Rotational Motion

Lessons

Senior 6 · Physics

Rotational Motion

4 lessons

Lesson 4 of 4

Angular Momentum and Its Conservation

25 min

The spin that speeds itself up

A figure skater begins a spin slowly with her arms and one leg stretched out. She draws them in towards her body and, with no push from anything outside, spins several times faster. A diver leaving the high board does the same trick by tucking into a ball. Neither pushes on anything, so what speeds them up? The answer is a conservation law as fundamental as the conservation of linear momentum you studied in S4.

Angular momentum

For a particle of mass $m$ moving with velocity $\vec{v}$ at position $\vec{r}$ from an axis point, the angular momentum is

\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v}, \qquad L = mvr\sin\varphi

For a particle moving in a circle of radius $r$ , $\varphi = 90^\circ$ and $L = mvr = mr^2\omega$ . Adding over all the particles of a rigid body turning about a fixed axis gives

L = \left(\sum m_i r_i^2\right)\omega = I\omega

Torque changes angular momentum

If $I$ is constant, $\tau = I\alpha = I\frac{d\omega}{dt} = \frac{d(I\omega)}{dt}$ . In general (and even when $I$ changes):

\sum \tau_{\text{ext}} = \frac{dL}{dt} \qquad \text{so} \qquad \tau\,\Delta t = \Delta L

This is the rotational form of Newton's second law, $F = \frac{dp}{dt}$ ; $\tau\,\Delta t$ is the angular impulse.

Definition

The angular momentum of a rigid body turning about a fixed axis is $L = I\omega$ ; for a particle it is $L = mvr\sin\varphi$ . Unit: $\mathrm{kg\,m^2\,s^{-1}}$ (equivalently $\mathrm{J\,s}$ ). Its direction lies along the axis, given by the right-hand rule.

Important

The principle of conservation of angular momentum. If the resultant external torque on a system is zero, its total angular momentum stays constant:

\sum \tau_{\text{ext}} = 0 \quad \Rightarrow \quad I_1\omega_1 = I_2\omega_2

Internal forces (a skater's muscles, the tension in a string pulled from the centre) can change $I$ , but they cannot change $L$ .

Worked example 1: the spinning skater

A skater spinning at 2.0 rev/s has a moment of inertia of $3.6\,\mathrm{kg\,m^2}$ with her arms out. She pulls them in, reducing it to $1.2\,\mathrm{kg\,m^2}$ . Find her new rate of spin and compare the kinetic energies. Friction from the ice is negligible.

I_1\omega_1 = I_2\omega_2 \quad \Rightarrow \quad \omega_2 = \frac{3.6 \times 2.0}{1.2} = 6.0\,\mathrm{rev\,s^{-1}}

K_1 = \tfrac{1}{2} \times 3.6 \times (2.0 \times 2\pi)^2 = 284\,\mathrm{J}, \qquad K_2 = \tfrac{1}{2} \times 1.2 \times (6.0 \times 2\pi)^2 = 853\,\mathrm{J}

Angular momentum is conserved, but kinetic energy is NOT: it has tripled ( $\frac{K_2}{K_1} = \frac{I_1}{I_2}$ ). The extra 569 J is the work her arm muscles do pulling the masses inwards against their tendency to fly outwards.

Worked example 2: jumping onto a roundabout

The playground roundabout of lesson 2 ( $I = 169\,\mathrm{kg\,m^2}$ , radius 1.5 m) spins freely at $2.0\,\mathrm{rad\,s^{-1}}$ . A 40 kg child running straight towards the centre jumps onto its edge. Find the new angular velocity and the kinetic energy lost.

The child moves radially, so she brings no angular momentum about the axis. Treating her as a point mass at the rim:

I_2 = 168.75 + 40 \times 1.5^2 = 259\,\mathrm{kg\,m^2}

\omega_2 = \frac{168.75 \times 2.0}{258.75} = 1.30\,\mathrm{rad\,s^{-1}}

\Delta K = \tfrac{1}{2} \times 168.75 \times 2.0^2 - \tfrac{1}{2} \times 258.75 \times 1.3043^2 = 337.5 - 220.1 = 117\,\mathrm{J} \ \text{lost}

Like a completely inelastic collision, this "rotational collision" conserves angular momentum but turns kinetic energy into heat and sound.

