Rotation and angular momentum

Put a sticker near the middle of a turntable and another near its edge, and let the table turn a little. Both stickers move along circles around the centre. The outer one travels farther, because its circle is bigger. Yet both stickers sit on one rigid table, so in an obvious sense they have turned by the same amount. We want a number for that amount, and the number should not depend on which sticker we watch.

One such number is the angle in degrees, which divides a full turn into $360$ equal parts. Degrees work, but $360$ is an arbitrary choice. For rotation there is a more useful measure built from lengths, which we met on the polar coordinates page. Let us build it again here.

Watch a sticker at distance $r$ from the centre while the table turns, and call the length of the curved path it travels the arc length $s$. Now look at a sticker twice as far out. Its circle is the first circle enlarged to twice the size, so for the same turn it travels twice as far: $r$ doubles and $s$ doubles. Their ratio $s/r$ therefore does not change. It is the same for every sticker on the table, so it measures the turn itself. We call this ratio the angle in radians and write it $\theta$ (the Greek letter theta):

How many radians make a full turn? In one full turn a sticker goes once round its circle. The number $\pi \approx 3.1416$ is defined as the distance round any circle divided by its diameter $2r$, so the distance round, the circumference, is $\pi \times 2r = 2\pi r$. Dividing by $r$ gives $\theta = 2\pi r / r = 2\pi$. Half a turn is therefore $\pi$ radians and a quarter turn is $\pi/2$. A turn of $d$ degrees is the fraction $d/360$ of a full turn, so in radians it is $\theta = \frac{d}{360} \times 2\pi$.

Move the slider and compare the two arcs. The blue sticker is $1\ \mathrm{m}$ from the centre and the orange one is $2\ \mathrm{m}$ from it. The orange arc is always twice as long as the blue arc, while $s/r$ is the same number for both.

For example, a quarter turn is $\theta = \pi/2 \approx 1.57$ radians. Multiplying both sides of $\theta = s/r$ by $r$ gives $s = r\theta$, so the blue sticker travels $s = 1\ \mathrm{m} \times 1.57 = 1.57\ \mathrm{m}$ and the orange sticker travels $s = 2\ \mathrm{m} \times 1.57 = 3.14\ \mathrm{m}$. The length ratio is dimensionless; radian is its named angle unit. That is why multiplying $r$ by the radian value of $\theta$ gives a length in metres.

On the Newton’s laws page, average velocity meant signed displacement divided by elapsed time; average speed meant distance travelled divided by elapsed time. A turn also needs a sign: here anticlockwise is positive. Divide the signed angular change $\Delta\theta$ by the interval $\Delta t$ to obtain the average angular velocity:

Here $\omega$ is the Greek letter omega, and $\Delta$ means “the change in”. The unit is radians per second, $\mathrm{rad/s}$. For changing rotation, successively shorter intervals give the angular velocity at one instant. For our steadily turning table, the average and instantaneous values are the same. Every sticker turns through the same angle in the same time, so the rigid table has one $\omega$. Starting from angle zero gives $\theta=\omega t$.

The stickers still move at different speeds, and now we can work them out. During $\Delta t$ the table turns through $\Delta\theta$, so a sticker at distance $r$ travels the arc length $s = r\,\Delta\theta$ (chapter 1). Its speed is that distance divided by the time:

In the third step we moved $r$ out of the fraction, which we may do because $r$ stays the same during the interval. So every point on the table shares one $\omega$, but its speed $v = r\omega$ grows in proportion to its distance from the centre. A point exactly on the axis, where $r = 0$, does not move at all.

Play the turntable and change $\omega$. Each arrow points along its sticker’s direction of motion, and its length shows the speed. The orange arrow is always twice as long as the blue one.

With $\omega = 1.2\ \mathrm{rad/s}$, after $t = 2\ \mathrm{s}$ the table has turned through $\theta = \omega t = 1.2 \times 2 = 2.4$ radians. The blue sticker has travelled $s = 1 \times 2.4 = 2.4\ \mathrm{m}$ and moves at $v = 1 \times 1.2 = 1.2\ \mathrm{m/s}$. The orange sticker has travelled $s = 2 \times 2.4 = 4.8\ \mathrm{m}$ and moves at $v = 2 \times 1.2 = 2.4\ \mathrm{m/s}$.

