Rotational Dynamics and Torque

Beginner

Push a door right next to its hinges and it hardly moves; push just as hard at the handle and it swings open. Turning follows rules much like Newton's laws for straight-line motion.

Step 1. The turning effect of a force is called torque. It depends on how hard you push and on how far from the pivot you push: the same force twice as far out gives twice the turning effect. That is why spanners have long handles. A push aimed straight at the hinges does not turn the door at all.

Step 2. A steady push makes a cart go faster and faster; in the same way, a steady torque makes a wheel spin faster and faster. In the animation a weight hangs from a string wound round the hub of a heavy wheel. As it falls, the string keeps twisting the wheel, and the blue bar shows the spin growing steadily.

Step 3. Mass resists changes of motion, and for turning what matters is where the mass sits. Mass far from the axle makes a wheel much harder to spin up, and harder to stop. That is why the weight in the animation falls so slowly, and why flywheels are built heavy at the rim.

Step 4. When the weight lands, the string goes slack and the twisting stops. The wheel does not stop: it keeps turning at a steady rate, just as a moving object keeps moving when no force acts. In a real machine, friction in the axle slowly brings it to rest.

Check yourself: two children push a roundabout equally hard, one at the rim and one halfway to the centre. Who gives the bigger turning effect? Answer: the one at the rim, because the same push acts twice as far from the pivot.

Intermediate

Describing a rotation

Every point of a turning wheel turns through the same angle, yet a point on the rim moves faster than a point near the axle. A rigid body turning about a fixed axis needs one number to say where it is: the angle $\theta$ through which it has turned, measured in radians (a full turn is $2\pi \approx 6.28\mathrm{rad}$). The angular velocity $\omega$ is the rate of change of that angle, in rad/s, and the angular acceleration $\alpha$ is the rate of change of $\omega$, in rad/s². In the animation the wheel turns clockwise, along the curved green arrow; take that sense as positive.

Every point of the body shares the same $\omega$ and $\alpha$, but a point at distance $r$ from the axis moves at $v=\omega r$, with tangential acceleration $a=\alpha r$. So a weight whose string unwinds from the hub without slipping moves exactly as fast as the hub's edge.

Torque: force times lever arm

Push a door at its handle, 0.80 m from the hinges, with 20 N at right angles to the door: the turning effect, or torque, is $\tau =0.80\cdot 20=16\mathrm{N}\cdot \mathrm{m}$. The same push 0.10 m from the hinges gives only $2\mathrm{N}\cdot \mathrm{m}$. In general, for a force $F$ applied at a distance $r$ from the axis,
$\tau =rF\sin \theta$
where $\theta$ is the angle between the force and the radius to the point where it acts (not the angle the body has turned).

There are two equivalent readings: only the part of the force perpendicular to the radius, $F\sin \theta$, turns the body, or the whole force acts with a lever arm $d=r\sin \theta$, the perpendicular distance from the axis to its line of action. Push the handle at 30° to the door and $\tau =16\cdot \sin 30^{\circ }=8\mathrm{N}\cdot \mathrm{m}$; push straight at the hinges and nothing turns.

Torque is measured in newton metres, never in joules: it is not an energy. Torques that turn opposite ways get opposite signs.

Newton's second law for rotation

A net torque causes an angular acceleration, as a net force causes an acceleration:
$\tau =I\alpha$

The moment of inertia $I$, in kg·m², plays the part of mass. A small piece of mass $m$ at distance $s$ from the axis has a tangential acceleration $\alpha s$, which needs a force $ms\alpha$ and therefore a torque $m{s}^2\alpha$. Adding up the pieces gives $\tau =(\sum m{s}^2)\alpha$, so $I=\sum m{s}^2$. Each kilogram counts with the square of its distance: twice as far out, four times the resistance.

A thin ring of mass $M$ and radius $R$ has all its mass at the rim, so $I=M{R}^2$; a solid sphere has $\frac{2}{5}M{R}^2$. The wheel in the animation is a uniform disc:
$I=\frac{1}{2}M{R}^2\text{(uniform disc)}$
With $M=10\mathrm{kg}$ and $R=0.40\mathrm{m}$, $I=\frac{1}{2}\cdot 10\cdot 0.40^2=0.80\mathrm{kg}\cdot {\mathrm{m}}^2$, half the value for a thin ring of the same mass and size.

