# 11.2 Angular momentum  (Page 2/8)

 Page 2 / 8

Here we have used the definition of $\stackrel{\to }{p}$ and the fact that a vector crossed into itself is zero. From Newton’s second law, $\frac{d\stackrel{\to }{p}}{dt}=\sum \stackrel{\to }{F},$ the net force acting on the particle, and the definition of the net torque, we can write

$\frac{d\stackrel{\to }{l}}{dt}=\sum \stackrel{\to }{\tau }.$

Note the similarity with the linear result of Newton’s second law, $\frac{d\stackrel{\to }{p}}{dt}=\sum \stackrel{\to }{F}$ . The following problem-solving strategy can serve as a guideline for calculating the angular momentum of a particle.

## Problem-solving strategy: angular momentum of a particle

1. Choose a coordinate system about which the angular momentum is to be calculated.
2. Write down the radius vector to the point particle in unit vector notation.
3. Write the linear momentum vector of the particle in unit vector notation.
4. Take the cross product $\stackrel{\to }{l}=\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{p}$ and use the right-hand rule to establish the direction of the angular momentum vector.
5. See if there is a time dependence in the expression of the angular momentum vector. If there is, then a torque exists about the origin, and use $\frac{d\stackrel{\to }{l}}{dt}=\sum \stackrel{\to }{\tau }$ to calculate the torque. If there is no time dependence in the expression for the angular momentum, then the net torque is zero.

## Angular momentum and torque on a meteor

A meteor enters Earth’s atmosphere ( [link] ) and is observed by someone on the ground before it burns up in the atmosphere. The vector $\stackrel{\to }{r}=25\phantom{\rule{0.2em}{0ex}}\text{km}\stackrel{^}{i}+25\phantom{\rule{0.2em}{0ex}}\text{km}\stackrel{^}{j}$ gives the position of the meteor with respect to the observer. At the instant the observer sees the meteor, it has linear momentum $\stackrel{\to }{p}=15.0\phantom{\rule{0.2em}{0ex}}\text{kg}\left(-2.0\text{km}\text{/}\text{s}\stackrel{^}{j}\right)$ , and it is accelerating at a constant $2.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}{\text{s}}^{2}\left(\text{−}\stackrel{^}{j}\right)$ along its path, which for our purposes can be taken as a straight line. (a) What is the angular momentum of the meteor about the origin, which is at the location of the observer? (b) What is the torque on the meteor about the origin? An observer on the ground sees a meteor at position r → with linear momentum p → .

## Strategy

We resolve the acceleration into x - and y -components and use the kinematic equations to express the velocity as a function of acceleration and time. We insert these expressions into the linear momentum and then calculate the angular momentum using the cross-product. Since the position and momentum vectors are in the xy -plane, we expect the angular momentum vector to be along the z -axis. To find the torque, we take the time derivative of the angular momentum.

## Solution

The meteor is entering Earth’s atmosphere at an angle of $90.0\text{°}$ below the horizontal, so the components of the acceleration in the x - and y -directions are

${a}_{x}=0,\phantom{\rule{0.5em}{0ex}}{a}_{y}=-2.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}{\text{s}}^{2}.$

We write the velocities using the kinematic equations.

${v}_{x}=0,\phantom{\rule{0.5em}{0ex}}{v}_{y}=-2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}-\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}{\text{s}}^{2}\right)t.$
1. The angular momentum is
$\begin{array}{cc}\hfill \stackrel{\to }{l}& =\stackrel{\to }{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{p}=\left(25.0\phantom{\rule{0.2em}{0ex}}\text{km}\stackrel{^}{i}+25.0\phantom{\rule{0.2em}{0ex}}\text{km}\stackrel{^}{j}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}15.0\phantom{\rule{0.2em}{0ex}}\text{kg}\left(0\stackrel{^}{i}+{v}_{y}\stackrel{^}{j}\right)\hfill \\ & =15.0\phantom{\rule{0.2em}{0ex}}\text{kg}\left[25.0\phantom{\rule{0.2em}{0ex}}\text{km}\left({v}_{y}\right)\stackrel{^}{k}\right]\hfill \\ & =15.0\phantom{\rule{0.2em}{0ex}}\text{kg[}2.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{m}\left(-2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}-\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}{\text{s}}^{2}\right)t\right)\stackrel{^}{k}\right].\hfill \end{array}$

At $t=0$ , the angular momentum of the meteor about the origin is
${\stackrel{\to }{l}}_{0}=15.0\phantom{\rule{0.2em}{0ex}}\text{kg}\left[2.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{m}\left(-2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}\right)\stackrel{^}{k}\right]=7.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\text{/}\text{s}\left(\text{−}\stackrel{^}{k}\right).$

This is the instant that the observer sees the meteor.
2. To find the torque, we take the time derivative of the angular momentum. Taking the time derivative of $\stackrel{\to }{l}$ as a function of time, which is the second equation immediately above, we have
$\frac{d\stackrel{\to }{l}}{dt}=-15.0\phantom{\rule{0.2em}{0ex}}\text{kg}\left(2.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}{\text{s}}^{2}\right)\stackrel{^}{k}.$

Then, since $\frac{d\stackrel{\to }{l}}{dt}=\sum \stackrel{\to }{\tau }$ , we have
$\sum \stackrel{\to }{\tau }=-7.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\text{N}·\text{m}\stackrel{^}{k}.$

The units of torque are given as newton-meters, not to be confused with joules. As a check, we note that the lever arm is the x -component of the vector $\stackrel{\to }{\text{r}}$ in [link] since it is perpendicular to the force acting on the meteor, which is along its path. By Newton’s second law, this force is
$\stackrel{\to }{F}=ma\left(\text{−}\stackrel{^}{j}\right)=15.0\phantom{\rule{0.2em}{0ex}}\text{kg}\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}{\text{s}}^{2}\right)\left(\text{−}\stackrel{^}{j}\right)=30.0\phantom{\rule{0.2em}{0ex}}\text{kg}·\text{m}\text{/}{\text{s}}^{2}\left(\text{−}\stackrel{^}{j}\right).$

The lever arm is
${\stackrel{\to }{r}}_{\perp }=2.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{m}\phantom{\rule{0.2em}{0ex}}\stackrel{^}{i}.$

Thus, the torque is
$\begin{array}{cc}\hfill \sum \stackrel{\to }{\tau }={\stackrel{\to }{r}}_{\perp }\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{F}& =\left(2.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{m}\phantom{\rule{0.2em}{0ex}}\stackrel{^}{i}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(-30.0\phantom{\rule{0.2em}{0ex}}\text{kg}·\text{m}\text{/}{\text{s}}^{2}\stackrel{^}{j}\right),\hfill \\ & =7.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}\left(\text{−}\stackrel{^}{k}\right).\hfill \end{array}$

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