The solid cylinder shown in figure is rolling without slipping between two moving planks P1 and P2 moving with velocities 2V and V respectively. If mass of the cylinder is m, then kinetic energy of the cylinder relative to ground frame is
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a
18 mV2
b
1116 mV2
c
716 mV2
d
58 mV2
answer is B.
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Detailed Solution
Angular velocity of the cylinder ω=2V+V2R=3V2R∴ Velocity of centre C is given by V→C = V→B + V→C B∴ Kinetic energy relative to ground frame=12 mVC2 + 12 IC ω2 = 12 . m V22 + 12 12 mR2. 3V2R2= 1116 mV2