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.

Explanation

(a) By angular momentum conservation, the common angular speed
`omega=(I_(1)omega_(1)+I_(2)omega_(2))//(I_(1)+I_(2))`
(b) The loss is due to energy dissipation in frictional contact which brings the two discs to a common angular speed `ω`. However, since frictional torques are internal to the system, angular momentum is unaltered.