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(A) The original rod has mass $m$ and length $2l$. After cutting a length $l/2$ (mass $m/4$),the remaining rod has length $3l/2$ and mass $3m/4$. The

Vedclass · Accepted Answer

$\frac{3g}{5l}$

Vedclass · Answer

(A) The original rod has mass $m$ and length $2l$. After cutting a length $l/2$ (mass $m/4$),the remaining rod has length $3l/2$ and mass $3m/4$. The cut piece of length $l/2$ is placed on top of the remaining rod. The pivot is at the center of the original rod.The moment of inertia $I$ about the pivot is:$I = I_{	ext{rod}} + I_{	ext{piece}} = \left[ \frac{1}{12} \left( \frac{3m}{4} ight) \left( \frac{3l}{2} ight)^2 + \left( \frac{3m}{4} ight) \left( \frac{l}{4} ight)^2 ight] + \left[ \frac{1}{12} \left( \frac{m}{4} ight) \left( \frac{l}{2} ight)^2 + \left( \frac{m}{4} ight) \left( \frac{l}{4} ight)^2 ight]$$= \frac{3m}{4} \left( \frac{3l^2}{16} + \frac{l^2}{16} ight) + \frac{m}{4} \left( \frac{l^2}{48} + \frac{l^2}{16} ight) = \frac{3m}{4} \left( \frac{4l^2}{16} ight) + \frac{m}{4} \left( \frac{4l^2}{48} ight) = \frac{3ml^2}{16} + \frac{ml^2}{48} = \frac{9ml^2 + ml^2}{48} = \frac{10ml^2}{48} = \frac{5ml^2}{24}$.The torque $	au$ about the pivot is:$	au = \left( \frac{3m}{4} ight) g \left( \frac{l}{4} ight) - \left( \frac{m}{4} ight) g \left( \frac{l}{4} ight) = \frac{2mg}{4} \cdot \frac{l}{4} = \frac{mgl}{8}$.Using $	au = I\alpha$:$\frac{mgl}{8} = \left( \frac{5ml^2}{24} ight) \alpha \implies \alpha = \frac{mgl}{8} \cdot \frac{24}{5ml^2} = \frac{3g}{5l}$.

$A$ uniform rod of mass $m$ and length $2l$ is balanced on a triangular prism. Now,a length of $l/2$ is cut from one end of the rod and placed over the shortened part such that the ends meet. The initial angular acceleration is:

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