Rotational Motion

A billiard ball is struck horizontally by a cue. At what height must it be struck so that it rolls without slipping immediately, with the cloth doing nothing at all?

Most say through the centre. Strike a ball through its centre and it skids — it slides forward with no spin, and friction has to drag it into rolling over the next several inches.

Let the impulse $J$ act at height $h$ above the centre. Two equations, one constraint:

$J=M{v}_{cm},Jh=I_{cm}\omega =\frac{2}{5}M{R}^2\omega ,v_{cm}=\omega R$

Substituting the first and third into the second:

$Jh=\frac{2}{5}M{R}^2\cdot \frac{v_{cm}}{R}=\frac{2}{5}R\left(M{v}_{cm}\right)=\frac{2}{5}RJ⟹h=\frac{2R}{5}$

Strike two fifths of a radius above the centre — that is $7R/5$ above the cloth — and the ball rolls from the first instant. This is why a player's cue tip sits noticeably above centre, and it is the same calculation that locates the sweet spot of a cricket bat.

Nothing here was about forces. It was about where the impulse acted, and about choosing the point to take moments about. That choice is what this chapter is really testing.

1. Angular momentum about a point, not just an axis

At Main level $L=I\omega$ about a fixed axis is enough. Advanced problems need angular momentum about an arbitrary point, and it splits exactly as kinetic energy does:

${\vec{L}}_O=\underset{\text{spin about the c.m.}}{\underbrace{I_{cm}\vec{\omega }}}+\underset{\text{orbital, c.m. about }O}{\underbrace{M\left({\vec{r}}_{cm}\times {\vec{v}}_{cm}\right)}}$

Illustration 1

A uniform disc of mass $M$ and radius $R$ rolls without slipping at $v_{cm}=\omega R$. Find its angular momentum about a point $O$ on the ground.

By the split:

$L_O=I_{cm}\omega +M{v}_{cm}R=\frac{1}{2}M{R}^2\omega +M(\omega R)R=\frac{3}{2}M{R}^2\omega$

By the instantaneous axis: the contact point is momentarily at rest, so the disc is in pure rotation about it, with $I_{contact}=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$ from the parallel axis theorem:

$L=I_{contact}\omega =\frac{3}{2}M{R}^2\omega$

The two agree, as they must. The orbital term is what candidates forget, and dropping it gives $\frac{1}{2}M{R}^2\omega$ — a third too small.

A body moving in a straight line has angular momentum about any off-line point. $L=Mvd$ with $d$ the perpendicular distance. Nothing has to be spinning.

2. Moment of inertia by integration

Advanced expects you to derive the standard values, and to handle bodies not in the table.

$I=\int r^{2}dm,r\perp \text{the axis}$

The technique is always the same: choose an element whose points are all at the same distance from the axis, express $dm$ through the density, and integrate.

Illustration 2

Derive the moment of inertia of a uniform disc of mass $M$ and radius $R$ about its central perpendicular axis.

Choose an annular ring of radius $r$ and thickness $dr$ — every point of it is at distance $r$ from the axis, which is exactly the requirement.

Surface density $\sigma =\frac{M}{\pi R^2}$, and the ring's area is $2\pi rdr$, so

$dm=\sigma \left(2\pi rdr\right)$

$I={\int }_0^R{r}^2\sigma \left(2\pi r\right)dr=2\pi \sigma {\int }_0^R{r}^{3}dr=2\pi \sigma \frac{R^4}{4}$

Substituting $\sigma$:

$I=\frac{2\pi R^4}{4}\cdot \frac{M}{\pi R^2}=\boxed{\frac{1}{2}M{R}^2}$

Why a ring and not a strip: a straight strip has its points at many different distances from the axis, so $r^2$ could not come out of the integral. Choosing the element correctly is the whole skill.

Illustration 3

Find the moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis through one end, making angle $\theta$ with the rod.

Take an element $dx$ at distance $x$ from the end. Its perpendicular distance from the axis is $x\sin \theta$, not $x$ — and that is the entire question.

$dm=\frac{M}{L}dx$, so

$I={\int }_0^L{\left(x\sin \theta \right)}^2\frac{M}{L}dx=\frac{M{\sin }^2\theta }{L}\cdot \frac{L^3}{3}$

$I=\frac{1}{3}M{L}^2{\sin }^2\theta$

Check both extremes. At $\theta =90°$ it gives $\frac{1}{3}M{L}^2$, the familiar perpendicular-through-the-end value. At $\theta =0$ the axis lies along the rod, every element sits on it, and $I=0$. Both correct.

3. Where you may take torques

This is the single most important structural fact in the chapter, and it is where most Advanced marks are lost.

${\tau }_{net}=I\alpha \text{is valid about}$

The centre of mass is privileged. You may take torques about it even when it is accelerating wildly, and no pseudo-force correction is needed. About any other accelerating point you must add the torque of the pseudo force $-M{\vec{a}}_{cm}$ acting at the centre of mass.

The contact point of a rolling body is a favourite trap: it is instantaneously at rest but it is accelerating, so torques about it are legitimate only in the special case of rolling on a stationary surface, where the correction happens to vanish.

