A rod of length L revolves in a horizontal pl | vedclass

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(C) The rotational kinetic energy of a rod is given by the formula:$K_{rot} = \frac{1}{2} I \omega^2$For a rod of mass $M$ and length $L$ rotating abo

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$\frac{1}{24} A L^3 \rho \omega^2$

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(C) The rotational kinetic energy of a rod is given by the formula:$K_{rot} = \frac{1}{2} I \omega^2$For a rod of mass $M$ and length $L$ rotating about an axis passing through its center and perpendicular to its length, the moment of inertia is:$I = \frac{M L^2}{12}$Substituting this into the kinetic energy formula:$K_{rot} = \frac{1}{2} \left( \frac{M L^2}{12} ight) \omega^2 = \frac{1}{24} M L^2 \omega^2$  $(i)$The mass $M$ of the rod can be expressed in terms of its volume and density:$M = 	ext{Volume} 	imes 	ext{Density} = (A 	imes L) 	imes ho = A L ho$  $(ii)$Substituting equation $(ii)$ into equation $(i)$:$K_{rot} = \frac{1}{24} (A L ho) L^2 \omega^2 = \frac{1}{24} A L^3 ho \omega^2$

$A$ rod of length $L$ revolves in a horizontal plane about an axis passing through its centre and perpendicular to its length. The angular velocity of the rod is $\omega$. If $A$ is the area of cross-section of the rod and $\rho$ is its density, then the rotational kinetic energy of the rod is

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