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Work done by wheel. |
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| Oct31-06, 06:15 PM | #1 |
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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?? |
| Oct31-06, 06:20 PM | #2 |
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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 |
| Oct31-06, 06:30 PM | #3 |
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Recognitions:
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For (b), simply use the fact that the work done equals the change of kinetic energy.
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