Angular Momentum and Its Conservation – IP

Read this to understand angular momentum, how torque plays a role, and how angular momentum is conserved without torque.

Conservation of Angular Momentum

Making Connections: Conservation Laws

Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero.

We can now understand why Earth keeps on spinning. As we saw in the previous example, $\Delta L=(\text { net } \tau) \Delta t$ . This equation means that, to change angular momentum, a torque must act over some period of time. Because Earth has a large angular momentum, a large torque acting over a long time is needed to change its rate of spin. So what external torques are there? Tidal friction exerts torque that is slowing Earth's rotation, but tens of millions of years must pass before the change is very significant. Recent research indicates the length of the day was 18 h some 900 million years ago. Only the tides exert significant retarding torques on Earth, and so it will continue to spin, although ever more slowly, for many billions of years.

What we have here is, in fact, another conservation law. If the net torque is zero, then angular momentum is constant or conserved. We can see this rigorously by considering $\text { net } \tau=\frac{\Delta L}{\Delta t}$ for the situation in which the net torque is zero. In that case,

$\operatorname{net} \tau=0$

implying that

$\frac{\Delta L}{\Delta t}=0$ .

If the change in angular momentum $\Delta L$ is zero, then the angular momentum is constant; thus,

$L=\text { constant }(\text { net } \tau=0)$

$L=L^{\prime}(\operatorname{net} \tau=0)$ .

These expressions are the law of conservation of angular momentum. Conservation laws are as scarce as they are important.

An example of conservation of angular momentum is seen in Figure 10.23, in which an ice skater is executing a spin. The net torque on her is very close to zero, because there is relatively little friction between her skates and the ice and because the friction is exerted very close to the pivot point. (Both $F$ and $r$ are small, and so $\tau$ is negligibly small.) Consequently, she can spin for quite some time. She can do something else, too. She can increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that

$L=L^{\prime} \text {. }$

Expressing this equation in terms of the moment of inertia,

$I \omega=I^{\prime} \omega^{\prime}$ ,

where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because $I^{\prime}$ is smaller, the angular velocity $\omega^{\prime}$ must increase to keep the angular momentum constant. The change can be dramatic, as the following example shows.

The image a shows an ice skater spinning on the tip of her skate with both her arms and one leg extended. The image b shows the ice skater spinning on the tip of one skate, with her arms crossed and one leg supported on another.

Figure 10.23 (a) An ice skater is spinning on the tip of her skate with her arms extended. Her angular momentum is conserved because the net torque on her is negligibly small. In the next image, her rate of spin increases greatly when she pulls in her arms, decreasing her moment of inertia. The work she does to pull in her arms results in an increase in rotational kinetic energy.

Example 10.14 Calculating the Angular Momentum of a Spinning Skater

Suppose an ice skater, such as the one in Figure 10.23, is spinning at 0.800 rev/ s with her arms extended. She has a moment of inertia of $2.34 \mathrm{~kg} \cdot \mathrm{m}^{2}$ with her arms extended and of $0.363 \mathrm{~kg} \cdot \mathrm{m}^{2}$ with her arms close to her body. (These moments of inertia are based on reasonable assumptions about a 60.0-kg skater.) (a) What is her angular velocity in revolutions per second after she pulls in her arms? (b) What is her rotational kinetic energy before and after she does this?

Strategy

In the first part of the problem, we are looking for the skater's angular velocity $\omega^{\prime}$ after she has pulled in her arms. To find this quantity, we use the conservation of angular momentum and note that the moments of inertia and initial angular velocity are given. To find the initial and final kinetic energies, we use the definition of rotational kinetic energy given by

$\mathrm{KE}_{\mathrm{rot}}=\frac{1}{2} I \omega^{2}$

Solution for (a)

Because torque is negligible (as discussed above), the conservation of angular momentum given in $I \omega=I^{\prime} \omega^{\prime}$ is applicable. Thus,

$L=L^{\prime}$

$I \omega=I^{\prime} \omega^{\prime}$

Solving for $\omega^{\prime}$ and substituting known values into the resulting equation gives

$\begin{aligned} \omega^{\prime} &=\frac{I}{I} \omega=\left(\frac{2.34 \mathrm{~kg} \cdot \mathrm{m}^{2}}{0.363 \mathrm{~kg} \cdot \mathrm{m}^{2}}\right)(0.800 \mathrm{rev} / \mathrm{s}) \\ &=5.16 \mathrm{rev} / \mathrm{s} \end{aligned}$

Solution for (b)

Rotational kinetic energy is given by

$\mathrm{KE}_{\mathrm{rot}}=\frac{1}{2} I \omega^{2}$

The initial value is found by substituting known values into the equation and converting the angular velocity to rad/s:

$\begin{aligned} \mathrm{KE}_{\mathrm{rot}} &=(0.5)\left(2.34 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)((0.800 \mathrm{rev} / \mathrm{s})(2 \pi \mathrm{rad} / \mathrm{rev}))^{2} \\ &=29.6 \mathrm{~J} \end{aligned}$

The final rotational kinetic energy is

$\mathrm{KE}_{\mathrm{rot}}^{\prime}=\frac{1}{2} I^{\prime} \omega^{2}$

Substituting known values into this equation gives

$\begin{aligned} K E_{\mathrm{rot}}^{\prime} &=(0.5)\left(0.363 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)[(5.16 \mathrm{rev} / \mathrm{s})(2 \pi \mathrm{rad} / \mathrm{rev})]^{2} \\ &=191 \mathrm{~J} \end{aligned}$

Discussion

In both parts, there is an impressive increase. First, the final angular velocity is large, although most world-class skaters can achieve spin rates about this great. Second, the final kinetic energy is much greater than the initial kinetic energy. The increase in rotational kinetic energy comes from work done by the skater in pulling in her arms. This work is internal work that depletes some of the skater's food energy.

There are several other examples of objects that increase their rate of spin because something reduced their moment of inertia. Tornadoes are one example. Storm systems that create tornadoes are slowly rotating. When the radius of rotation narrows, even in a local region, angular velocity increases, sometimes to the furious level of a tornado. Earth is another example. Our planet was born from a huge cloud of gas and dust, the rotation of which came from turbulence in an even larger cloud. Gravitational forces caused the cloud to contract, and the rotation rate increased as a result. (See Figure 10.24.)

Figure 10.24 The Solar System coalesced from a cloud of gas and dust that was originally rotating. The orbital motions and spins of the planets are in the same direction as the original spin and conserve the angular momentum of the parent cloud.

In case of human motion, one would not expect angular momentum to be conserved when a body interacts with the environment as its foot pushes off the ground. Astronauts floating in space aboard the International Space Station have no angular momentum relative to the inside of the ship if they are motionless. Their bodies will continue to have this zero value no matter how they twist about as long as they do not give themselves a push off the side of the vessel.

Check Your Understanding

Is angular momentum completely analogous to linear momentum? What, if any, are their differences?

Solution

Yes, angular and linear momentums are completely analogous. While they are exact analogs they have different units and are not directly inter-convertible like forms of energy are.