With a constant net torque, $\alpha$ is constant and the equations of straight-line motion carry over, with $\theta$, $\omega$ and $\alpha$ in place of $x$, $v$ and $a$:

• $\omega ={\omega }_0+\alpha t$
• $\theta ={\theta }_0+{\omega }_{0}t+\frac{1}{2}\alpha t^2$
• ${\omega }^2={\omega }_0^2+2\alpha \Delta \theta$

Adding up $\frac{1}{2}m{v}^2$ over the pieces, with $v=\omega s$, gives the energy stored in the spin, the rotational kinetic energy
$K=\frac{1}{2}I{\omega }^2$

Worked example: the wheel in the animation

A string wound round the hub, of radius $r=0.12\mathrm{m}$, carries a weight $m=1\mathrm{kg}$. The tension $F$ is unknown, so write one equation for the weight (downward positive), one for the wheel and one for the string, which does not slip:
$mg-F=ma,Fr=I\alpha ,a=\alpha r$
Substituting $F=I\alpha /r$ and $a=\alpha r$ into the first equation gives
$\alpha =\frac{mgr}{I+m{r}^2}=\frac{1\cdot 9.81\cdot 0.12}{0.80+1\cdot 0.12^2}=\frac{1.177}{0.814}\approx 1.45rad/{\mathrm{s}}^2$

So the weight accelerates at only $a=\alpha r\approx 0.173m/{\mathrm{s}}^2$, the tension is $F=m(g-a)\approx 9.64\mathrm{N}$, a little less than the weight's 9.81 N, and the torque is $\tau =rF\approx 1.16\mathrm{N}\cdot \mathrm{m}$. That is the steady value on the red $\tau$ readout; the red arrow $F$ at the edge of the hub and the red line $r$ show where it comes from.

The weight falls $h=0.88\mathrm{m}$ in $t=\sqrt{2h/a}=\sqrt{2\cdot 0.88/0.173}\approx 3.19\mathrm{s}$. Meanwhile the blue $\omega$ readout climbs by 1.45 rad/s every second, up to $\omega =\alpha t\approx 4.60rad/s$, and the wheel turns through $0.88/0.12\approx 7.33\mathrm{rad}$, just over one turn. Then the string goes slack: the arrow $F$ vanishes, the $\tau$ readout drops to zero, $\alpha =0$, and $\omega$ stays at 4.60 rad/s, as the green label points out.

Worked example: braking a flywheel

An exercise bike's flywheel, a uniform disc of 8.0 kg and radius 0.20 m, spins at 30 rad/s (about 287 rpm) when the rider stops pedalling, and a felt pad rubs its rim with a friction force of 4.0 N. How long does it take to stop, and how many turns does it make?

The friction acts along the rim, so its lever arm is the full radius:
$I=\frac{1}{2}\cdot 8.0\cdot 0.20^2=0.16\mathrm{kg}\cdot {\mathrm{m}}^2,\tau =0.20\cdot 4.0=0.80\mathrm{N}\cdot \mathrm{m}$
Then $\alpha =-0.80/0.16=-5.0rad/{\mathrm{s}}^2$, negative because the torque opposes the spin. The wheel stops after $30/5.0=6.0\mathrm{s}$, and ${\omega }^2={\omega }_0^2+2\alpha \Delta \theta$ gives $\Delta \theta =30^2/(2\cdot 5.0)=90\mathrm{rad}$, about 14.3 turns.

Energy confirms it: the wheel starts with $K=\frac{1}{2}\cdot 0.16\cdot 30^2=72\mathrm{J}$, and friction does work $\tau \Delta \theta =0.80\cdot 90=72\mathrm{J}$ against the spin, heating the pad and the rim. A ring of the same mass and radius would store 144 J and need 12 s to stop.

