# 11.2 Angular momentum  (Page 6/8)

 Page 6 / 8

Check Your Understanding Which has greater angular momentum: a solid sphere of mass m rotating at a constant angular frequency ${\omega }_{0}$ about the z -axis, or a solid cylinder of same mass and rotation rate about the z -axis?

${I}_{\text{sphere}}=\frac{2}{5}m{r}^{2},\phantom{\rule{0.5em}{0ex}}{I}_{\text{cylinder}}=\frac{1}{2}m{r}^{2}$ ; Taking the ratio of the angular momenta, we have:
$\frac{{L}_{\text{cylinder}}}{{L}_{\text{sphere}}}=\frac{{I}_{\text{cylinder}}{\omega }_{0}}{{I}_{\text{sphere}}{\omega }_{0}}=\frac{\frac{1}{2}m{r}^{2}}{\frac{2}{5}m{r}^{2}}=\frac{5}{4}$ . Thus, the cylinder has $25%$ more angular momentum. This is because the cylinder has more mass distributed farther from the axis of rotation.

## Summary

• The angular momentum $\stackrel{\to }{l}=\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{p}$ of a single particle about a designated origin is the vector product of the position vector in the given coordinate system and the particle’s linear momentum.
• The angular momentum $\stackrel{\to }{l}=\sum _{i}{\stackrel{\to }{l}}_{i}$ of a system of particles about a designated origin is the vector sum of the individual momenta of the particles that make up the system.
• The net torque on a system about a given origin is the time derivative of the angular momentum about that origin: $\frac{d\stackrel{\to }{L}}{dt}=\sum \stackrel{\to }{\tau }$ .
• A rigid rotating body has angular momentum $L=I\omega$ directed along the axis of rotation. The time derivative of the angular momentum $\frac{dL}{dt}=\sum \tau$ gives the net torque on a rigid body and is directed along the axis of rotation.

## Conceptual questions

Can you assign an angular momentum to a particle without first defining a reference point?

For a particle traveling in a straight line, are there any points about which the angular momentum is zero? Assume the line intersects the origin.

All points on the straight line will give zero angular momentum, because a vector crossed into a parallel vector is zero.

Under what conditions does a rigid body have angular momentum but not linear momentum?

If a particle is moving with respect to a chosen origin it has linear momentum. What conditions must exist for this particle’s angular momentum to be zero about the chosen origin?

The particle must be moving on a straight line that passes through the chosen origin.

If you know the velocity of a particle, can you say anything about the particle’s angular momentum?

## Problems

A 0.2-kg particle is travelling along the line $y=2.0\phantom{\rule{0.2em}{0ex}}\text{m}$ with a velocity $5.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}$ . What is the angular momentum of the particle about the origin?

A bird flies overhead from where you stand at an altitude of 300.0 m and at a speed horizontal to the ground of 20.0 m/s. The bird has a mass of 2.0 kg. The radius vector to the bird makes an angle $\theta$ with respect to the ground. The radius vector to the bird and its momentum vector lie in the xy -plane. What is the bird’s angular momentum about the point where you are standing?

The magnitude of the cross product of the radius to the bird and its momentum vector yields $rp\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta$ , which gives $r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta$ as the altitude of the bird h . The direction of the angular momentum is perpendicular to the radius and momentum vectors, which we choose arbitrarily as $\stackrel{^}{k}$ , which is in the plane of the ground:
$\stackrel{\to }{L}=\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{p}=hmv\stackrel{^}{k}=\left(300.0\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(2.0\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(20.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}\right)\stackrel{^}{k}=12,000.0\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\text{/}\text{s}\stackrel{^}{k}$

A Formula One race car with mass 750.0 kg is speeding through a course in Monaco and enters a circular turn at 220.0 km/h in the counterclockwise direction about the origin of the circle. At another part of the course, the car enters a second circular turn at 180 km/h also in the counterclockwise direction. If the radius of curvature of the first turn is 130.0 m and that of the second is 100.0 m, compare the angular momenta of the race car in each turn taken about the origin of the circular turn.

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l stands for linear displacement and l>>r
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