Calculating Torque & Work on a Flywheel of Mass 183kg & Radius 0.63m

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SUMMARY

The discussion focuses on calculating the torque and work done on a flywheel with a mass of 183 kg and a radius of 0.63 m. The required torque to accelerate the flywheel from rest to 120 rpm in 34.3 seconds is determined to be 26.65 N·m. The work done during this time was initially calculated using the formula W = torque * theta, yielding an incorrect result of 11.5 kJ. The correct approach involves applying the principle that work done equals the change in kinetic energy of the rotating mass.

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bobby3280
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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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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
 
For (b), simply use the fact that the work done equals the change of kinetic energy.
 

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