A simple pendulum of length ell is displaced | vedclass

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(A) For a simple pendulum of length $\ell$, the angular velocity $\omega_1$ at an angle $	heta$ below the horizontal is derived from energy conservat

Vedclass · Accepted Answer

$\frac{3 \ell}{2}$

Vedclass · Answer

(A) For a simple pendulum of length $\ell$, the angular velocity $\omega_1$ at an angle $	heta$ below the horizontal is derived from energy conservation: $mg(\ell \sin 	heta) = \frac{1}{2} m v^2 = \frac{1}{2} m (\omega_1 \ell)^2$. Thus, $\omega_1 = \sqrt{\frac{2g \sin 	heta}{\ell}}$.For a uniform bar of length $L$ pivoted at one end, the moment of inertia is $I = \frac{1}{3} m L^2$. The potential energy lost when it rotates by angle $	heta$ is $mg(\frac{L}{2} \sin 	heta)$. By energy conservation: $mg(\frac{L}{2} \sin 	heta) = \frac{1}{2} I \omega_2^2 = \frac{1}{2} (\frac{1}{3} m L^2) \omega_2^2$. Simplifying gives $\omega_2 = \sqrt{\frac{3g \sin 	heta}{L}}$.Since the motions are synchronous, $\omega_1 = \omega_2$, so $\frac{2}{\ell} = \frac{3}{L}$.Therefore, $L = \frac{3}{2} \ell$.

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