# 11.2 Angular momentum  (Page 5/8)

 Page 5 / 8

## Angular momentum of a robot arm

A robot arm on a Mars rover like Curiosity shown in [link] is 1.0 m long and has forceps at the free end to pick up rocks. The mass of the arm is 2.0 kg and the mass of the forceps is 1.0 kg. See [link] . The robot arm and forceps move from rest to $\omega =0.1\pi \phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}$ in 0.1 s. It rotates down and picks up a Mars rock that has mass 1.5 kg. The axis of rotation is the point where the robot arm connects to the rover. (a) What is the angular momentum of the robot arm by itself about the axis of rotation after 0.1 s when the arm has stopped accelerating? (b) What is the angular momentum of the robot arm when it has the Mars rock in its forceps and is rotating upwards? (c) When the arm does not have a rock in the forceps, what is the torque about the point where the arm connects to the rover when it is accelerating from rest to its final angular velocity?

## Strategy

We use [link] to find angular momentum in the various configurations. When the arm is rotating downward, the right-hand rule gives the angular momentum vector directed out of the page, which we will call the positive z -direction. When the arm is rotating upward, the right-hand rule gives the direction of the angular momentum vector into the page or in the negative z- direction. The moment of inertia is the sum of the individual moments of inertia. The arm can be approximated with a solid rod, and the forceps and Mars rock can be approximated as point masses located at a distance of 1 m from the origin. For part (c), we use Newton’s second law of motion for rotation to find the torque on the robot arm.

## Solution

1. Writing down the individual moments of inertia, we have
Robot arm: ${I}_{\text{R}}=\frac{1}{3}{m}_{\text{R}}{r}^{2}=\frac{1}{3}\left(2.00\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(1.00\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}=\frac{2}{3}\phantom{\rule{0.1em}{0ex}}\text{kg}·{\text{m}}^{2}.$
Forceps: ${I}_{\text{F}}={m}_{\text{F}}{r}^{2}=\left(1.0\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(1.0\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}=1.0\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}.$
Mars rock: ${I}_{\text{MR}}={m}_{\text{MR}}{r}^{2}=\left(1.5\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(1.0\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}=1.5\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}.$
Therefore, without the Mars rock, the total moment of inertia is
${I}_{\text{Total}}={I}_{\text{R}}+{I}_{\text{F}}=1.67\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}$
and the magnitude of the angular momentum is
$L=I\omega =1.67\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\left(0.1\pi \phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}\right)=0.17\pi \phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\text{/}\text{s}.$

The angular momentum vector is directed out of the page in the $\stackrel{^}{k}$ direction since the robot arm is rotating counterclockwise.
2. We must include the Mars rock in the calculation of the moment of inertia, so we have
${I}_{\text{Total}}={I}_{\text{R}}+{I}_{\text{F}}+{I}_{\text{MR}}=3.17\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}$

and
$L=I\omega =3.17\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\left(0.1\pi \phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}\right)=0.32\pi \phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\text{/}\text{s}\text{.}$

Now the angular momentum vector is directed into the page in the $\text{−}\stackrel{^}{k}$ direction, by the right-hand rule, since the robot arm is now rotating clockwise.
3. We find the torque when the arm does not have the rock by taking the derivative of the angular momentum using [link] $\frac{d\stackrel{\to }{L}}{dt}=\sum \stackrel{\to }{\tau }.$ But since $L=I\omega$ , and understanding that the direction of the angular momentum and torque vectors are along the axis of rotation, we can suppress the vector notation and find
$\frac{dL}{dt}=\frac{d\left(I\omega \right)}{dt}=I\frac{d\omega }{dt}=I\alpha =\sum \tau ,$

which is Newton’s second law for rotation. Since $\alpha =\frac{0.1\pi \phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}}{0.1\phantom{\rule{0.2em}{0ex}}\text{s}}=\pi \phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}{\text{s}}^{2}$ , we can calculate the net torque:
$\sum \tau =I\alpha =1.67\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\left(\pi \phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}{\text{s}}^{2}\right)=1.67\pi \phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}.$

## Significance

The angular momentum in (a) is less than that of (b) due to the fact that the moment of inertia in (b) is greater than (a), while the angular velocity is the same.

V=E½-P-½ where v; velocity, P; density and E; constant. Find dimension and it's units of E (constant)
ML-3
LAWAL
derivation of simple harmonic equation
if an equation is dimensionally correct does this mean that equation must be true?
how do I calculate angular velocity
w=vr where w, angular velocity. v; velocity and r; radius of a circle
michael
sorry I meant Maximum positive angular velocity of
Priscilla
Can any one give me the definition for Bending moment plz...
I need a question for moment
what is charge
An attribution of particle that we have thought about to explain certain things like Electomagnetism
Nikunj
please what is the formula instantaneous velocity in projectile motion
A computer is reading from a CD-ROM that rotates at 780 revolutions per minute.What is the centripetal acceleration at a point that is 0.030m from the center of the disc?
change revolution per minute by multiplying from 2pie and devide by 60.and take r=.030 and use formula centripital acceleration =omega sqare r.
Kumar
OK thank you
Rapqueen
observation of body boulded
a gas is compressed to 1/10 0f its original volume.calculate the rise temperature if the original volume is 400k. gamma =1.4
the specific heat of hydrogen at constant pressure and temperature is 14.16kj|k.if 0.8kg of hydrogen is heated from 55 degree Celsius to 80 degree Celsius of a constant pressure. find the external work done .
Celine
hi
shaik
hy
Prasanna
g
what is imaginary mass and how we express is
what is imaginary mass how we express it
Yash
centre of mass is also called as imaginary mass
Lokmani
l'm from Algeria and fell these can help me
Many amusement parks have rides that make vertical loops like the one shown below. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if: (a) The car goes over the top at faster than this speed? (b) The car goes over the top at slower than this speed?
how can I convert mile to meter per hour
1 mile * 1609m
Boon
hey can someone show me how to solve the - "Hanging from the ceiling over a baby bed ...." question
i wanted to know the steps
Shrushti
sorry shrushti..
Rashid
which question please write it briefly
Asutosh