A Kathak dancer ends a piece with a run of chakkars, rapid spins on the heel of one foot. She begins a spin with her arms spread wide and turns fairly slowly. Then she draws her arms in close to her body, and she whirls round far faster. To end the spin slowly, she spreads her arms again. Nobody pushes her, and the floor is smooth under her heel. Yet her rate of spin changes at will, by a factor of two or more.
When no torque acts
The torque law of lesson 2 says \(\tau_{\text{ext}} = dL/dt\). If the net external torque about an axis is zero, then \(dL/dt = 0\), and the angular momentum about that axis cannot change. This is the law of conservation of angular momentum. It holds for any system: rigid or not, spinning or not, and whatever the internal forces do. For a body that changes shape while it spins about a fixed axis, \(L = I\omega\) stays constant, so a smaller \(I\) means a larger \(\omega\).
If the net external torque on a system about a fixed axis is zero, the total angular momentum about that axis stays constant. This holds whatever happens inside the system.
Read the condition carefully. It asks for zero external torque about the axis, not zero external force. A turntable on a vertical axle feels its weight and the push of the axle, and both are large. Both act along the axis itself, so their moment arms about the axis are zero, and they add no torque. The internal forces never matter either: they come in pairs along one line, and their torques cancel. The law can even hold about one axis while failing about another.
Conservation of angular momentum fixes the product \(I\omega\), never \(\omega\) alone. When \(I\) changes, \(\omega\) must change. A spinner who starts turning faster has not received a torque; she has changed her moment of inertia.
A skater spinning on smooth ice pulls her arms in, and she spins faster. Which quantity stays constant while she does it?
The energy is not conserved
A physics laboratory often has a stool that turns freely on a smooth bearing. A student sits on it, holding a dumbbell in each hand. Someone sets the stool turning slowly with the arms held out. Then the student pulls the dumbbells in towards the chest.
The student on the stool pulls the dumbbells in. The angular momentum stays constant. What happens to the kinetic energy of rotation?
In the figure above, the student and the stool alone have \(I_0 = 0.8\,\text{kg m}^2\). Each dumbbell has a mass of \(2.5\,\text{kg}\). With the arms out, both are \(0.8\,\text{m}\) from the axis, so \(I_1 = 0.8 + 5 \times 0.8^2 = 4.0\,\text{kg m}^2\). With the arms in, they are \(0.2\,\text{m}\) from the axis, and \(I_2 = 0.8 + 5 \times 0.2^2 = 1.0\,\text{kg m}^2\). The stool starts at \(1.5\,\text{rad/s}\), so \(L = 6\,\text{kg m}^2\text{/s}\). It ends at \(\omega_2 = 6/1.0 = 6\,\text{rad/s}\), four times as fast.
Now compare the energies. At the start \(K_1 = \tfrac12 \times 4.0 \times 1.5^2 = 4.5\,\text{J}\). At the end \(K_2 = \tfrac12 \times 1.0 \times 6^2 = 18\,\text{J}\). The kinetic energy has grown by \(13.5\,\text{J}\). At constant \(L\) this is no surprise. Since \(\omega = L/I\), the kinetic energy is \(K = \tfrac12 I\omega^2 = L^2/2I\), and a smaller \(I\) means a larger \(K\).
The extra \(13.5\,\text{J}\) is not free. Each dumbbell moves on a circle, and it needs an inward pull to keep doing so. Pulling it inward means pulling through a distance along that force, so the arms do positive work. The student's muscles spend their stored chemical energy doing it, as the chapter on Work, Kinetic Energy and Power described for internal work. Angular momentum is conserved, but the kinetic energy is bought with work.
A skater spins at \(2\,\text{rad/s}\) with her arms out, when her moment of inertia is \(3.6\,\text{kg m}^2\). She pulls her arms in, and her moment of inertia becomes \(1.2\,\text{kg m}^2\). Find her new angular velocity, in rad/s.
A spinning skater pulls her arms in. Her angular momentum stays constant, yet her kinetic energy grows. Where does the extra energy come from?
The hands and arms move on circles, so they need an inward pull to keep moving round. When she draws them inward, her muscles pull in the direction the arms move, so the muscles do positive work. That work, taken from the chemical energy stored in her muscles, becomes the extra kinetic energy. The work is internal, so it changes the kinetic energy without any external torque, and \(L\) stays the same.
A child, a clutch and an orbit
The same law works where nothing is rigid. A merry-go-round turning freely carries a child at its rim. The child walks to the centre. The push between the child's feet and the platform is internal to the system of platform and child. So the angular momentum about the axle is conserved. As the child comes inward, the moment of inertia falls, and the platform turns faster.
A merry-go-round with a moment of inertia of \(150\,\text{kg m}^2\) turns freely at \(1\,\text{rad/s}\). A \(30\,\text{kg}\) child stands at its rim, \(2\,\text{m}\) from the axle. The child walks to the centre. Find the new angular velocity.
