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A flywheel of moment of inertia `5.0kg-m^(2)` is rotated at a speed of `10rad//s` because of the friction at the axis it comes to rest in 10s. Find the average torque of the friction. |
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Answer» `0=omega_(0)-alphat` `impliesalpha=(omega_(0))/(t)=(10)/(10)` `=1rad//s^(2)` `tau=Ialpha=5N-m` |
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