The wheel's free-body diagram

Three forces act on the wheel: its weight $Mg=98.1\mathrm{N}$ at the centre, the 9.64 N tension at the edge of the hub, and the axle's push, which must be $98.1+9.64\approx 107.7\mathrm{N}$ upward because the centre stays put. The net force is zero, yet the net torque is not: the weight and the axle force pass through the axis and have no lever arm, so only the tension turns the wheel.

The reverse happens too. Pull a steering wheel up with 10 N on one side and down with 10 N on the other, both 0.18 m from the centre: the forces cancel, but the torques add up to $2\cdot 10\cdot 0.18=3.6\mathrm{N}\cdot \mathrm{m}$. Such a pair of forces is called a couple.

One $\omega$ means many speeds: at landing the wheel's rim moves at $\omega R\approx 1.84m/s$, the weight at only $\omega r\approx 0.55m/s$.

Common mistakes

• Using the wheel's radius as the lever arm. The string pulls on the hub, so the lever arm is $r=0.12\mathrm{m}$; using $R=0.40\mathrm{m}$ gives 3.85 N·m, more than three times the real torque.
• Taking the tension equal to the weight $mg$. The weight could not then accelerate; the tension is 9.64 N, and the missing 0.17 N is exactly $ma$.
• Thinking that only the mass matters. Moving the wheel's 10 kg out to the rim doubles $I$ to $1.60\mathrm{kg}\cdot {\mathrm{m}}^2$ and cuts the angular acceleration from 1.45 to about 0.73 rad/s².
• Using revolutions per minute or degrees in $v=\omega r$ or $K=\frac{1}{2}I{\omega }^2$. These need radians: 4.60 rad/s is 44 revolutions per minute, and putting 44 into $K$ gives about 770 J instead of 8.5 J.

Where you meet it

Spanners, door handles and bicycle pedals give a long lever arm. Engines and electric motors are rated by the torque at their shaft, a few hundred newton metres for a car engine, and gears trade angular speed for torque. Flywheels in engines and potter's wheels carry their mass in a heavy rim to smooth out an uneven drive.

This lesson builds on Circular Motion. Set the net torque to zero and you reach Static Equilibrium and Levers; follow $L=I\omega$ and you reach Angular Momentum and Its Conservation. Simple Machines: Levers, Pulleys and Gears puts these torques to work.

Advanced

Torque defined through work

A force can push hard on a wheel and still fail to turn it, if it points straight at the axle. To describe turning we need a quantity that captures not only how hard a force pulls but also where and in which direction it acts. The cleanest way to find it is to ask how much work the force does when the body turns. Throughout, the numbers refer to the animation: a uniform disc of mass 10 kg and radius 0.40 m turns on a fixed, frictionless axle, driven by a string wound round a light hub of radius 0.12 m and carrying a 1 kg weight that falls 0.88 m.

Legend

$O$	The fixed axis, seen end-on: it points out of the page.
$r$	Distance from the axis to the point where the force acts.	$\mathrm{m}$
$d\theta$	The small angle the body turns through.	$\mathrm{rad}$
$d\vec{l}$	Step of the point: along the tangent, at right angles to the radius, with length $dl=rd\theta$.	$\mathrm{m}$
$\vec{F}$	Force acting at the point.	$\mathrm{N}$
$F_{\perp }$	Part of $\vec{F}$ along the step, at right angles to the radius: the only part that does work.	$\mathrm{N}$
$F_{\parallel }$	Part of $\vec{F}$ along the radius, at right angles to the step: it does no work.	$\mathrm{N}$

Figure 1 shows a body that can turn only about a fixed axis, seen end-on at $O$ and pointing out of the page. Its position is then given by a single number, the angle $\theta$ through which it has turned. A force $\vec{F}$ acts at a point a distance $r$ from the axis. Let the body turn through a small angle $d\theta$. The point moves along its circle by the short step $d\vec{l}$. Because the turn is small, the step lies along the tangent, at right angles to the radius $r$, and its length is a short arc of the circle: $dl=rd\theta$, with $d\theta$ in radians.