A trolley full of bottles is harder to start moving than an empty one. On the Newton’s laws page this resistance to a change of motion is measured by mass: a force $F$ gives a mass $m$ the acceleration $a = F/m$. Is mass also what makes a body hard to set turning? Not quite. Hold a broom by the middle and it is easy to twirl. Hold it by the end, with the heavy head far from your hand, and the same broom is much harder to swing round. The mass has not changed. Where the mass sits matters.

To find out exactly how, we use energy, because energy adds up piece by piece. The work and energy page showed that a body of mass $m$ moving at speed $v$ has the kinetic energy $\tfrac{1}{2}mv^2$: this is the work we must do to bring it from rest to the speed $v$. Now take two small masses $m$ on a light rod, each at distance $r$ from the axis, and set the rod turning at $\omega$. From chapter 2, each mass moves at $v = r\omega$, so each has the kinetic energy

The rod is light, so it carries almost no energy itself, and the total is the energy of the two masses added together:

Every piece of any turning body can be treated the same way. A piece of mass $m$ at distance $r$ contributes $\tfrac{1}{2} m r^2 \omega^2$, and all the pieces share the same $\omega$. So we can always take $\tfrac{1}{2}\omega^2$ outside the sum and write $K = \tfrac{1}{2} I \omega^2$, where $I$ is the sum of $m r^2$ over all the pieces. This $I$ is called the moment of inertia:

Compare $K = \tfrac{1}{2} I \omega^2$ with $\tfrac{1}{2} m v^2$: $I$ plays the part that mass plays in straight-line motion. To bring a body from rest to the angular velocity $\omega$ we must supply the work $\tfrac{1}{2} I \omega^2$, so a body with a larger $I$ needs more work for the same turning: it is harder to spin up. Each piece counts with $r^2$, so mass far from the axis counts for much more than mass close to it. That is the broom.

Move the two masses along the rod and change their size. The rod turns at $\omega = 2\ \mathrm{rad/s}$ throughout. Each arrow shows a mass’s speed, and the bar shows the total kinetic energy.

With $m = 1\ \mathrm{kg}$ at $r = 0.5\ \mathrm{m}$, each mass moves at $v = 0.5 \times 2 = 1\ \mathrm{m/s}$ and has $\tfrac{1}{2} \times 1 \times 1^2 = 0.5\ \mathrm{J}$, so together $K = 1\ \mathrm{J}$. The formula agrees: $I = 2 \times 1 \times 0.5^2 = 0.5\ \mathrm{kg\,m^2}$ and $\tfrac{1}{2} \times 0.5 \times 2^2 = 1\ \mathrm{J}$. Now move the masses out to $r = 1\ \mathrm{m}$. The distance has doubled, $r^2$ has become four times as large, and so has the energy: $K = 4\ \mathrm{J}$.

Later two flywheels with the same mass $M$ and the same outer radius $R$ race each other, so we need their moments of inertia. Start with a thin ring, whose mass all sits at the rim. Every piece has $r = R$, so $I = \sum m R^2 = R^2 \sum m = M R^2$. A solid disk of the same mass spreads it from the centre out to the rim. Most of its pieces are closer than $R$, so its $I$ must be smaller. Adding up the thin rings that make up the disk gives exactly half: $I = \tfrac{1}{2} M R^2$. A thick ring, called an annulus, whose mass lies between an inner radius $R_{\text{in}}$ and $R$, gives $I = \tfrac{1}{2} M (R^2 + R_{\text{in}}^2)$, which lies between the other two.

To open a heavy door, we usually push far from its hinges. With the same perpendicular push closer to the hinge, the turning effect is smaller. The levers page measured this with a moment: force times the perpendicular distance from the axis to the force’s line of action. That perpendicular distance is the moment arm. In rotation we also call the moment torque.

Look at the door from above, with the hinge as the axis. Suppose we push with the force $F$ at distance $r$ from the hinge. If we push at right angles to the door, all of the push works to turn it, and the torque is $\tau = rF$ ($\tau$ is the Greek letter tau). If we push at an angle $\varphi$ to the door, we split the push into two parts, one along the door and one across it. The part along the door only presses on the hinge and cannot turn the door. The part across the door has size $F\sin\varphi$, so

Here $\sin\varphi$ is the fraction of the push that points across the door. It is $1$ when $\varphi = 90^\circ$ and $0$ when the push points straight along the door. (The unit circle page shows where $\sin$ comes from: on a circle of radius $1$, it is the height of the point at angle $\varphi$. On a force arrow of length $F$ drawn at angle $\varphi$ to the door, the part across the door is $F$ times that height.)

