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

#### Questions & Answers

when I click on the links in the topics noting shows. what should I do.
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Torque is only referred a force to rotate objects.
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thanks
Aarti
You are welcome
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thnx
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Sharath
density or relative density
Shaibu
density
Ian
Upthrust =720-34.3=685.7N mass of water displayed = 685.7/g vol of water displayed = 685.7/g/997 hence, density of man = 720/g / (685.7/g/997) =1046.6 kg/m3
1046.8
R.d=weight in air/upthrust in water =720/34.3=20.99 R.d=density of substance/density of water 20.99=x/1 x=20.99g/cm^3
Shaibu
Kg /cubic meters
Shaibu
Upthrust = 720-34.3=685.7N vol of water = 685.7/g/density of water = 685.7/g/997 so density of man = 720/g /(685.7/g/997) =1046.8 kg/m3
is there anyway i can see your calculations
Ian
Upthrust =720-34.3=685.7
Upthrust 720-34.3
=685.7N
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
=1046.8 kg/m3
so the density of water is 997
Shaibu
Yes
Okay, thanks
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
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okay i got it thanks
Ian
u welcome
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Same here, the function looks very ambiguous. please restate the question properly.
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thank you so much Sharath Kumar
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No issues at all. I love the subject and teaching it is fun. Cheers!
Sharath
cheers!
Liteboho
cheers too
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hii
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hii too
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o
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Govindsingh By   By  By Lakeima Roberts    By Qqq Qqq