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Force in relation to angular momentum

by valvan1
Tags: angular, force, momentum, relation
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Oct28-10, 02:52 AM
P: 3
1. The problem statement, all variables and given/known data

so a student is sitting on a spinning stool and has a 2kg dumbbell in each hand angular velocity is 3rad/sec arms stretched out is a radius of 80cm and he pulls in his arms to 20cm. for this problem your ignoring the students weight.

from other problems i have figured out
so angular velocity to start with is 3 rad/sec
angular velocity final is 48 rad/sec

kinetic energy initial is 11.52( dont know what this dimensions is so im guessing J)
kinetic energy final is 184.32
angular momentum = 7.68

and i need to find the force required to pull one of the dumbbells in at a constant speed is equal to F=((initial angular momentum of 1 block)^2)/(4*Mass of one weight*Radius^3))

2. Relevant equations

3. The attempt at a solution
do i say Torque=F*d and Torque=I*alpha and then go Force = Ia/d? or is there another way to solve this that im not seeing ?
or do i use the change in kinetic energy equation 1/2*I*[tex]\omega[/tex]^2Final-1/2*I*[tex]\omega[/tex]^2initial = F*d
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Oct28-10, 12:20 PM
HW Helper
P: 6,202
I would think the change in kinetic energy would be equal to the work done in moving the dumbell from 80cm to 20 cm.
Oct28-10, 01:09 PM
P: 60
Yes, KE would increase when moved from 80cm to 20cm. It would increase 16 times!

The KE is gained at the expense of work done overcoming the radial inertial force (m * w^2 * r) along a distance 60 cms. w = angular velocity. As the dumbell is moving from 80 to 20, the angular velocity increases to satisfy the angular momentum conservation. So "w" at any radial distance is given by w(r) = w(R) * (R/r)^2

where w(R) is the angular velocity at radius R. If we put this inertial force and work done in moving infinitesimal distance "dr" then we get a definite integral that gives the work done to increase the KE from bringing the dumbell from 80cms to 20cms

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