A thin,uniform metal rod of mass M and length | vedclass

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(A) The rod swings about a horizontal axis passing through its end. By the principle of conservation of energy,the maximum rotational kinetic energy a

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The rod swings about a horizontal axis passing through its end. By the principle of conservation of energy,the maximum rotational kinetic energy at the lowest point is equal to the maximum gravitational potential energy gained by the centre of mass at the highest point.$1$. The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^{2}$.$2$. The moment of inertia of a rod of mass $M$ and length $L$ about an axis passing through its end is $I = \frac{ML^{2}}{3}$.$3$. The potential energy gained by the centre of mass is $U = Mgh$,where $h$ is the maximum height reached by the centre of mass.$4$. Equating kinetic energy and potential energy: $Mgh = \frac{1}{2} I \omega^{2}$.$5$. Substituting the value of $I$: $Mgh = \frac{1}{2} \left( \frac{ML^{2}}{3} \right) \omega^{2}$.$6$. Simplifying the equation: $Mgh = \frac{ML^{2} \omega^{2}}{6}$.$7$. Solving for $h$: $h = \frac{L^{2} \omega^{2}}{6g}$.

$A$ 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 $(g = \text{acceleration due to gravity})$

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