A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy will be …………..

Options

A. `(R^(2))/(k^(2)+R^(2))`
B. `(k^(2)+R^(2))/(R^(2))`
C. `(k^(2))/(R^(2))`
D. `(k^(2))/(k^(2)+R^(2))`

Answer

D

Explanation

Centre of mass of rolling ball also moves with linear velocity along rotational motion of ball,
Rotational kinetic energy = `(1)/(2)Iomega^(2)`
`=(1)/(2)mk^(2)omega^(2)(because I=mK^(2))`
and linear kinetic energy
`=(1)/(2)mv^(2)`
`=(1)/(2)mR^(2)omega^(2)(because v=Romega)`
`therefore` Fraction associated with rotational motion
= `("Rotational kinetic energy")/("Total kinetic energy")`
`=("Rotational kinetic energy")/("Rotational kinetic energy + linear kinetic energy")`
`=((1)/(2)mk^(2)omega^(2))/((1)/(2)mk^(2)omega^(2)+(1)/(2)mR^(2)omega^(2))`
`=(k^(2))/(k^(2)+R^(2))`