A thin rod of length L is suspended from one | vedclass

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(C) The rotational kinetic energy is given by $K_{rot} = \frac{1}{2} I \omega^2$.For a thin rod of mass $m$ and length $L$ rotating about one end,the

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$\frac{2}{3} \ mL^2 \pi^2 n^2$

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(C) The rotational kinetic energy is given by $K_{rot} = \frac{1}{2} I \omega^2$.For a thin rod of mass $m$ and length $L$ rotating about one end,the moment of inertia is $I = \frac{mL^2}{3}$.The angular velocity $\omega$ in terms of revolutions per second $n$ is $\omega = 2 \pi n$.Substituting these values into the formula:$K_{rot} = \frac{1}{2} \left( \frac{mL^2}{3} \right) (2 \pi n)^2$$K_{rot} = \frac{1}{2} \left( \frac{mL^2}{3} \right) (4 \pi^2 n^2)$$K_{rot} = \frac{2}{3} mL^2 \pi^2 n^2$.

$A$ thin rod of length $L$ is suspended from one end and rotates with $n$ revolutions per second. What will be the rotational kinetic energy of the rod?

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