Angular Momentum and Its Conservation – IP

Example 10.11 Calculating Angular Momentum of the Earth

Strategy

No information is given in the statement of the problem; so we must look up pertinent data before we can calculate $L=I \omega$. First, according to Figure 10.12, the formula for the moment of inertia of a sphere is

$I=\frac{2 M R^{2}}{5}$

so that

$L=I \omega=\frac{2 M R^{2} \omega}{5}$

Earth's mass $M$ is $5.979 \times 10^{24} \mathrm{~kg}$ and its radius $R$ is $R \text { is } 6.376 \times 10^{6} \mathrm{~m} \text {. }$. The Earth's angular velocity ω is, of course, exactly one revolution per day, but we must covert $\omega$ to radians per second to do the calculation in SI units.

Solution

Substituting known information into the expression for $L$  and converting $\omega$ to radians per second gives

$\begin{aligned} L &=0.4\left(5.979 \times 10^{24} \mathrm{~kg}\right)\left(6.376 \times 10^{6} \mathrm{~m}\right)^{2}\left(\frac{1 \mathrm{rev}}{\mathrm{d}}\right) \\ &=9.72 \times 10^{37} \mathrm{~kg} \cdot \mathrm{m}^{2} \cdot \mathrm{rev} / \mathrm{d} \end{aligned}$

Substituting $2π$ rad for $1$  rev and $8.64×104s$ for 1 day gives

$\begin{aligned} L &=\left(9.72 \times 10^{37} \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left(\frac{2 \pi \mathrm{rad} / \mathrm{rev}}{8.64 \times 10^{4} \mathrm{~s} / \mathrm{d}}\right)(1 \mathrm{rev} / \mathrm{d}) \\ &=7.07 \times 10^{33} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s} \end{aligned}$

Discussion

This number is large, demonstrating that Earth, as expected, has a tremendous angular momentum. The answer is approximate, because we have assumed a constant density for Earth in order to estimate its moment of inertia.

When you push a merry-go-round, spin a bike wheel, or open a door, you exert a torque. If the torque you exert is greater than opposing torques, then the rotation accelerates, and angular momentum increases. The greater the net torque, the more rapid the increase in $L$ . The relationship between torque and angular momentum is

$\text { net } \tau=\frac{\Delta L}{\Delta t}$

This expression is exactly analogous to the relationship between force and linear momentum, $F=\Delta p / \Delta t$ . The equation $\text { net } \tau=\frac{\Delta L}{\Delta t}$ is very fundamental and broadly applicable. It is, in fact, the rotational form of Newton's second law.

Example 10.13 Calculating the Torque Putting Angular Momentum Into a Lazy Susan

Figure 10.21 shows a Lazy Susan food tray being rotated by a person in quest of sustenance. Suppose the person exerts a 2.50 N force perpendicular to the lazy Susan's 0.260-m radius for 0.150 s. (a) What is the final angular momentum of the lazy Susan if it starts from rest, assuming friction is negligible? (b) What is the final angular velocity of the lazy Susan, given that its mass is 4.00 kg and assuming its moment of inertia is that of a disk?

Figure 10.21 A partygoer exerts a torque on a lazy Susan to make it rotate. The equation $\text { net } \tau=\frac{\Delta L}{\Delta t}$ gives the relationship between torque and the angular momentum produced.

Strategy

We can find the angular momentum by solving $\tau=\frac{\Delta L}{\Delta t} \text { for } \Delta L$ , and using the given information to calculate the torque. The final angular momentum equals the change in angular momentum, because the lazy Susan starts from rest. That is, $\Delta L=L$. To find the final velocity, we must calculate ω from the definition of $L$ in $L=I \omega$.

