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A small solid sphere of mass m is released from a pt A at a height h above the bottom of a rough track as shown in fig.If the sphere rolls down the track without slipping ,its rotational kinetic energy when it comes to the bottom of track is

Asked by m.nilu 20th October 2018, 4:59 PM
Initial energy = only potential energy = m×g×h  ...............................(1)

final energy E when the ball reaches bottom = translational kinetic energy + rotational kinetic energy

E= .........................(2)
v is linear speed, ω is angular speed of rotation and r is radius of ball

By conservation of energy, if we equate (1) and (2), we get ω2 × r2 = (10/7)g×h ...........................(3)

Only rotational kinetic energy = (1/2)×I×ω2 = (1/2)×(2/5)m×r2×ω2 = (2/7)m×g×h ..............(4)
Eqn.(3) is used to substitute ω2 × r2 in (4)
Answered by Expert 21st October 2018, 10:11 PM
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