Now look at the two arrows drawn head to tail from the point. They split $\vec{F}$ into a part $F_{\parallel }$ along the radius and a part $F_{\perp }$ along the step. The work done by a force is its component along the displacement times the displacement. $F_{\parallel }$ is at right angles to the step, so it does no work; $F_{\perp }$ lies along the step and does all of it:
$dW=\vec{F}\cdot d\vec{l}=F_{\perp }rd\theta$

The factor in front of $d\theta$ is the quantity we were looking for. Define the torque about the axis as
$\tau =r{F}_{\perp }$
Then the work done in a small turn is $dW=\tau d\theta$: torque times angle is work, just as force times distance is.

The same product can be read another way. If $\phi$ is the angle between the radius and the force, then $F_{\perp }=F\sin \phi$, and $r\sin \phi$ is the lever arm, the perpendicular distance from the axis to the line along which the force acts. So torque is also the whole force times its lever arm. A force aimed straight at the axis has $F_{\perp }=0$ and no lever arm, so it has no torque: that is the wheel that refused to turn.

Coordinates give the same result as a formula that is easy to use. Put the axis along $z$ through the origin. When the body turns anticlockwise through $d\theta$, a point at $(x,y)$ moves by $dx=-yd\theta$ and $dy=xd\theta$. This is the step $d\vec{l}$ of Figure 1 written in components: it is at right angles to the radius and has length $rd\theta$. A force $(F_x,F_y)$ acting there does the work
$dW=F_{x}dx+F_{y}dy=(x{F}_y-y{F}_x)d\theta$
so $\tau =x{F}_y-y{F}_x$. This is the $z$ component of the cross product $\vec{r}\times \vec{F}$, positive when the force tends to turn the body anticlockwise as seen from the $+z$ side. Because $r$, $x$ and $y$ are all measured from the axis, the value of a torque depends on the axis chosen, so always say which axis a torque is about.

Dividing $dW=\tau d\theta$ by the time $dt$ the turn takes gives the power delivered to a turning body. With the angular velocity $\omega =d\theta /dt$,
$P=\tau \omega$
the rotational twin of $P=Fv$.

The animation's numbers bear this out. Just before the weight lands, the string pulls with a tension $T\approx 9.64\mathrm{N}$. On the hub, at a lever arm of 0.12 m, this is a torque $\tau =9.64\cdot 0.12\approx 1.16\mathrm{N}\cdot \mathrm{m}$, and the wheel turns at 4.60 rad/s, so the string delivers $\tau \omega \approx 1.16\cdot 4.60\approx 5.3\mathrm{W}$ to the wheel. At its other end, the string pulls up on the weight, which moves down at 0.553 m/s, so it takes energy from the weight at the rate $Tv\approx 9.64\cdot 0.553\approx 5.3\mathrm{W}$. The two rates agree: a light string stores no energy and only passes it along.

Newton's second law for a turning body

Torque says how hard a body is being turned. We still need the law that says how quickly it responds, the rotational form of $F=ma$. We get it by applying Newton's second law to every piece of the body and adding up the results.

Legend

$O$	The axis through the centre, seen end-on.
$R$	Radius of the disc.	$\mathrm{m}$
$s$	Distance of the gold ring from the axis.	$\mathrm{m}$
$ds$	Width of the ring.	$\mathrm{m}$
$\ell$	Length of the ring cut open and straightened: its circumference $2\pi s$.	$\mathrm{m}$
$dm$	Mass of the ring: its area $2\pi sds$ times the mass per area $M/(\pi R^2)$.	$\mathrm{kg}$
${\omega }^{2}s$	Centripetal acceleration of each piece of the ring, aimed at the axis.	$m/{\mathrm{s}}^2$
$\alpha s$	Tangential acceleration of each piece of the ring, along the circle.	$m/{\mathrm{s}}^2$

Figure 2 shows the pieces. Group together all the matter at the same distance $s$ from the axis: in a disc, that is a thin ring like the gold one, of width $ds$ and mass $dm$. Every piece of the ring moves on a circle of radius $s$, with the angular velocity $\omega$ and the angular acceleration $\alpha =d\omega /dt$ of the whole body. Its acceleration therefore has two parts, both drawn at the top of the ring: a centripetal part ${\omega }^{2}s$, aimed at the axis $O$, which keeps the piece on its circle, and a tangential part $\alpha s$ along the circle, which speeds it up.

