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 torques). 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.

Prove the result that the velocity v of translation of a rolling body

1.	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 torques). 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» Solution :(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.