Consider the rotation of a rod of mass m and length l from position AB to AB'. Which of the following statements are correct?
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a
Centre of gravity of the rod is lowered by l/2.
b
Loss of gravitational potential energy is l/2 mgl.
c
Angular velocity is 3g/l.
d
Rotational kinetic energy is ml2ω26
answer is A.
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Detailed Solution
Equating loss of gravitational potential energy with the gain of rotational kinetic energy, we get12Iω2=12mgl12×ml23×ω2=12mglω=3glRotational kinetic energy =12ml23ω2=ml2ω26