• # question_answer A torque of SON m is acted on a 5kg wheel of moment of inertia $2\text{ }kg\text{ }{{m}^{2}}$for 10 second. Then the angle rotated by wheel in 10 second                      A)  750 radB)  1500 rad               C)  3000 rad          D)  6000 rad

Solution :

$\alpha =\frac{\tau }{I}=\frac{30}{2}=15rad/{{\sec }^{2}}$ Now $\theta =0+\frac{1}{2}\alpha {{t}^{2}}$ $=\frac{1}{2}\times 15\times {{(10)}^{2}}=750\,\,rad$

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