Change where you push and the direction of the push. Orange shows the full $10\ \mathrm{N}$ force, blue shows its component perpendicular to the door, and gold marks the perpendicular moment arm. The door is held in one position so we can compare the applied torque without also changing its orientation; the holding constraint supplies a balancing torque.

A push of $F = 10\ \mathrm{N}$ at right angles, $0.8\ \mathrm{m}$ from the hinge, gives $\tau = 0.8 \times 10 = 8\ \mathrm{N\,m}$. The same push $0.2\ \mathrm{m}$ from the hinge gives only $0.2 \times 10 = 2\ \mathrm{N\,m}$. Back at $0.8\ \mathrm{m}$ but at $\varphi = 30^\circ$, where $\sin 30^\circ = 0.5$, the torque is $0.8 \times 10 \times 0.5 = 4\ \mathrm{N\,m}$.

Now we can say how a torque changes the turning. Take one small mass $m$ on a light rod at distance $r$ from the axis, and push it with a force $F$ at right angles to the rod. Newton’s second law says the mass speeds up along its path with the acceleration $a = F/m$, that is, $F = ma$. Its speed is $v = r\omega$, so when $\omega$ changes by $\Delta\omega$, the speed changes by $\Delta v = r\,\Delta\omega$. Dividing both by the time taken gives $a = r\alpha$, where $\alpha = \Delta\omega/\Delta t$ is called the angular acceleration ($\alpha$ is the Greek letter alpha). Putting $a = r\alpha$ into $F = ma$:

In the second line we multiplied both sides by $r$ to turn the force into a torque, and in the third we recognised $rF = \tau$ and $m r^2 = I$. A whole body is made of many such pieces. Each piece is also pushed and pulled by its neighbours, but by Newton’s third law these forces come in equal and opposite pairs acting along the same line, so their torques cancel when we add up all the pieces. What remains is the torque from outside the body, and the sum of $m r^2$, which is $I$:

For this rigid body rotating about a fixed axis with constant $I$ — a body whose parts keep the same distances, so it does not deform — this is Newton’s second law for turning ($\\tau = I\\alpha$). Here $\\tau$ is the net outside torque about that axis and $\\alpha$ is the instantaneous angular acceleration; finite-interval averages also obey the equation when $I$ stays constant. Larger $\\tau$ gives larger $\\alpha$, while larger $I$ gives smaller $\\alpha$ at the same torque. If the mass distribution changes, we must use angular momentum rather than keep applying this constant-$I$ form.

Here is a machine that tests $\tau = I\alpha$. Two flywheels with the same mass $M = 2\ \mathrm{kg}$ and the same outer radius $0.88\ \mathrm{m}$ turn on fixed axles. On the left is a solid disk; on the right is a thick ring with an inner radius of $0.69\ \mathrm{m}$. From chapter 3, their moments of inertia are $I = \tfrac{1}{2} \times 2 \times 0.88^2 = 0.774\ \mathrm{kg\,m^2}$ and $I = \tfrac{1}{2} \times 2 \times (0.88^2 + 0.69^2) = 1.251\ \mathrm{kg\,m^2}$. A string wound round each axle holds a weight of mass $m = 0.35\ \mathrm{kg}$, and both weights are released together.

Before you press Play: the wheels have equal masses and the weights are equal. Which wheel spins up faster?

Let us work it out. Call the string’s pull the tension $T$ and the axle’s radius $b$. Two things move, the weight and the wheel, so we write Newton’s law for each. The weight is pulled down by gravity $mg$ and up by the string $T$, so $mg - T = ma$. The string pulls the axle at distance $b$, which gives the wheel the torque $Tb$, so $Tb = I\alpha$. The string does not slip, so the weight falls exactly as fast as the axle’s surface moves round: by chapter 2 that is $v = b\omega$, and for the changes it gives $a = b\alpha$. Now we solve:

In the first line we divided $Tb = I\alpha$ by $b$ and used $\alpha = a/b$. In the second we put this $T$ into the weight’s equation. In the third we collected the two terms containing $a$, giving $mg = a\,(m + I/b^2)$, and divided. For the solid disk, with $g = 9.81\ \mathrm{m/s^2}$ and $b = 0.2\ \mathrm{m}$, we get $a = \frac{0.35 \times 9.81}{0.35 + 0.774/0.04} = 0.174\ \mathrm{m/s^2}$. For the thick ring the same formula gives $a = 0.109\ \mathrm{m/s^2}$. The disk wins, because its mass sits closer to the axle and its moment of inertia is smaller.

