A uniform rod of mass 400 g and length 1 m | vedclass

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(D) The center of mass of the rod is located at its midpoint,which is at a distance of $l/2$ from the pivot.When the rod is displaced by an angle $	h

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$1$

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(D) The center of mass of the rod is located at its midpoint,which is at a distance of $l/2$ from the pivot.When the rod is displaced by an angle $	heta = 60^o$,the center of mass rises by a vertical height $h$.From the geometry of the figure,the vertical distance of the center of mass from the pivot in the displaced position is $(l/2) \cos 	heta$.The initial vertical distance of the center of mass from the pivot is $l/2$.Therefore,the rise in height $h$ is given by:$h = \frac{l}{2} - \frac{l}{2} \cos 	heta = \frac{l}{2}(1 - \cos 	heta)$The increase in potential energy $\Delta U$ is given by $\Delta U = mgh$.Given: $m = 400 \ g = 0.4 \ kg$,$l = 1 \ m$,$	heta = 60^o$,and taking $g = 10 \ m/s^2$.Substituting the values:$\Delta U = 0.4 	imes 10 	imes \frac{1}{2}(1 - \cos 60^o)$$\Delta U = 4 	imes 0.5 	imes (1 - 0.5)$$\Delta U = 2 	imes 0.5 = 1 \ J$.

$A$ uniform rod of mass $400 \ g$ and length $1 \ m$ is pivoted at one end. It is displaced by an angle of $60^o$. The increase in its potential energy is ....... $J$.

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