A small sphere S of radius ,r and mass m rolls without slipping, inside a large hemispherical bowl B of radius R as shown in figure. S starts from rest at the top point of the hemisphere

Options

A. The fraction of translational kinetic energy at the sphere is 5/7.
B. The fraction of rotational kinetic energy of the sphere is 2/7.
C. The fraction of rotational kinetic energy of the sphere is `2/7mv^(2)`.
D. The total kinetic energy of the sphere at the bottom of the bowl is `2/7mv^(2)`

Answer

A, B, C

Explanation

Let the linear and angular velocities of the small sphere S be v and co respectively when it reaches the bottom of the bowl B.
At this position (bottom of B), the translational kinetic energy of the sphere is given by
`E_(1)=1/2mv^(2)`, where m is the mass of the sphere.
As the sphere rolls down without slipping, we have, `v = romega`, r is the radius of the sphere.
`thereforeE_(2)=1/5mr^(2)omega^(2)=1/5mv^(2)`
Therefore, total kinetic energy of the sphere at the bottom of the bowl is
`E=E_(1)+E_(2)=1/2mv^(2)+1/5mv^(2)=7/10mv^(2)`
Fraction of translational kinetic energy of the sphere
`=(E_(1))/E=(1/2mv^(2))/(7/10mv^(2))=5/7`
Fraction of rotational kinetic energy of the sphere
`=(E_(2))/E=(1/5mv^(2))/(7/10mv^(2))=2/7`