Rotational Dynamics/moment of inertia/frictional torque

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  • #1
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Homework Statement



The combination of an applied force and a constant frictional force produces a constant total torque of 36.8 N·m on a wheel rotating about a fixed axis. The applied force acts for 5.92 s. During this time the angular speed of the wheel increases from 0 to 10.2 rad/s. The applied force is then removed, and the wheel comes to rest in 59.5 s.

a)find the moment of inertia

b)find the magnitude of the frictional torque

Homework Equations



torque=(inertia) (angular acceleration)
angular acceleration=change in angular speed/change in time


The Attempt at a Solution



for (a) I know what I have to do but for some reason I can't figure it out. I set the torque (36.8 N m) equal to the product of the moment of inertia the angular acceleration. however, I'm just stuck on finding the angular acceleration. Do I use an equation for constant angular acceleration? (e.g. omega(final)=omega(initial) + (angular acceleration)(time)) or do I have to integrate it? If so, how?
 

Answers and Replies

  • #2
674
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They said constant total torque, so that means constant angular acceleration since the inertia doesn't change.
 
  • #3
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Use the relation between angular impulse and angular momentum.
 
  • #4
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Awesome. I figured out the moment of inertia for the wheel by using the constant angular acceleration formulas, as well as the number of revolutions for the wheel.
First, I used Omega(final)=Omega(initial)+Angular Acceleration*time
Then I was able to find the moment of inertia by plugging it into
Torque=I*Angular Acceleration.

To find how many times the wheel revolved I used Omega(final)^2 - Omega(initial)^2= 2*Angular Acceleration*Total amount angle rotated

But now I just don't know how to find the magnitude of the frictional torque. Help!!!
 
  • #5
tiny-tim
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But now I just don't know how to find the magnitude of the frictional torque.
Yes you do :smile:
… 10.2 rad/s. The applied force is then removed, and the wheel comes to rest in 59.5 s.
 

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