Question

Prove the result that the velocity `v` of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by `v^(2) = (2gh)/((1 + k^(2)//R^(2))` using dynamical consideration (i.e. by consideration of forces and torque). Note k is the radius of gyration of the body about its symmetry axis, and `R` is the radius of the body. The body starts from rest at the top of the plane.

Answer 1

Let a rolling body `(l=Mk^(2))` rolls down an inclined plane with an initial velocity u=0, When it reaches the bottom of inclined plane, let its linear velocity be v. Then from conservation ofmechanical energy, we have Loss in P.E. =Gain in translationa K.E. +Gain in rotational K.E.

` therefore " " Mgh=(1)/(2)mv^(2)+(1)/(2)Iomega^(2)`
`=(1)/(2)mv^(2)+(1)/(2)(mk^(2))((v^(2))/(R^(2)))`
`therefore " " Mgh =(1)/(2)mv^(2)(1+(k^(2))/(R^(2)))`
` implies " " v^(2)=(2gh)/((1+(k^(2))/(R^(2)))`