Question

# A cylinder of mass m and radius R is rolling without slipping on a horizontal surface with angular velocity ω0. The velocity of center of mass cylinder is ω0R. The cylinder comes across a step of height R4. Then the angular velocity of cylinder just after the collision is (Assume cylinder remains in contact and no slipping occurs on the edge of the step)

A
5ωo6
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B
ω0
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C
2ωo
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D
6ω0
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Solution

## The correct option is B 5ωo6Given,Mass of cylinder =mRadius of cylinder =RAngular velocity =ω0Velocity of center of mass =ω0R Height h=R4Using conservation of angular momentumInitial angular momentum = Final angular momentumThe angular momentum is the product of the moment of inertia and angular velocity.L=rcmmvcm+Icmω34mR2ω=(R−R4)mv0+12mR2(v0R)34Rω=34v0+12v0ω=56×v0Rω=5ω06

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