A small sphere of radius r at rest begins to slide down the surface of a hemispherical bowl from the brim of the bowl. When the sphere reaches the bottom of the bowl, what fraction of its total energy will be converted into translational kinetic energy and what fraction into rotational kinetic energy ?

Explanation

Let the initial position of the small sphere be A Fig. Velocity of the sphere when it reaches the point B=V.
`:.` At the point B, translational kinetic energy of the sphere `= K_(t)=(1)/(2)mV^(2)` and rotational kinetic energy of the sphere
`K_(r)=(1)/(2)Iomega^(2)`
`=(1)/(2)xx(2)/(5) mr^(2)xxomega`
`=(1)/(5) mV^(2)`
`:.` Total energy of the sphere,
`K=K_(t) + K_(r)`
`=(1)/(2)mV^(2)+(1)/(5)mV^(2)=(7)/(10) mV^(2)`
`:.` The ratio of the translational kinetic energy to the total kinetic energy,
`(K_(t))/(K)=((1)/(2)mV^(2))/((7)/(10)mV^(2))=(5)/(7).`
Again the ratio of the rotational kinetic energy to the total kinetic energy,
`(K_(r))/(K)=((1)/(5)mV^(2))/((7)/(10)mV^(2))=(2)/(7).`
So, `(5)/(7)` part of the total energy will be converted into translational kinetic energy and `(2)/(7)` part into rotational kinetic energy.