By Newton's second law, the net force on the ring has the same two parts: its mass times each acceleration. The centripetal force points at the axis, so it has no lever arm and no torque. Only the tangential force $\alpha sdm$ has a torque, with lever arm $s$:
$d\tau =s\cdot (\alpha sdm)=s^2\alpha dm$

Now add these torques over the whole body. The net force on each piece is made of external forces and of internal forces from the other pieces. The internal forces cancel in pairs: two pieces act on each other with equal and opposite forces along the line joining them, so the two forces have the same lever arm and turn in opposite senses. Only the external torques survive, and since every piece shares the same $\alpha$, it comes out of the sum:
${\tau }_{\mathrm{ext}}=\left(\int s^{2}dm\right)\alpha =I\alpha$
The integral $I=\int s^{2}dm$ is the moment of inertia. It plays the part that mass plays in $F=ma$, but it depends on where the mass sits, not only on how much there is: each piece counts with the square of its distance from the axis.

For the uniform disc of Figure 2, with radius $R$ and mass $M$, the integral is a sum over rings. Cut the gold ring open and straighten it, as drawn below the disc: it becomes a strip of length $\ell =2\pi s$ and width $ds$, so its area is $2\pi sds$. The disc's mass is spread evenly over its area $\pi R^2$, so the mass per area is $M/(\pi R^2)$ and the ring's mass is
$dm=\frac{M}{\pi R^2}\cdot 2\pi sds=\frac{2M}{R^2}sds$
Adding up the rings from the centre to the edge,
$I={\int }_0^R{s}^2\frac{2M}{R^2}sds=\frac{2M}{R^2}\cdot \frac{R^4}{4}=\frac{1}{2}M{R}^2$
This is half the value $M{R}^2$ of a thin hoop of the same mass and radius, because much of the disc's mass lies close to the axis. For the animation's disc, $I=\frac{1}{2}\cdot 10\cdot 0.40^2=0.80\mathrm{kg}\cdot {\mathrm{m}}^2$.

The moment of inertia also depends on which axis is used. For an axis parallel to one through the centre of mass, at a distance $d$ from it, the parallel-axis theorem gives
$I=I_{\mathrm{cm}}+M{d}^2$
To see why, write each piece's squared distance from the new axis as its squared distance from the old one, plus $d^2$, plus a cross term. Added over the body, the cross term is proportional to the mass-weighted mean position measured from the centre of mass, and that mean is zero by the very definition of the centre of mass. Hung from a pin at its rim, the animation's disc has $d=0.40\mathrm{m}$ and $I=0.80+10\cdot 0.40^2=2.40\mathrm{kg}\cdot {\mathrm{m}}^2$, three times its value about the centre.

Angular momentum and the rotational dictionary

In straight-line motion, Newton's law is most general in the form $F=dp/dt$. Rotation has the same form once angular momentum is defined, and with it every result of one-dimensional dynamics carries over.

Each piece of the ring in Figure 2 moves at speed $\omega s$, so its momentum is $\omega sdm$ along the circle. Its angular momentum about the axis is that momentum times its lever arm $s$, which is $s\cdot (\omega s)dm$. Adding over the body gives the total angular momentum, $L=I\omega$. For a rigid body $I$ does not change, so $I\alpha =d(I\omega )/dt$ and the rotation law takes the form
$L=I\omega ,\tau =\frac{dL}{dt}$
the twin of $F=dp/dt$: torque is the rate of change of angular momentum.

Kinetic energy is built the same way. A piece moving at $\omega s$ carries $\frac{1}{2}(\omega s{)}^{2}dm$, and adding up gives
$K=\frac{1}{2}I{\omega }^2$
The work done by the torque, $\int \tau d\theta$, equals the change of $K$, just as $\int Fdx$ equals the change of $\frac{1}{2}m{v}^2$.