System: platform and child. The axle's push acts along the axis, and the push of the child's feet is internal, so \(L\) about the axle is conserved.
At the rim, the child counts as a particle: \(I_1 = 150 + 30 \times 2^2 = 270\,\text{kg m}^2\). At the centre the child adds nothing, so \(I_2 = 150\,\text{kg m}^2\).
\(270 \times 1 = 150\,\omega_2\) gives \(\omega_2 = 1.8\,\text{rad/s}\). The system was never one rigid body, and the law never needed it to be.
Now lock two spinning bodies together. A disc spins on an axle, and a second disc at rest is dropped onto it, on the same axle. Friction between them makes them slip at first, and then they turn together. That friction is internal, so the angular momentum is shared and conserved. The kinetic energy is another story. While the discs slip, friction turns part of it into heat. This is the clutch of a motorcycle. It is the rotational twin of a collision in which two bodies stick, from the chapter on Collisions.
A disc with \(I_1 = 0.3\,\text{kg m}^2\) spins at \(50\,\text{rad/s}\) on a vertical axle. A disc with \(I_2 = 0.2\,\text{kg m}^2\), at rest, is dropped onto it on the same axle, and friction locks them together. Find the common angular velocity and the kinetic energy lost.
System: both discs. The friction between them is internal. Gravity and the axle's push act along the axis, so there is no external torque about it.
\(I_1\omega_1 + I_2 \times 0 = (I_1 + I_2)\,\omega\).
\(\omega = 0.3 \times 50/0.5 = 30\,\text{rad/s}\). Before, \(K = \tfrac12 \times 0.3 \times 50^2 = 375\,\text{J}\). After, \(K = \tfrac12 \times 0.5 \times 30^2 = 225\,\text{J}\).
The discs turn together at \(30\,\text{rad/s}\), and \(150\,\text{J}\) of kinetic energy, two fifths of it, became heat.
The Earth moving round the Sun obeys the same law. The Sun's pull on the Earth points straight at the Sun. So its moment arm about the Sun is zero, and it has no torque there. So the Earth's angular momentum about the Sun is constant. At the nearest point, the perihelion, and the farthest point, the aphelion, the velocity is at right angles to the line from the Sun. So \(L = mvr\) at both. With \(r = 1.47 \times 10^{11}\,\text{m}\) and \(v = 30.3\,\text{km/s}\) at the nearest point, and \(r = 1.52 \times 10^{11}\,\text{m}\) at the farthest, the speed there is \(30.3 \times 1.47/1.52 \approx 29.3\,\text{km/s}\). The chapter on Gravitational Energy and Orbits turns the same fact into Kepler's law of equal areas.
A platform with a moment of inertia of \(200\,\text{kg m}^2\) turns freely at \(1.2\,\text{rad/s}\). A \(40\,\text{kg}\) child walks from \(2\,\text{m}\) to \(1\,\text{m}\) from the axle. Treat the child as a particle. Find the new angular velocity, in rad/s.
A spinning disc and a disc at rest are locked together on one axle by friction. Which statement is correct?
The engine of a helicopter turns the big main rotor one way. By Newton's third law the rotor exerts an opposite torque on the engine, and so on the body. With nothing else acting, the body would spin opposite to the rotor, keeping the total angular momentum zero. The small tail rotor pushes sideways on air at the end of a long boom. Its large moment arm supplies the external torque that keeps the cabin pointing straight.
In 1894 the French physiologist Étienne-Jules Marey filmed a cat dropped upside down with his fast camera. The cat turned over in the air and landed on its feet. The Paris Academy of Sciences was puzzled. The cat began with no angular momentum and had nothing to push against, so how could it turn? The answer is that it changes its shape. It turns its front half one way with the legs drawn in, and its back half the other way with the legs spread. The total angular momentum stays zero throughout.
Return to the Kathak dancer. Her heel spins on a smooth floor, so the floor exerts almost no torque about her vertical axis, and her weight acts along that axis. So her angular momentum is nearly constant through each spin. With her arms spread, her moment of inertia is perhaps \(2.4\,\text{kg m}^2\); with her arms close, perhaps \(1.2\,\text{kg m}^2\). Halving \(I\) doubles her spin. The extra kinetic energy comes from her arm muscles, which pull her arms inward against their tendency to fly outward.
- Assuming \(\omega\) stays constant when no torque acts.The law fixes \(L = I\omega\). When \(I\) changes, \(\omega\) changes the other way.
- Assuming kinetic energy is conserved whenever angular momentum is.Pulling mass inward raises \(K\); locking two discs lowers it. Only \(L\) is guaranteed.
- Thinking that zero torque needs zero force.Forces along the axis, or through the chosen point, have no moment arm and so no torque.