A unifrom rod of length `l` and mass `m` is free to rotate in a vertical plane about `A`, Fig. The rod initially in horizontal position is released. The initial angular acceleration of the rod is `(MI "of rod about" A "is" (ml^(2))/(3))` A. `mgl//2`B. `3 g//2l`C. `2l//3g`D. `3g//2l^(2)`

A unifrom rod of length `l` and mass `m` is free to rotate in a vertic

1.	A unifrom rod of length `l` and mass `m` is free to rotate in a vertical plane about `A`, Fig. The rod initially in horizontal position is released. The initial angular acceleration of the rod is `(MI "of rod about" A "is" (ml^(2))/(3))` A. `mgl//2`B. `3 g//2l`C. `2l//3g`D. `3g//2l^(2)`
Answer» Correct Answer - B Torque on rod = moment of weight of the rod about `A` `tau = mg xx (l)/(2)` As `tau = I alpha = mg (l)/(2) :. (ml^(2))/(3) alpha = mg (l)/(2)` `alpha = (3g)/(2l)`

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A unifrom rod of length `l` and mass `m` is free to rotate in a vertical plane about `A`, Fig. The rod initially in horizontal position is released. The initial angular acceleration of the rod is `(MI "of rod about" A "is" (ml^(2))/(3))` A. `mgl//2`B. `3 g//2l`C. `2l//3g`D. `3g//2l^(2)`

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