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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

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Asked by m.nilu 20th October 2018, 4:59 PM
Answered by Expert
Answer:
Initial energy = only potential energy = m×g×h  ...............................(1)
 
final energy E when the ball reaches bottom = translational kinetic energy + rotational kinetic energy

E= begin mathsize 12px style 1 half m cross times v squared space plus space 1 half I cross times omega squared space equals fraction numerator begin display style 1 end style over denominator begin display style 2 end style end fraction m cross times open parentheses omega r close parentheses squared space plus space fraction numerator begin display style 1 end style over denominator begin display style 2 end style end fraction open parentheses 2 over 5 cross times m cross times r squared close parentheses cross times omega squared space space equals space 7 over 10 m cross times omega squared cross times r squared end style.........................(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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