The dictionary between straight-line and rotational motion is now complete: $x\leftrightarrow \theta$, $v\leftrightarrow \omega$, $a\leftrightarrow \alpha$, $m\leftrightarrow I$, $F\leftrightarrow \tau$, $p\leftrightarrow L$, $\frac{1}{2}m{v}^2\leftrightarrow \frac{1}{2}I{\omega }^2$ and $Fv\leftrightarrow \tau \omega$. The expert panel of the animation lists four of these pairs. Every result of one-dimensional dynamics carries over, the constant-acceleration equations included.

One entry has no exact twin. A body's mass is fixed, but its moment of inertia changes whenever the body changes shape, because $I$ depends on where the mass sits. Then $\tau =dL/dt$ still holds while $\tau =I\alpha$ fails. That is how a skater spins faster when she pulls her arms in, although no torque acts on her: $L=I\omega$ stays the same, $I$ falls, so $\omega$ rises.

Reading the angular-velocity graph

A graph of angular velocity against time shows the whole motion of the wheel at a glance. It is read exactly like a graph of velocity against time: the slope is the angular acceleration and the area underneath is the angle turned.

Legend

$t$	Time since the weight was released.	$\mathrm{s}$
$\omega$	Angular velocity of the wheel, $d\theta /dt$.	$rad/s$
$t_1$	Time at which the weight lands.	$\mathrm{s}$
${\omega }_1$	Angular velocity at the landing, the largest value reached.	$rad/s$
$\Delta t$	A time interval: the run of the slope triangle.	$\mathrm{s}$
$\alpha \Delta t$	Rise of $\omega$ in the time $\Delta t$: the slope $\alpha$ times $\Delta t$.	$rad/s$
$\tau$	Torque of the string on the wheel: constant while the string pulls, zero after the landing.	$\mathrm{N}\cdot \mathrm{m}$
$\theta$	Angle turned from the release to the landing: the shaded area under the line.	$\mathrm{rad}$

Figure 3 is the graph the expert layer of the animation plots: $\omega$ against the time $t$ since the weight was released. While the string pulls, its torque $\tau$ is constant, so by $\tau =I\alpha$ the angular acceleration is constant too, and the graph is a straight line. Look at the small slope triangle: in any time $\Delta t$, $\omega$ rises by $\alpha \Delta t$, so the slope of the line is
$\alpha =\tau /I=1.16/0.80\approx 1.45rad/{\mathrm{s}}^2$

Now look at the shaded triangle. Each short interval $dt$ adds $\omega dt$ to the angle turned, so the area under the line is the angle $\theta$. The weight lands at $t_1=3.19\mathrm{s}$, when the wheel has reached ${\omega }_1=4.60rad/s$, so
$\theta =\frac{1}{2}\cdot 3.19\mathrm{s}\cdot 4.60rad/s\approx 7.3\mathrm{rad}$
The string gives an independent check. It unwinds from the hub without slipping, so 0.88 m of string off a hub of radius 0.12 m means $\theta =0.88/0.12\approx 7.33\mathrm{rad}$, the same angle. The $L$ readout in the animation is simply $I$ times this graph: it climbs by 1.16 kg·m²/s every second, because the slope of $L(t)$ is the torque.

At $t_1$ the weight lands, the string stops pulling and $\tau =0$, as marked on the flat part of the graph. Then $dL/dt=0$: the line turns flat and $L$ stays at $0.80\cdot 4.60\approx 3.68\mathrm{kg}\cdot {\mathrm{m}}^2/s$, since there is no friction to slow the wheel. The area under the graph keeps growing, now by 4.60 rad every second. At the corner the slope jumps from 1.45 rad/s² to zero, but $\omega$ itself does not jump: a jump in $\omega$ would change $L$ by a finite amount in no time, and that would take an infinite torque.

Energy and the effective mass

Energy conservation checks the whole calculation independently. It also gives a useful way to picture the wheel: as extra mass that the falling weight has to drag along.

