Rotational motion and angular momentum
70 Angular Momentum and Its Conservation
Learning Objectives
- Understand the analogy between angular momentum and linear momentum.
- Observe the relationship between torque and angular momentum.
- Apply the law of conservation of angular momentum.
Why does Earth keep on spinning? What started it spinning to begin with? And how does an ice skater manage to spin faster and faster simply by pulling her arms in? Why does she not have to exert a torque to spin faster? Questions like these have answers based in angular momentum, the rotational analog to linear momentum.
By now the pattern is clear—every rotational phenomenon has a direct translational analog. It seems quite reasonable, then, to define angular momentum [latex]L[/latex] as
This equation is an analog to the definition of linear momentum as [latex]p=\text{mv}[/latex]. Units for linear momentum are [latex]\text{kg}\cdot \text{m}\text{/s}[/latex] while units for angular momentum are [latex]\text{kg}\cdot {\text{m}}^{2}\text{/s}[/latex]. As we would expect, an object that has a large moment of inertia [latex]I[/latex], such as Earth, has a very large angular momentum. An object that has a large angular velocity [latex]\omega[/latex], such as a centrifuge, also has a rather large angular momentum.
Making Connections
Angular momentum is completely analogous to linear momentum, first presented in Uniform Circular Motion and Gravitation. It has the same implications in terms of carrying rotation forward, and it is conserved when the net external torque is zero. Angular momentum, like linear momentum, is also a property of the atoms and subatomic particles.
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 [latex]L=\mathrm{I\omega }[/latex]. First, according to Figure 69.3, the formula for the moment of inertia of a sphere is
so that
Earth’s mass [latex]M[/latex] is [latex]5\text{.}\text{979}×{\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex] and its radius [latex]R[/latex] is [latex]6\text{.}\text{376}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m}[/latex]. The Earth’s angular velocity [latex]\omega[/latex] is, of course, exactly one revolution per day, but we must covert [latex]\omega[/latex] to radians per second to do the calculation in SI units.
Solution
Substituting known information into the expression for [latex]L[/latex] and converting [latex]\omega[/latex] to radians per second gives
Substituting [latex]2\pi[/latex] rad for [latex]1[/latex] rev and [latex]8\text{.}\text{64}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{s}[/latex] for 1 day gives
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