A solid sphere rolls on horizontal surface without slipping. What is the ratio of its rotational to translation kinetic energy.

Explanation

Consider a symmetric rigid body, like a sphere or a whell or a disc, rolling without slipping on a horizontal plane surface with friction along a straight path. The rolling motion of the body can be considered as a combination of translation of the centre of mass and rotation about the centre of mass. Hence, the kinetic nergy of a rolling body is
`E=E_(tran)+E_(rot)" "`.....(1)
where `E_(tran)` and `E_(rot)` are the kinetic energies associated with translation of the centre of mass and rotation about an axis through the centre of mass, respectively.
Let M and R be the mass and radius of the body. Let `omega`, k and I be the angular speed, radius of gyration and moment of inertia for rotation about an axis thorugh its centre of mass, and v be the translational speed of the centre of mass.
`:. v=omegaR` and `I=Mk^(2)" "`.........(2)
`:. E_(tran)=(1)/(2)Mv^(2)` and `E_(rot)=(1)/(2)Iomega^(2)" "`........(3)
`:. E=(1)/(2)Mv^(2)+(1)/(2)Iomega^(2)=(1)/(2)Mv^(2)+(1)/(2)Mk^(2)(v^(2))/(R^(2))`
`=(1)/(2)Mv^(2)(1+(k^(2))/(R^(2)))" "`......(4)
For a solid sphere, `k=sqrt((2)/(5))R" "`......(5)
`:. E=(1)/(2)Mv^(2)(1+(2)/(5))=(1)/(2)Mv^(2)((7)/(5))=(7)/(10)Mv^(2)" "`....(6)