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

A solid sphere is rolling down an inclined plane. Thenthe ratio of its

1.	A solid sphere is rolling down an inclined plane. Thenthe ratio of its translational kinetic energy to its rotational kinetic energy is
Answer» 2.5 1.5 1 0.4 Solution :Let M and R be the mass and RADIUS of the solid sphere . As it is rolling down an inclinedplane , `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`