A uniform rod of length l,hinged at the lower | vedclass

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Question

(A) The torque $	au$ acting on the rod about the hinge is given by $	au = mg \cdot r_{\perp}$,where $r_{\perp}$ is the perpendicular distance from t

Vedclass · Accepted Answer

$\frac{3g}{2\sqrt{2}l}$

Vedclass · Answer

(A) The torque $	au$ acting on the rod about the hinge is given by $	au = mg \cdot r_{\perp}$,where $r_{\perp}$ is the perpendicular distance from the hinge to the line of action of gravity.When the rod makes an angle $	heta$ with the horizontal,the angle with the vertical is $(90^{\circ} - 	heta)$. The center of mass is at $l/2$ from the hinge.The perpendicular distance is $r_{\perp} = \frac{l}{2} \cos(90^{\circ} - 	heta) = \frac{l}{2} \sin 	heta$.Using the rotational analog of Newton's second law,$	au = I\alpha$,where $\alpha$ is the angular acceleration.$mg \left( \frac{l}{2} \sin 	heta ight) = I \alpha$.Given $I = \frac{ml^2}{3}$,we have:$mg \frac{l}{2} \sin 	heta = \left( \frac{ml^2}{3} ight) \alpha$.Solving for $\alpha$:$\alpha = \frac{mg \frac{l}{2} \sin 	heta}{\frac{ml^2}{3}} = \frac{3g \sin 	heta}{2l}$.For $	heta = 45^{\circ}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$.$\alpha = \frac{3g}{2l} \cdot \frac{1}{\sqrt{2}} = \frac{3g}{2\sqrt{2}l}$.

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