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A solid sphere is rolling on a frictionless plane surface about its axis of symmetry. Find ratio of its rotational energy to its total energy

Answer»

Solution :Rot K.E. `=1/2 Iomega^(2) = 1/2 xx 2/5 MR^(2) xx V^(2)/R^(2)`
(As `omega = V/R, I = 2/5 MR^(2)`)
`=1/5 MV^(2)`
TOTAL energy = Translation K.E. + Rot. K.E.
`=1/2mv^(2) + 1/5 mv^(2) = 7/10 mv^(2)`
`THEREFORE ("Rot. K.E.")/("Total Energy") =(1/5 mv^(2))/(7/10 mv^(2)) = 2/7`


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