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Work done by wheel.

 
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Oct31-06, 06:15 PM   #1
 

Work done by wheel.


A flywheel of mass 183 kg has an effective radius of 0.63 m (assume the mass is concentrated along a circumference located at the effective radius of the flywheel).

(a) What torque is required to bring this wheel from rest to a speed of 120 rpm in a time interval of 34.3 s?

(b) How much work is done during the 34.3 s?

Alright I figured out (a) using the moment of Inertia and the angular acceleration to be 26.65 N * m

So for part (b) I tried using W = torque * theta
to find theta i used many ways all coming up with right around 432 rad
so W = 26.65 * 432
= 11.5 kj
But this answer isn't correct any suggestions as to where i went wrong??
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Oct31-06, 06:20 PM   #2
 
All the work done on the flywheel is converted to rotational energy. Do you know an equation that will give you the energy of a rotating mass?

Hope this helps,
Sam
Oct31-06, 06:30 PM   #3
 
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For (b), simply use the fact that the work done equals the change of kinetic energy.
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