planar bodies only.
• $I$ is minimum about an axis through the centre of mass.
• $\tau =rF\sin \theta =F\times$ moment arm. A force aimed at the axis gives no torque.
• Equilibrium needs zero net force and zero net torque.
• Take torques about a point where an unknown acts and it drops out. Ladder on a smooth wall: ${\mu }_{min}=\frac{1}{2}\cot \theta$, independent of mass and length.
• A particle moving in a straight line has $L=mvd$ about an off-line point — non-zero and conserved, with nothing spinning.
• ${\tau }_{net}=dL/dt$, so zero external torque conserves $L$. Shrink $I$ and $\omega$ rises.
• Rolling: $v_{cm}=\omega R$, contact point at rest, top point at $2v$.
• $a=\frac{g\sin \theta }{1+k^2/R^2}$ — mass and radius cancel, only shape matters. Sphere beats disc beats ring.
• Rolling friction does no work; below ${\mu }_{min}$ the body slips and it does.