The angular momentum of a body rotating about | vedclass

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(C) The rotational kinetic energy $K$ is related to angular momentum $L$ and moment of inertia $I$ by the formula: $K = \frac{L^2}{2I}$.Since the mome

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$21$

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(C) The rotational kinetic energy $K$ is related to angular momentum $L$ and moment of inertia $I$ by the formula: $K = \frac{L^2}{2I}$.Since the moment of inertia $I$ is constant for a body rotating about a fixed axis, we have $K \propto L^2$.Let the initial angular momentum be $L_1 = L$ and the final angular momentum be $L_2 = L + 0.10L = 1.10L$.The ratio of kinetic energies is $\frac{K_2}{K_1} = \left( \frac{L_2}{L_1} ight)^2 = (1.10)^2 = 1.21$.This means $K_2 = 1.21 K_1$.The percentage increase in kinetic energy is given by $\frac{K_2 - K_1}{K_1} 	imes 100\% = (1.21 - 1) 	imes 100\% = 0.21 	imes 100\% = 21\%$.

The angular momentum of a body rotating about a fixed axis is increased by $10\%$. What is the percentage increase in its rotational kinetic energy (in $\%$)?

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