Experiment 4.1: the rotating stool

Apparatus: a stool or platform that turns freely on a bearing, two dumbbells of 1 to 2 kg, a stopwatch, and a partner to spin you gently.

Figure 4.1: Pulling the dumbbells in reduces $I$ , so $\omega$ rises to keep $I\omega$ constant.

Sit on the stool holding the dumbbells at arm's length. A partner turns you gently and lets go.
Your partner times three revolutions. Pull the dumbbells to your chest and time three more.
Stretch your arms out again: you slow down. With the arms out $I \approx 3.5\,\mathrm{kg\,m^2}$ and in $I \approx 1.5\,\mathrm{kg\,m^2}$ (illustrative), 1.0 rev/s becomes $\frac{3.5 \times 1.0}{1.5} = 2.3\,\mathrm{rev\,s^{-1}}$ .

Warning

Safety. Spin slowly: a gentle push is enough. Keep the area around the stool clear, have a spotter standing by to steady you, keep your feet off the floor while turning and put them down gently to stop. Hold the dumbbells firmly and never let go of them while spinning. Anyone who feels dizzy stops at once and stays seated until it passes.

A particle on a string

Figure 4.2: The string pulls towards the centre, so its torque about the axis is zero and $m v r$ stays constant.

Worked example 3: pulling in the string

A 0.10 kg ball moves in a horizontal circle of radius 0.50 m at $4.0\,\mathrm{m\,s^{-1}}$ on a string through a tube (Figure 4.2). The string is pulled down until the radius is 0.25 m. Find the ball's angular momentum, its new speed and the work done in pulling the string.

L = mvr = 0.10 \times 4.0 \times 0.50 = 0.20\,\mathrm{kg\,m^2\,s^{-1}}

v_2 = \frac{L}{mr_2} = \frac{0.20}{0.10 \times 0.25} = 8.0\,\mathrm{m\,s^{-1}}

W = \Delta K = \tfrac{1}{2} \times 0.10 \times (8.0^2 - 4.0^2) = 2.4\,\mathrm{J}

Halving the radius doubles the speed. The same law explains why a planet moves fastest when it is closest to the Sun (Kepler's second law).

Worked example 4: angular impulse

In lesson 2, a child pushed the roundabout ( $I = 169\,\mathrm{kg\,m^2}$ ) with a torque of 90 N m for 5.0 s from rest. Use angular impulse to find its final angular velocity.

\Delta L = \tau\,\Delta t = 90 \times 5.0 = 450\,\mathrm{kg\,m^2\,s^{-1}}

\omega = \frac{\Delta L}{I} = \frac{450}{168.75} = 2.7\,\mathrm{rad\,s^{-1}}

This agrees with the answer from $\tau = I\alpha$ in lesson 2.

Applications

Sport: divers and gymnasts tuck to spin faster and open out to slow down before entering the water or landing; a cat twists itself by turning its front and back halves in opposite directions.
Helicopters: when the engine turns the main rotor one way, the body tends to turn the other way to keep the total angular momentum zero; the small tail rotor provides the torque that stops this.
Stability: a spinning wheel resists changes in the direction of its axis, because changing the direction of $\vec{L}$ needs a torque. This steadies a rolling bicycle wheel, a spinning top and the gyroscopes used in navigation.
Turbines and generators: the heavy rotating parts of a hydropower set carry a large angular momentum, so sudden changes in the electrical load change their speed only slowly, which helps keep the grid frequency steady.