Notice that the string does not pull with the weight $mg$. From $mg - T = ma$ we get $T = m(g - a)$, so the disk’s string pulls with $0.35 \times (9.81 - 0.174) = 3.37\ \mathrm{N}$, a little less than the weight $0.35 \times 9.81 = 3.43\ \mathrm{N}$. If the string pulled with the full weight, nothing would be left over to speed up the weight itself.

Press Play and watch the two weights fall. Each wheel turns a brighter cyan as it gains kinetic energy, and each weight fades as it loses height. Then change the axle’s radius $b$ and play again: a larger $b$ gives the same string a longer lever arm.

The momentum page described motion along a line with the momentum $p = mv$, and wrote Newton’s second law as “force equals the rate at which momentum changes”, $F = \Delta p/\Delta t$. When no outside force acts, momentum stays the same. Is there a version of this for turning?

Start from $\tau = I\alpha$ for a body whose moment of inertia does not change. Write $\alpha$ as $\Delta\omega/\Delta t$. Because $I$ is constant, $I$ times the change in $\omega$ is the same as the change in $I\omega$:

The quantity $I\omega$ is called the angular momentum and written $L$. So torque is the rate at which angular momentum changes, $\tau = \Delta L/\Delta t$, just as force is the rate at which momentum changes. For one small piece, $L = m r^2 \omega = r \times m(r\omega) = r \times mv$: the piece’s momentum $mv$ times its distance from the axis, in the same way that a torque is a force times its distance.

If no outside torque acts, $\Delta L = 0$ and the angular momentum stays the same. This is the conservation of angular momentum. Its most useful feature is that it still holds when the body changes shape and $I$ changes, as long as the forces inside the body cannot produce a torque about the axis. Then $I\omega$ stays fixed, so if $I$ becomes smaller, $\omega$ must become larger. That is the spinning chair.

This machine tests the claim. Two identical rotors start at the same $\omega = 1.6\ \mathrm{rad/s}$. Each carries two weights with a total mass of $M = 2\ \mathrm{kg}$ on rails, $1.25\ \mathrm{m}$ from the axle. On rotor A (left) a small winder at the centre pulls the weights in to $0.42\ \mathrm{m}$ and later lets them out again. On rotor B (right) the weights stay where they are, so B shows what A would do if nothing changed. The winder’s cords pull straight towards the axle, so they give the rotor no torque about it.

Most of each rotor’s moment of inertia comes from the weights, $M r^2 = 2 \times 1.25^2 = 3.125\ \mathrm{kg\,m^2}$. The rest of the rotor adds a fixed amount that does not depend on where the weights sit: the hub and rails, and the spread of each weight about its own centre, together $0.2656\ \mathrm{kg\,m^2}$. So $I = 3.125 + 0.2656 = 3.3906\ \mathrm{kg\,m^2}$ at the start. With the weights at $0.42\ \mathrm{m}$, $M r^2 = 2 \times 0.42^2 = 0.3528$ and $I = 0.3528 + 0.2656 = 0.6184\ \mathrm{kg\,m^2}$.

Before you press Play: when A’s weights move in, what happens to A’s turning? And to B’s?

No outside torque acts about the axle, so A keeps the angular momentum it started with: $L = I\omega = 3.3906 \times 1.6 = 5.425\ \mathrm{kg\,m^2/s}$. When the weights reach $0.42\ \mathrm{m}$, $\omega = L/I = 5.425/0.6184 = 8.77\ \mathrm{rad/s}$, more than five times as fast. When they go back out, $I$ returns to $3.3906$ and $\omega$ returns to $1.6\ \mathrm{rad/s}$. B never changes.

Press Play, or drag the time slider to stop at any moment and compare the readouts: $I$ falls, $\omega$ rises, and their product $L$ stays put. You can also change how far in the weights are pulled.