Illustration 4

A solid cylinder rolls without slipping down an incline of angle $\theta$. Find its acceleration, twice — once about the centre of mass and once about the contact point.

About the centre of mass. Only friction has a moment arm about the centre.

$Mg\sin \theta -f=M{a}_{cm}$

$fR=I_{cm}\alpha =\frac{1}{2}M{R}^2\cdot \frac{a_{cm}}{R}$

From the second, $f=\frac{1}{2}M{a}_{cm}$. Substituting:

$Mg\sin \theta =\frac{3}{2}M{a}_{cm}\Rightarrow a_{cm}=\frac{2}{3}g\sin \theta$

About the contact point. Friction and the normal force both pass through it, so only gravity has a torque:

$Mg\sin \theta R=I_{contact}\alpha =\frac{3}{2}M{R}^2\cdot \frac{a_{cm}}{R}$

$a_{cm}=\frac{2}{3}g\sin \theta$

Same answer, and the second route never needed the friction at all. That is why the contact point is worth using — but only because the surface here is stationary.

General result: $a_{cm}=\frac{g\sin \theta }{1+k^2/R^2}$, and the friction required is $f=\frac{Mg\sin \theta }{1+R^2/k^2}$.

4. Rolling, slipping, and the transition

The velocity of the contact point decides everything:

$v_{contact}=v_{cm}-\omega R$

In both slipping cases friction pushes the body toward $v_{cm}=\omega R$. Rolling is an attractor, which is why a skidding ball always ends up rolling.

Illustration 5

A solid sphere is projected along a rough floor with speed $v_0$ and no spin. Find when pure rolling begins and the speed then.

Route 1 — forces and kinematics.

$v=v_0-\mu gt$ and $\omega =\frac{\mu MgR}{I_{cm}}t=\frac{5\mu g}{2R}t$

Setting $v=\omega R$:

$v_0-\mu gt=\frac{5}{2}\mu gt\Rightarrow t=\frac{2v_0}{7\mu g},v=\frac{5}{7}{v}_0$

Route 2 — angular momentum about a point on the ground, in one line.

Friction acts at the contact point, which lies on the ground line, and friction is horizontal. Its moment about any point $O$ on that line is therefore zero, and gravity and the normal force cancel. So $L_O$ is conserved.

$M{v}_{0}R=I_{cm}\omega +MvR=\frac{2}{5}MvR+MvR=\frac{7}{5}MvR$

$v=\frac{5}{7}{v}_0$

The friction coefficient never appeared. It sets when rolling starts, not at what speed.

Illustration 6

Find the fraction of kinetic energy lost in that process.

$K_i=\frac{1}{2}M{v}_0^2$

$K_f=\frac{1}{2}M{v}^2+\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{v}{R}\right)}^2=\frac{1}{2}M{v}^2\left(1+\frac{2}{5}\right)=\frac{7}{10}M{v}^2$

With $v=\frac{5}{7}{v}_0$:

$K_f=\frac{7}{10}M\left(\frac{25}{49}{v}_0^2\right)=\frac{5}{14}M{v}_0^2$

$\frac{K_f}{K_i}=\frac{5/14}{1/2}=\frac{5}{7}\Rightarrow \text{lost}=\frac{2}{7}$

Two sevenths of the energy is burned as heat, and again independently of $\mu$ — a rougher floor simply burns it over a shorter distance.

5. Angular impulse and the point of strike

$\int \vec{\tau }dt=\Delta \vec{L},\text{and for an impulse }J\text{ at perpendicular distance }d:\Delta L=Jd$

The hook's calculation generalises. For a body of radius of gyration $k$ struck horizontally at height $h$ above the centre, immediate rolling requires

$h=\frac{k^2}{R}$

Strike below that height and the body skids forward; strike above it and it over-spins. A ring must be struck at its topmost point, which is why a hoop is so hard to start rolling cleanly.

Illustration 7

A uniform rod of mass $M$ and length $L$ is hinged at one end and lies at rest. It is struck by a horizontal impulse $J$ at distance $x$ from the hinge. Find $x$ such that the hinge experiences no impulsive reaction.

If the hinge feels nothing, the only horizontal impulse is $J$ itself, so

$J=M{v}_{cm}$

The rod turns about the hinge, so $v_{cm}=\omega \frac{L}{2}$, giving $J=\frac{M\omega L}{2}$.

Taking moments about the centre of mass, where the impulse acts at distance $\left(x-\frac{L}{2}\right)$:

$J\left(x-\frac{L}{2}\right)=I_{cm}\omega =\frac{M{L}^2}{12}\omega$

Substituting $J$:

$\frac{M\omega L}{2}\left(x-\frac{L}{2}\right)=\frac{M{L}^2\omega }{12}\Rightarrow x-\frac{L}{2}=\frac{L}{6}$

$x=\frac{2L}{3}$

This is the centre of percussion — the sweet spot. A cricket bat or tennis racket struck there transmits no jarring impulse to the hands, and every bat is designed so that the sweet spot sits where the ball is normally met.

