Three bodies, a ring, a soild cylinder and a soild 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

Let us assume that there is no loss of energy due to friction when a mass m starting from rest rolls down an inclined plane.
Now, using law of conservation of energy,
P.E. lost by the body = K.E. gained by the body = Translational K.E+Rotational K.E.
`impliesmgh=(1)/(2)mv^(2)+(1)/(2)Iomega^(2)=(1)/(2)mv^(2)+(1)/(2)mk^(2)((v)/(R))^(2)`
`impliesmgh=(1)/(2)mv^(2)(1+(k^(2))/(R^(2)))`
`impliesv=sqrt((2gh)/(1+k^(2)//R^(2)))`
where k is the radius of gyration.
From the formula obtained, we can see that the velocity v attained by the rolling body at the bottom of the inclined plane is independent of its mass.
Now, 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_("cylinder")=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))`
Sphere has the greatest velocity on reaching the bottom.