A solid sphere is rolling down an inclined plane. Then the ratio of its translational kinetic energy to its rotational kinetic energy is

Options

A. 2.5
B. 1.5
C. 1
D. 0.4

Answer

A

Explanation

Let M and R be the mass and radius of the solid sphere .
As it is rolling down an inclined plane ,
`therefore` Its translational kinetic energy is `K_(t) = (1)/(2) M v_(cm)^(2)`
(where `v_(cm)` is the velocity of centre of mass) and its rotational kinetic energy is `K_(r) = (1)/(2) I omega^(2)`
But `I = (2)/(5) MR^(2)` (For solid sphere)
and `omega = (v_(cm)) /(R) therefore K_(r) = (1)/(2) ((2)/(5) MR^(2)) ((v_(cm))/(R))^(2) = (1)/(5) M v_(cm)^(2)`
Their corresponding ratio is `(K_(i))/(K_(r)) = ((1)/(2) M v_(cm)^(2))/((1)/(5) M v_(cm)^(2)) = (5)/(2) = 2.5`