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A nearly massless rod is pivoted at one end so that it can swing freely as a pendulum. Two masses 2m andm are attached to it at distance b and 3b respectively from the pivot. The rod is held horizontal and then released. The angular acceleration of the rod at the instant it is released is : |
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Answer» Solution :Here torque DUE to two masses 2m and m is `tau=2mgxxb+mgxx3b` `=5mgb` M.I. of the two masses `I=2mb^(2)+9mb^(2)` `=11mb^(2)` Now `alpha=(tau)/(I)=(5mgb)/(11mb^(2))` `=(5g)/(11b)` |
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