A slender uniform rod of mass M and length l | vedclass

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Question

(A) The torque $	au$ acting on the rod about the pivot point is due to the gravitational force $Mg$ acting at the center of mass of the rod,which is

Vedclass · Accepted Answer

$\frac{3 g}{2 l} \sin 	heta$

Vedclass · Answer

(A) The torque $	au$ acting on the rod about the pivot point is due to the gravitational force $Mg$ acting at the center of mass of the rod,which is at a distance $l/2$ from the pivot.When the rod makes an angle $	heta$ with the vertical,the perpendicular distance from the pivot to the line of action of the weight is $(l/2) \sin 	heta$.Therefore,the torque is $	au = Mg \cdot (l/2) \sin 	heta$.The moment of inertia $I$ of a uniform rod of mass $M$ and length $l$ about an axis passing through one end is $I = \frac{Ml^2}{3}$.Using the rotational analog of Newton's second law,$	au = I \alpha$,where $\alpha$ is the angular acceleration:$Mg \frac{l}{2} \sin 	heta = \left( \frac{Ml^2}{3} ight) \alpha$Solving for $\alpha$:$\alpha = \frac{Mg (l/2) \sin 	heta}{Ml^2 / 3} = \frac{Mg l \sin 	heta}{2} \cdot \frac{3}{Ml^2} = \frac{3g \sin 	heta}{2l}$.

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