A huge Francis turbine runner being lifted by a crane at the Grand Coulee Dam, United States, in 1947; the people show its size. Water flowing through the curved blades makes the runner turn and drive a generator. Photo: U.S. Bureau of Reclamation, public domain, via Wikimedia Commons

Warning

Common mistakes. (1) Saying kinetic energy is conserved when a skater pulls her arms in: only angular momentum is. (2) Applying conservation when there IS an external torque, for example a motor still driving the system. (3) Forgetting to add the new mass's $m r^{2}$ to $I$ after something lands on a turning body. (4) Mixing rev/s and rad/s between the two sides of an equation (either is fine if both sides use the same one).

Tip

Match the pairs: $p = mv \leftrightarrow L = I\omega$ ; $F = \frac{dp}{dt} \leftrightarrow \tau = \frac{dL}{dt}$ ; no external force $\Rightarrow p$ constant $\leftrightarrow$ no external torque $\Rightarrow L$ constant.

Angular momentum demo: platform and dumbbells (Physics Demos, Utah State University). Watch the rate of spin change as the dumbbells are pulled in and pushed out, as in Experiment 4.1.

✦

Practice

Find the angular momentum of a solid disc of mass 2.0 kg and radius 0.10 m turning at 300 rev/min.
A diver leaves the board turning at 1.5 rev/s with $I = 12\,\mathrm{kg\,m^2}$ . She tucks and reduces $I$ to $3.6\,\mathrm{kg\,m^2}$ . Find her new rate of turning.
A turntable ( $I = 0.020\,\mathrm{kg\,m^2}$ ) turns freely at $3.5\,\mathrm{rad\,s^{-1}}$ . A 0.10 kg lump of clay is dropped onto it 0.12 m from the axis and sticks. Find the new angular velocity.
A constant torque of 4.0 N m acts on a wheel for 3.0 s. By how much does its angular momentum change?
A ball on a string through a tube moves at $3.0\,\mathrm{m\,s^{-1}}$ in a circle of radius 0.60 m. The string is pulled until the radius is 0.20 m. Find the new speed.
Explain why a skater's kinetic energy increases when she pulls in her arms, although no external work is done on her.
A student claims: "The Earth spins because the Sun exerts a torque on it every day." Evaluate this claim using the ideas of this lesson.

Show answersHide answers

$I = \frac{1}{2} \times 2.0 \times 0.10^2 = 0.010\,\mathrm{kg\,m^2}$ ; $\omega = 31.4\,\mathrm{rad\,s^{-1}}$ ; $L = 0.31\,\mathrm{kg\,m^2\,s^{-1}}$ .
$\omega_2 = \frac{12 \times 1.5}{3.6} = 5.0\,\mathrm{rev\,s^{-1}}$
$I_2 = 0.020 + 0.10 \times 0.12^2 = 0.0214\,\mathrm{kg\,m^2}$ ; $\omega_2 = \frac{0.020 \times 3.5}{0.0214} = 3.3\,\mathrm{rad\,s^{-1}}$
$\Delta L = \tau\Delta t = 12\,\mathrm{kg\,m^2\,s^{-1}}$
$v_2 = \frac{v_1 r_1}{r_2} = \frac{3.0 \times 0.60}{0.20} = 9.0\,\mathrm{m\,s^{-1}}$
With $L$ constant, $K = \frac{L^2}{2I}$ , so a smaller $I$ means a larger $K$ . The energy comes from her own muscles: they do work (internal work) pulling her arms in against the outward tendency of the moving masses.
The claim is wrong. A body keeps spinning with no torque at all, because its angular momentum is conserved; the Earth has kept the spin it had when it formed. A torque would be needed to CHANGE its spin (tidal torques from the Moon and Sun do slow it very slightly).

Key points

$L = I\omega$ (rigid body), $L = mvr\sin\varphi$ (particle); $\sum\tau = \frac{dL}{dt}$ , $\tau\Delta t = \Delta L$ .
With no external torque, $I_1\omega_1 = I_2\omega_2$ ; kinetic energy need not be conserved.
Applications: skaters and divers, helicopters, gyroscopes and the steady spin of turbines.

1The spin that speeds itself up

2Angular momentum

3Torque changes angular momentum

4Experiment 4.1: the rotating stool

5A particle on a string

6Applications