But where does the extra speed come from? The kinetic energy is $K = \tfrac{1}{2} I \omega^2$ (chapter 3). While $L$ is fixed we can replace $\omega$ by $L/I$:

With $L$ fixed, a smaller $I$ means more rotational kinetic energy. Initially $K=5.425^2/(2\times3.3906)=4.34\ \mathrm{J}$; after the weights stop at the inner position, $K=5.425^2/(2\times0.6184)=23.80\ \mathrm{J}$. The extra $19.46\ \mathrm{J}$ comes from work done by the winder. Without a force, a weight would continue along its tangent, not turn towards the axle. Pulling it inward requires a force through an inward displacement, which transfers energy. During the inward or outward slide, radial velocity $\dot r$ is perpendicular to tangential velocity $r\omega$. Thus $v^2=\dot r^2+(r\omega)^2$, and the readout also includes radial energy $M\dot r^2/2$. That term vanishes at both stationary-radius endpoints. The reversible model returns work to the winder during the outward stroke; real mechanisms also dissipate energy.

Angular momentum can also pass from one body to another. Two equal disks, each with $I = \tfrac{1}{2} \times 2 \times 0.88^2 = 0.7744\ \mathrm{kg\,m^2}$, share one axle. The lower disk turns at $1.6\ \mathrm{rad/s}$. The upper disk starts at rest and is lowered onto it. Friction between the two faces slows the lower disk and speeds up the upper one until they turn together.

Where the disks touch, the lower disk drags the upper one forward and the upper disk drags the lower one back. By Newton’s third law these friction forces are equal and opposite, and they act at the same distance from the axle, so their torques are equal and opposite too. In every interval the angular momentum one disk gains is exactly what the other loses, so the total angular momentum does not change. Call the starting rates $\omega_a$ (lower) and $\omega_b$ (upper), and the common final rate $\omega_f$:

Here $\omega_a = 1.6\ \mathrm{rad/s}$ and $\omega_b = 0$, so $\omega_f = (1.6 + 0)/2 = 0.8\ \mathrm{rad/s}$: the turning is shared equally.

Use the slider to choose the upper disk’s starting rate, from $-1$ to $1$ times the lower disk’s, and press Play. Compare the final rate with $(\omega_a + \omega_b)/2$, and the heat with $\tfrac{1}{4} I (\omega_a - \omega_b)^2$.

Energy is not shared so neatly. Before contact, $K = \tfrac{1}{2} \times 0.7744 \times 1.6^2 = 0.99\ \mathrm{J}$. Afterwards both disks turn at $0.8\ \mathrm{rad/s}$, so $K = 2 \times \tfrac{1}{2} \times 0.7744 \times 0.8^2 = 0.50\ \mathrm{J}$. Half of the kinetic energy has gone. The rubbing faces turned it into internal energy: they became warmer, which the orange glow at the contact shows. For any two equal disks the loss, $\Delta K = K_{\text{before}} - K_{\text{after}}$, is

In the second line we substituted $\omega_f = (\omega_a + \omega_b)/2$, so that $\omega_f^2 = (\omega_a + \omega_b)^2/4$. In the third we wrote both terms over $4$ and multiplied out the bracket, and in the last we recognised $(\omega_a - \omega_b)^2$. The loss is zero only when the disks already turn at the same rate, so that nothing rubs.

Angular momentum has a direction: turning one way counts as positive and the other way as negative. Set the upper disk to start at $-1$ times the lower disk’s rate, the same speed in the opposite direction. The total is $0.7744 \times 1.6 + 0.7744 \times (-1.6) = 0$, so the disks must end at $\omega_f = 0$: both stop. Yet both were turning, and kinetic energy, which depends on $\omega^2$, is never negative. All of it, $\tfrac{1}{4} \times 0.7744 \times (1.6 - (-1.6))^2 = 1.98\ \mathrm{J}$, becomes internal energy. Angular momenta can cancel; energies cannot.

Now return to the chair. We consider you and the chair together, about the swivel axis, neglecting bearing friction for this ideal comparison. Pulling the arms in moves mass towards the axis and reduces $I$. With no net outside torque, $L=I\omega$ stays fixed and $\omega$ rises. Muscles supply work while the arms move inward; holding them at a fixed radius is not further mechanical work along the radius, even though real muscles consume energy. Spreading the arms raises $I$ and lowers $\omega$. The comparison below treats the two hand-held weights as point masses and keeps the chair-and-body inertia fixed. It is a controlled model of the mechanism, not a measurement of a particular person.