A metre stick of mass 400 g is pivoted at on | vedclass

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(D) The centre of mass of the metre stick is at its midpoint,which is at a distance of $L/2 = 0.5 \ m$ from the pivot.When the stick is displaced by a

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$1$

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(D) The centre of mass of the metre stick is at its midpoint,which is at a distance of $L/2 = 0.5 \ m$ from the pivot.When the stick is displaced by an angle $	heta = 60^o$,the vertical height $h$ by which the centre of mass rises is given by:$h = \frac{L}{2} - \frac{L}{2} \cos 	heta$Given $L = 1 \ m$ and $	heta = 60^o$:$h = 0.5 - 0.5 \cos 60^o = 0.5 - 0.5 	imes 0.5 = 0.5 - 0.25 = 0.25 \ m$The increase in potential energy $\Delta U$ is given by:$\Delta U = mgh$Given mass $m = 400 \ g = 0.4 \ kg$ and taking $g = 10 \ m/s^2$:$\Delta U = 0.4 	imes 10 	imes 0.25 = 4 	imes 0.25 = 1 \ J$.

$A$ metre stick of mass $400 \ g$ is pivoted at one end and displaced through an angle $60^o$. The increase in its potential energy is ..... $J$.

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