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Angular Momentum problem

  • Thread starter BARBARlAN
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  • #1
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Homework Statement


A student sits on a rotating stool holding two 2 kg masses. When his arms are extended horizontally, the masses are 0.78 m from the axis of rotation, and he rotates with an angular velocity of 1.7 rad/sec. The student then pulls the weights horizontally to a shorter distance 0.25 m from the rotation axis and his angular velocity increases to omega2. For simplicity, assume the student himself plus the stool he sits on have constant combined momentum of inertia I(subscript s) = 1.8kg*m^2. Find the new angular velocity omega2 of the student after he has pulled in the weights. Answer in units of rad/s.

When the student pulls the weights in, he performs mechanical work - which increases the kinetic energy of the rotating system. Calculate the increase in the kinetic energy. Answer in units of J.


Homework Equations


I1*Omega1=I2*Omega2
KE=1/2*m*v^2+1/2*I*omega^2


The Attempt at a Solution


I'm not sure of I*omega is the right formula, because it seems the moment of inertia stays constant throughout. So I'm stuck.
 

Answers and Replies

  • #2
gneill
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I'm not sure of I*omega is the right formula, because it seems the moment of inertia stays constant throughout. So I'm stuck.
What's the moment of inertia of a 2kg mass at a radius of 0.78 m from the axis of rotation?
How about when it is at 0.25 m?
 
  • #3
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1.2168 kg*m^2 at .78m, .125 kg*m^2 at .25m. How does this help though when I don't know the mass of the stool or the person?
 
  • #4
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ok i got part 1 right. any tips on how to do part 2?
 
  • #5
gneill
Mentor
20,793
2,773
1.2168 kg*m^2 at .78m, .125 kg*m^2 at .25m. How does this help though when I don't know the mass of the stool or the person?
You're given their moment of inertia. For angular motion this plays the same roll as mass.

Try to determine the rotational kinetic energy for the first situation: weights held at 0.78m and rotation rate 1.7 rad/sec. Show your work!
 
  • #6
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ok I got it right! Thank you all very much :)
 

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