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.