A rod of length `l` is pivoted about a horizontal , frictionless pin through one end. The rod is released from ret in a vertical position. Find the velocity of the `CM` of the rod when the rod is inclined at an angle `theta` from the vertical.
Text Solution
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The fall in postion of the `CM` of the rod, `h=1/2(1-costheta)` In the process, decrease in `E` is equal to increase in rotational `KE` of the rod, so `mgh=1/2Iomega^(2)` or `mg1/2((3g)/l(1-costheta))` The velocity of the CM of the rod `V_(CM)=omegar` is `sqrt((3g)/l(1-costheta))xxl/2=sqrt((3gl(1-costheta))/4`
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