After falling $h=0.88\mathrm{m}$, the weight of mass $m=1\mathrm{kg}$ has released $mgh\approx 8.63\mathrm{J}$. The wheel now holds $\frac{1}{2}I{\omega }^2\approx 8.48\mathrm{J}$ and the weight $\frac{1}{2}m{v}^2\approx 0.15\mathrm{J}$, so 98 % of the energy has gone into rotation. Because the string does not slip, the weight moves at the speed of the edge of the hub, $v=\omega r$, where $r=0.12\mathrm{m}$ is the hub radius. Writing $\omega =v/r$, the energy balance reads
$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(m+\frac{I}{r^2}\right)v^2$

Seen from the string, then, the wheel behaves like an extra mass $I/r^2=0.80/0.12^2\approx 55.6\mathrm{kg}$ that the weight has to drag along. Differentiating the balance with respect to time (the rate of change of $h$ is $v$, and that of $v$ is $a$) gives $mgv=(m+I/r^2)va$, so
$a=\frac{mg}{m+I/r^2}$
The 1 kg weight has to accelerate 56.6 kg of effective mass, which is why it creeps down at only 0.17 m/s².

The hub radius is a design choice. Halve it to 0.06 m and $I/r^2$ grows fourfold, to 222 kg, so the weight accelerates about four times more slowly. The wheel's angular acceleration $\alpha =a/r$ falls less, because a smaller hub turns further for each metre of string, but it still halves, to about 0.73 rad/s². In general
$\alpha =\frac{mgr}{I+m{r}^2}$
which grows with $r$ as long as $m{r}^2<I$, that is up to $r=\sqrt{I/m}\approx 0.89\mathrm{m}$. That is beyond the wheel's own radius of 0.40 m, so for this wheel a wider hub always spins it up faster: the longer lever arm outweighs the weight's own growing share of the inertia, $m{r}^2$.

Worked problem: the wheel as a yo-yo

So far the axle was fixed. A yo-yo shows the same laws at work when the axis itself moves. Tie the free end of the string to the ceiling and let the wheel fall, unwinding from its hub like a large yo-yo, with $M=10\mathrm{kg}$, $I=0.80\mathrm{kg}\cdot {\mathrm{m}}^2$ and hub radius $r=0.12\mathrm{m}$. How fast does it accelerate, how hard does the string pull, and how is the energy shared after a drop of 0.88 m?

Legend

$Mg$	Weight of the wheel, acting at its centre.	$\mathrm{N}$
$T$	Pull of the string, acting upward at the edge of the hub.	$\mathrm{N}$
$r$	Radius of the hub: the lever arm of $T$ about the centre.	$\mathrm{m}$
$a$	Acceleration of the centre, downward.	$m/{\mathrm{s}}^2$
$\alpha$	Angular acceleration of the wheel, anticlockwise because the string pulls up to the right of the centre.	$rad/{\mathrm{s}}^2$

Figure 4 shows the two forces on the wheel, drawn to scale. Its weight $Mg$ acts at the centre. The pull $T$ of the string acts upward at the right-hand edge of the hub, a distance $r$ from the centre, and is a little shorter than $Mg$. Their difference, $Mg-T$, accelerates the centre downward at $a$, the straight arrow on the left. About the centre the weight has no lever arm, so only $T$ turns the wheel. Because it pulls up to the right of the centre, it turns the wheel anticlockwise, with the angular acceleration $\alpha$ shown by the curved arrow.

Can $\tau =I\alpha$ still be used when the axis falls? Yes, for an axis through the centre of mass that keeps its direction. In a frame falling with the centre, the wheel simply turns about a fixed axis, and the only extra force in that frame, the inertial force $-M\vec{a}$, acts at the centre of mass, where it has no lever arm.

The two motions are linked. The string is fixed to the ceiling, so the point of the hub where the string leaves it is momentarily at rest. The wheel turns about that point, so the centre, a distance $r$ away, falls by $r$ for every radian turned: $a=\alpha r$.

Translation, rotation about the centre and this rolling condition give three equations for $a$, $T$ and $\alpha$:
$Mg-T=Ma,Tr=I\alpha ,a=\alpha r$
Put $\alpha =a/r$ into the second equation to get $T=Ia/r^2$, then put that into the first:
$a=\frac{g}{1+I/(M{r}^2)}=\frac{9.81}{1+5.56}\approx 1.50m/{\mathrm{s}}^2,T=M(g-a)\approx 83.1\mathrm{N}$
and $\alpha =a/r\approx 12.5rad/{\mathrm{s}}^2$. This is the effective-mass result again: the rotation adds $I/r^2\approx 55.6\mathrm{kg}$ to the 10 kg that gravity pulls on. After the 0.88 m drop, $v=\sqrt{2ah}\approx 1.62m/s$ and $\omega =v/r\approx 13.5rad/s$, almost three times the 4.60 rad/s reached in the animation, because now the wheel's whole weight drives it rather than a 1 kg weight.

