Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?

Explanation

We assume conservation of energy of the rolling body, i.e. there is no loss of energy due to friction etc. The potential energy lost by the body in rolling down the inclined plane (= mgh) must, therefore, be equal to kinetic energy gained. (See Fig.7.38) Since the bodies start from rest the kinetic energy gained is equal to the final kinetic energy of the bodies. From `K=(1)/(2)mv^(2)(1+(k^(2))/(R^(2)))`, where v is the final velocity of (the centre of mass of) the body. Equating K and mgh,

`mgh=(1)/(2)mv^(2)(1+(k^(2))/(R^(2)))`
`or v^(2)=((2gh)/(1+k^(2)//R^(2)))`
Note is independent of the mass of the rolling body,
For a ring, `k^(2) = R^(2)`
`v_("ring")=sqrt((2gh)/(1+1))`
`=sqrt(gh)`
For a solid cylinder `k^(2) = R^(2)//2`
`v_("disc")=sqrt((2gh)/(1+1//2))`
`=sqrt((4gh)/(3))`
For a solid sphere `k^(2) = 2R^(2)//5`
`v_("sphere")=sqrt((2gh)/(1+2//5))`
`=sqrt((10gh)/(7))`
From the results obtained it is clear that among the three bodies the sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.
Suppose the bodies have the same mass. Which body has the greatest rotational kinetic energy while reaching the bottom of the inclined plane?