6. Collisions that involve rotation

When a moving body strikes an extended one, linear momentum alone is not enough — you need angular momentum about a well-chosen point.

Choose the point where the unknown impulsive force acts. At a hinge, that means taking moments about the hinge, so the hinge reaction contributes nothing.

Illustration 8

A rod of mass $1\text{kg}$ and length $1\text{m}$ hangs vertically, hinged at its top end. A bullet of mass $10\text{g}$ travelling horizontally at $200\text{m/s}$ strikes the lower end and embeds itself. Find the angular velocity just afterwards.

The hinge exerts an unknown impulsive force, so take angular momentum about the hinge, where its moment is zero.

Before: only the bullet contributes, at perpendicular distance $L$:

$L_i=mvL=0.01(200)(1)=2{\text{kg m}}^2\text{/s}$

After: the rod rotates about the hinge with the bullet embedded at the end:

$I=\frac{M{L}^2}{3}+m{L}^2=0.333+0.01=0.343{\text{kg m}}^2$

$\omega =\frac{2}{0.343}=5.83\text{rad/s}$

Note what is not conserved. Linear momentum is destroyed by the hinge, and kinetic energy falls from $200$ J to $\frac{1}{2}(0.343)(5.83{)}^2=5.8$ J — over $97\%$ lost. Using either of those as a conservation law here would be a serious error.

7. Toppling versus sliding

A block of width $b$ and height $h$ on an incline has two independent ways to fail, and the question is always which happens first.

Sliding begins when friction is exhausted:

$\tan \theta =\mu$

Toppling begins when the weight's line of action passes outside the base — that is, beyond the lower edge:

$\tan \theta =\frac{b}{h}$

Compare $\mu$ with $b/h$. If $\mu <b/h$ the block slides first; if $\mu >b/h$ it topples first. A tall thin block on a grippy surface tips; a squat block on a slippery one slides.

Illustration 9

A block $0.3\text{m}$ wide and $0.9\text{m}$ tall sits on an incline with $\mu =0.5$. As the incline is slowly tilted, does it slide or topple, and at what angle?

$\frac{b}{h}=\frac{0.3}{0.9}=0.333$, and $\mu =0.5$.

Since $\mu >b/h$, toppling wins:

$\tan \theta =0.333\Rightarrow \theta =18.4°$

Sliding would not have begun until $\tan \theta =0.5$, i.e. $26.6°$ — by which point the block has long since tipped over.

Change one number and the answer flips. With $\mu =0.25$ the block slides at $14.0°$ and never topples at all.

Illustration 10

A block of mass $M$, width $b$ and height $h$ rests on a rough floor. A horizontal force $F$ is applied at height $y$. Find the condition on $y$ for the block to topple rather than slide.

Sliding starts at $F=\mu Mg$.

Toppling about the far bottom edge starts when the applied torque overcomes the restoring torque of the weight:

$Fy=Mg\frac{b}{2}\Rightarrow F=\frac{Mgb}{2y}$

Toppling happens first if it requires the smaller force:

$\frac{Mgb}{2y}<\mu Mg⟹y>\frac{b}{2\mu }$

Push high and it tips; push low and it slides. This is precisely why you push a heavy wardrobe near the floor, and why a lorry's load is stowed as low as possible.

8. Choosing the point: a checklist

The centre of mass never needs justification. Every other choice has to be earned, and the earning is usually the observation that some unknown force has zero moment there.

Summary

• ${\vec{L}}_O=I_{cm}\vec{\omega }+M\left({\vec{r}}_{cm}\times {\vec{v}}_{cm}\right)$ — spin plus orbital. Dropping the orbital term is the standard error.
• A rolling disc has $L=\frac{3}{2}M{R}^2\omega$ about a ground point, matching $I_{contact}\omega$.
• $I=\int r^{2}dm$ with $r$ perpendicular to the axis. Choose an element all of whose points share one $r$.
• Rod at angle $\theta$ about an end: $\frac{1}{3}M{L}^2{\sin }^2\theta$.
• $\tau =I\alpha$ holds about a fixed axis or about the centre of mass, and nowhere else without a correction.
• Rolling condition $v_{cm}=\omega R$. Kinetic friction always drives a slipping body toward it.
• Sphere projected without spin: rolling begins at $v=\frac{5}{7}{v}_0$, losing $\frac{2}{7}$ of the energy — both independent of $\mu$.
• Angular momentum about a point on the ground line is conserved while friction acts there, which solves such problems in one line.
• Immediate rolling from a horizontal strike needs $h=k^2/R$: $0.4R$ for a sphere, $R$ for a ring.
• Centre of percussion of a rod hinged at one end is at $2L/3$ — the sweet spot.
• In a rotational collision at a hinge, angular momentum about the hinge is conserved; linear momentum and energy are not.
• On an incline, sliding needs $\tan \theta =\mu$ and toppling needs $\tan \theta =b/h$. The smaller angle wins.
• A horizontal push topples rather than slides when applied above $y=\frac{b}{2\mu }$.