A wheel whose moment of inertia is 0.03kgm2 is accelerated from rest to 20rad/s in 5s. When the external torque is removed, the wheel stops in 1 min. Find x, when external torque is x×10−2Nm.
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Solution
∴τft=Iω or τf=Iωt=0.03×2060=0.01N.m
During acceleration (τe−τf)t=Iω ∴τe=Iωt+τf=0.03×205+0.01 =13×10−2Nm