When a body of mass m slides down from the top of an inclined plane and reaches the bottom its velocity becomes v. When a circular disc of the same mass is rolled down the inclined plane it acquires a velocity `v_(1)`. Show that ,` v_(1) = sqrt((2)/(3))v`.

Explanation

Let the vertical height of the inclined plane be h. In case of the first body,
`(1)/(2)mv^(2)=mgh or, v = sqrt(2gh) " "cdots(1)`
If the body is a circular disc then it possesses both translational kinetic energy and rotational kinetic energy while rolling down the inclined plane. If `I` be moment of inertia of the circular disc and `omega` be angular velocity of the disc at the bottom of the inclined plan,
`(1)/(2)"mv"_(1)^(2)+(1)/(2)Iomega^(2)=mgh`
or, `(1)/(2)"mv"_(1)^(2)+(1)/(2).(1)/(2)"mr"^(2).(v_(1)^(2))/(r^(2))=mgh`
[ r= radius of the disc , `I = (1)/(2)mr^(2), omega= (v^(1))/(r)]`
or, `((1)/(2)+(1)/(4))v_(1)^(2)=ghor,(3)/(4)v_(1)^(2)gh or, v_(1)=sqrt((4)/(3)gh)`
`:." " (v_(1))/(v)=sqrt(((4)/(3)gh)/(2gh))=sqrt((2)/(3))or, v_(1)=sqrt((2)/(3))v`.