A stick of mass m and length l is pivoted at | vedclass

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Question

(A) The potential energy of the stick is associated with its center of mass.Initially,the center of mass of the stick is at a distance $h_1 = \frac{l}

Vedclass · Accepted Answer

$m g \frac{l}{2}(1-\cos 	heta)$

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(A) The potential energy of the stick is associated with its center of mass.Initially,the center of mass of the stick is at a distance $h_1 = \frac{l}{2}$ from the pivot point.When the stick is displaced by an angle $	heta$ from the vertical,the vertical distance of the center of mass from the pivot point becomes $h_2 = \frac{l}{2} \cos 	heta$.The change in the height of the center of mass is $\Delta h = h_1 - h_2 = \frac{l}{2} - \frac{l}{2} \cos 	heta = \frac{l}{2}(1 - \cos 	heta)$.The increase in potential energy is given by $\Delta U = mg \Delta h = mg \frac{l}{2}(1 - \cos 	heta)$.

$A$ stick of mass $m$ and length $l$ is pivoted at one end and is displaced through an angle $\theta$. The increase in potential energy is ............

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