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Question

(A) The moment of inertia of a uniform thin rod of mass $m$ and length $l$ about an axis passing through one of its ends is given by $I = \frac{1}{3}m

Vedclass · Accepted Answer

$\frac{2}{3}{\pi ^2}{f^2}m{l^2}$

Vedclass · Answer

(A) The moment of inertia of a uniform thin rod of mass $m$ and length $l$ about an axis passing through one of its ends is given by $I = \frac{1}{3}ml^2$.The angular velocity $\omega$ in terms of frequency $f$ is given by $\omega = 2\pi f$.The rotational kinetic energy $K$ is given by the formula $K = \frac{1}{2}I\omega^2$.Substituting the values of $I$ and $\omega$ into the formula:$K = \frac{1}{2} 	imes (\frac{1}{3}ml^2) 	imes (2\pi f)^2$$K = \frac{1}{6}ml^2 	imes 4\pi^2 f^2$$K = \frac{2}{3}\pi^2 f^2 ml^2$.Thus,the correct option is $A$.

$A$ uniform thin rod of length $l$ and mass $m$ is suspended from one of its ends and is rotated at $f$ rotations per second. The rotational kinetic energy of the rod will be

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