A thin uniform rod of mass m and length l is hinged at the upper end and free to rotate about its upper end in vertical plane . When it is at rest, it receives an impulse J at its lowest point, normal to its length. Immediately after impact,
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a
the angular momentum of the rod is JI
b
the angular velocity of the rod is 3J/ml
c
the kinetic energy of the rod is 3J2/2m
d
the linear velocity of the midpoint of the rod is 3J/2m
answer is A.
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Detailed Solution
Angular momentum = linear momentum x perpendicular distance from the point of rotation During the course of collision angular momentum of the rod about the axis of rotation remains conserved . Therefore L=JI Also, I=ml23 ω=LI=JIml2/3=3Jml Kinetic energy =L22l=J2l22ml23=3J22m vc=ωl2=3J2m