The angular momentum of a wheel having a rotational inertia of 0.2 kgm² about its symmetric axis decreases from 4 to 2kgm²s^-1 in 4s. The average power of the wheel is

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Detailed Solution

$\begin{array}{l} L_{i} = {Iw}_{1} \Rightarrow w_{1} = \frac{Li}{I} \\ L_{t} = {Iw}_{2} \Rightarrow w_{2} = \frac{Lf}{I} \\ p = \frac{w}{t} = \frac{\frac{1}{2} {Iw}_{2}^{2} - \frac{1}{2} {Iw}_{1}^{2}}{t} \end{array}$

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