A thin uniform rod of mass m and length L is | vedclass

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Question

(A) The rod is pivoted at one end. The moment of inertia of the rod about the pivot is $I = \frac{mL^2}{3}$.The gravitational force $mg$ acts at the c

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The rod is pivoted at one end. The moment of inertia of the rod about the pivot is $I = \frac{mL^2}{3}$.The gravitational force $mg$ acts at the center of mass of the rod,which is at a distance $L/2$ from the pivot.The torque $	au$ about the pivot due to gravity is $	au = mg \cdot (L/2) \sin 	heta$.Using the rotational analog of Newton's second law,$	au = I \alpha$,where $\alpha$ is the angular acceleration:$mg \frac{L}{2} \sin 	heta = \left( \frac{mL^2}{3} ight) \alpha$Solving for $\alpha$:$\alpha = \frac{mg(L/2) \sin 	heta}{mL^2/3} = \frac{mgL/2}{mL^2/3} \sin 	heta = \frac{3g}{2L} \sin 	heta$.Thus,the angular acceleration is $\frac{3g \sin 	heta}{2L}$.

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