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(C) By the principle of conservation of energy,the rotational kinetic energy at the lowest point is equal to the gravitational potential energy at the

Vedclass · Accepted Answer

$\frac{L^2 \omega^2}{6g}$

Vedclass · Answer

(C) By the principle of conservation of energy,the rotational kinetic energy at the lowest point is equal to the gravitational potential energy at the maximum height.$\frac{1}{2} I \omega^2 = Mgh$$	herefore h = \frac{I \omega^2}{2Mg} \dots (i)$The moment of inertia $(I)$ of a uniform rod about an axis passing through its end is given by the parallel axis theorem:$I = I_{cm} + Md^2 = \frac{ML^2}{12} + M\left(\frac{L}{2}ight)^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2}{3} \dots (ii)$Substituting equation $(ii)$ into equation $(i)$:$h = \frac{(\frac{ML^2}{3}) \omega^2}{2Mg} = \frac{ML^2 \omega^2}{6Mg} = \frac{L^2 \omega^2}{6g}$

$A$ thin uniform metal rod of mass $M$ and length $L$ is swinging about a horizontal axis passing through its end. Its maximum angular velocity is $\omega$. Its centre of mass rises to a maximum height of (where $g$ is the acceleration due to gravity):

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