A uniform rod of length 2L is placed with one | vedclass

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(A) Let the mass of the rod be $m$ and its length be $l = 2L$.The rod rotates about the end in contact with the ground. The moment of inertia of the r

Vedclass · Accepted Answer

$\omega = \sqrt{\frac{3g \sin \alpha}{2L}}$

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(A) Let the mass of the rod be $m$ and its length be $l = 2L$.The rod rotates about the end in contact with the ground. The moment of inertia of the rod about this end is $I = \frac{m l^2}{3} = \frac{m(2L)^2}{3} = \frac{4mL^2}{3}$.Using the principle of conservation of mechanical energy,the loss in gravitational potential energy equals the gain in rotational kinetic energy.The initial height of the center of mass is $h = \frac{l}{2} \sin \alpha = L \sin \alpha$.Loss in potential energy = $mgh = mg(L \sin \alpha)$.Gain in rotational kinetic energy = $\frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{4mL^2}{3} \right) \omega^2 = \frac{2mL^2}{3} \omega^2$.Equating the two: $mgL \sin \alpha = \frac{2mL^2}{3} \omega^2$.Solving for $\omega$: $\omega^2 = \frac{3g \sin \alpha}{2L} \Rightarrow \omega = \sqrt{\frac{3g \sin \alpha}{2L}}$.

$A$ uniform rod of length $2L$ is placed with one end in contact with a horizontal surface. The other end is released from an angle $\alpha$ with the horizontal,such that the end in contact does not slip. What will be its angular velocity when it becomes horizontal?

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