A uniform rod of mass M and length L is hinge | vedclass

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Question

(B) When the string is cut,the only torque acting on the rod about the hinge is due to the gravitational force acting at the center of mass of the rod

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) When the string is cut,the only torque acting on the rod about the hinge is due to the gravitational force acting at the center of mass of the rod.The torque $	au$ about the hinge is given by $	au = Mg \cdot \frac{L}{2}$.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 analogue of Newton's second law,$	au = I \alpha$,where $\alpha$ is the angular acceleration:$Mg \cdot \frac{L}{2} = \left( \frac{ML^2}{3} ight) \alpha$Solving for $\alpha$:$\alpha = \frac{MgL/2}{ML^2/3} = \frac{3g}{2L}$.

$A$ uniform rod of mass $M$ and length $L$ is hinged at one end as shown in the figure. Initially,the rod is kept horizontal by a massless string. When the string is cut,the angular acceleration of the rod is:

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