Solution for (a)

Solving $\text { net } \tau=\frac{\Delta L}{\Delta t} \text { for } \Delta L$ gives

$\Delta L=(\operatorname{net} \tau) \Delta \mathrm{t}$

Because the force is perpendicular to $r$, we see that $\text { net } \tau=r F \text {, }$, so that

$\begin{aligned} L &=\mathrm{rF} \Delta t=(0.260 \mathrm{~m})(2.50 \mathrm{~N})(0.150 \mathrm{~s}) \\ &=9.75 \times 10^{-2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s} \end{aligned}$

Solution for (b)

The final angular velocity can be calculated from the definition of angular momentum,

$L=I \omega$

Solving for $\omega$ and substituting the formula for the moment of inertia of a disk into the resulting equation gives

$\omega=\frac{L}{I}=\frac{L}{\frac{1}{2} M R^{2}}$

And substituting known values into the preceding equation yields

$\omega=\frac{9.75 \times 10^{-2} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}}{(0.500)(4.00 \mathrm{~kg})(0.260 \mathrm{~m})^{2}}=0.721 \mathrm{rad} / \mathrm{s}$

Discussion

Note that the imparted angular momentum does not depend on any property of the object but only on torque and time. The final angular velocity is equivalent to one revolution in 8.71 s (determination of the time period is left as an exercise for the reader), which is about right for a lazy Susan.

Examp[le 10.13 Calculating the Torque in a Kick

The person whose leg is shown in Figure 10.22 kicks his leg by exerting a 2000-N force with his upper leg muscle. The effective perpendicular lever arm is 2.20 cm. Given the moment of inertia of the lower leg is $1.25 \mathrm{~kg} \cdot \mathrm{m}^{2}$, (a) find the angular acceleration of the leg. (b) Neglecting the gravitational force, what is the rotational kinetic energy of the leg after it has rotated through $57.3^{\circ}$ (1.00 rad)?

Figure 10.22 The muscle in the upper leg gives the lower leg an angular acceleration and imparts rotational kinetic energy to it by exerting a torque about the knee. $F$ is a vector that is perpendicular to $r$. This example examines the situation.

Strategy

The angular acceleration can be found using the rotational analog to Newton's second law, or $\alpha=\text { net } \tau / I$ . The moment of inertia I is given and the torque can be found easily from the given force and perpendicular lever arm. Once the angular acceleration $\alpha$ is known, the final angular velocity and rotational kinetic energy can be calculated.

Solution to (a)

From the rotational analog to Newton's second law, the angular acceleration $\alpha$ is

$\alpha=\frac{\text { net } \tau}{I}$

Because the force and the perpendicular lever arm are given and the leg is vertical so that its weight does not create a torque, the net torque is thus

$\begin{aligned} \text { net } \tau &=r_{\perp} F \\ &=(0.0220 \mathrm{~m})(2000 \mathrm{~N}) \\ &=44.0 \mathrm{~N} \cdot \mathrm{m} \end{aligned}$

Substituting this value for the torque and the given value for the moment of inertia into the expression for $\alpha$ gives

$\alpha=\frac{44.0 \mathrm{~N} \cdot \mathrm{m}}{1.25 \mathrm{~kg} \cdot \mathrm{m}^{2}}=35.2 \mathrm{rad} / \mathrm{s}^{2}$

Solution to (b)

The final angular velocity can be calculated from the kinematic expression

$\omega^{2}=\omega_{0}^{2}+2 \alpha \theta$

or

$\omega^{2}=2 \alpha \theta$

because the initial angular velocity is zero. The kinetic energy of rotation is

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

so it is most convenient to use the value of $\omega^{2}$ just found and the given value for the moment of inertia. The kinetic energy is then

$\begin{aligned} \mathrm{KE}_{\mathrm{rot}} &=0.5\left(1.25 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left(70.4 \mathrm{rad}^{2} / \mathrm{s}^{2}\right) \\ &=44.0 \mathrm{~J} \end{aligned}$

Discussion

These values are reasonable for a person kicking his leg starting from the position shown. The weight of the leg can be neglected in part (a) because it exerts no torque when the center of gravity of the lower leg is directly beneath the pivot in the knee. In part (b), the force exerted by the upper leg is so large that its torque is much greater than that created by the weight of the lower leg as it rotates. The rotational kinetic energy given to the lower leg is enough that it could give a ball a significant velocity by transferring some of this energy in a kick.

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.

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,

$L=L^{\prime}$

or

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

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.