The thin rod shown below has mass M and lengt | vedclass

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

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$\frac{3 F}{2 M L}$

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(A) The torque $	au$ acting on the rod about the hinge point is given by $	au = r_{\perp} F$,where $r_{\perp}$ is the perpendicular distance from the axis of rotation to the line of action of the force.From the geometry,$r_{\perp} = L \sin 30^{\circ} = L 	imes \frac{1}{2} = \frac{L}{2}$.Thus,the initial torque is $	au = F 	imes \frac{L}{2} = \frac{FL}{2}$.The moment of inertia $I$ of a thin rod of mass $M$ and length $L$ about an axis passing through one end is $I = \frac{ML^2}{3}$.Using the relation $	au = I \alpha$,the initial angular acceleration $\alpha$ is given by:$\alpha = \frac{	au}{I} = \frac{FL/2}{ML^2/3} = \frac{FL}{2} 	imes \frac{3}{ML^2} = \frac{3F}{2ML}$.Therefore,the correct option is $A$.

The thin rod shown below has mass $M$ and length $L$. $A$ force $F$ acts at one end as shown,and the rod is free to rotate about the other end in the plane of the force. The initial angular acceleration of the rod is ........

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