The released energy $Mgh\approx 86.3\mathrm{J}$ splits into $\frac{1}{2}M{v}^2\approx 13.2\mathrm{J}$ of translation and $\frac{1}{2}I{\omega }^2\approx 73.2\mathrm{J}$ of rotation. Their ratio is fixed at $I/(M{r}^2)\approx 5.56$, whatever the height, because $\omega =v/r$ at every moment. The string pulls with 83.1 N, not the full 98.1 N weight, so a spring balance holding the string reads less than the yo-yo's weight while it falls. And at the bottom, with the string fully unwound, the wheel is still spinning, so it winds itself back up the string: that is how a yo-yo returns.

Limits of validity

The model assumes a rigid wheel, a light string that neither stretches nor slips, and a frictionless axle. Friction matters more for coasting than for spinning up. A bearing torque of 0.02 N·m is under 2 % of the string's 1.16 N·m, so it would lower $\alpha$ by less than 2 %. Once the weight has landed, however, it is the only torque, and it would stop the coasting wheel in $I\omega /\tau =0.80\cdot 4.60/0.02\approx 184\mathrm{s}$, about three minutes.

Rigidity fails at high speed. The stress in a thin spinning rim is about $\rho v^2$, where $\rho$ is the density and $v$ the rim speed, so the rim bursts when $\rho v^2$ reaches the strength of the material. A steel rim (density 7800 kg/m³, strength near $10^9\mathrm{Pa}$) bursts at roughly 360 m/s. The rim in the animation moves at $4.60\cdot 0.40\approx 1.84m/s$.

The scalar law $\tau =I\alpha$ needs a fixed axis, or an axis through the centre of mass that keeps its direction. In three dimensions torque is a vector, $\vec{\tau }=\vec{r}\times \vec{F}$, and $\vec{L}$ is tied to $\vec{\omega }$ not by a single number but by the inertia tensor, a set of nine numbers that takes the place of $I$.

Spun about a symmetry axis, a wheel has $\vec{L}$ parallel to $\vec{\omega }$. A lopsided wheel has $\vec{L}$ tilted away from the axle, so as the wheel turns, $\vec{L}$ sweeps round a cone, and the bearings must supply a torque $d\vec{L}/dt$ even at constant speed. That is why car wheels are balanced, and it is where gyroscopes and precession begin.

Relativistic corrections, of order $v^2/c^2\approx 4\times 10^{-17}$ for the rim, play no part. In quantum mechanics angular momentum comes in steps of order $\hslash \approx 1.05\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$. The wheel's $3.68\mathrm{kg}\cdot {\mathrm{m}}^2/s$ is about $3.5\times 10^{34}$ such steps, so its spin looks perfectly continuous, but a molecule's rotation comes in discrete levels that show up in microwave spectra.

Where this leads

Set the net torque and the net force to zero and you have Static Equilibrium and Levers. Let the moment of inertia change, or let the axis tilt, and you reach Angular Momentum and Its Conservation, with skaters, Kepler's law of equal areas and the precessing gyroscope. A body hanging from a pivot feels a restoring torque from gravity and swings as a physical pendulum, with a period of $2\pi \sqrt{I/(Mgd)}$ for small swings, where $d$ is the distance from the pivot to the centre of mass. The disc hung from its rim, with $I=2.40\mathrm{kg}\cdot {\mathrm{m}}^2$ and $d=0.40\mathrm{m}$, swings with a period of about 1.55 s; Simple Harmonic Motion develops this. The torque of a magnetic field on a current loop drives every electric motor, the subject of Motors, Generators and Transformers, and the power formula $P=\tau \omega$ joins this lesson to Work, Energy and Power.