A solid ball of mass `'m'` is released on a rough fixed inclined plane from height `H`. The ball will perform pure rolling on the inclined plane. Then

Options

A. Kinetic energy of the ball at `O` will be less than `mgH`
B. Translational kinetic energy of the ball at `O` will be `(5mgH)/(7)`
C. Rotational kinetic energy of the ball at` O` will be `(2mgH)/(7)`
D. Angular momentum of the ball will be conserved about a point on the inclined plane.

Answer

B, C, D

Explanation

In pure rolling static friction acts so energy remains conserved . So kinetic energy of ball at `O=mgh`
`mgh=(1)/(2)mv^(2)+(1)/(2)xx(2)/(5)mr^(2)(v^(2))/(r^(2)`
translational Rotational
kinetic energy kinetic
`mgh =(1)/(2) mv^(2)+(1)/(5)mv^(2)`
Translational kinetic energy`=(mghxx(1)/(2))/((1)/(2)+(1)/(2))`
`=(mgh xx(1)/(2))/((7)/(10))=(5mgh)/(7)`
Rotational kinetic enrgy `=(2mgh)/(7)`