When a sphere, ring, disc and a spherical shell are coming down on an inclined plane, rolling without slipping, which will reach the ground first?

Three bodies - a ring, a solid cylinder and a solid sphere of same mass and radii roll down an inclined plane from the same height. Which of the three will reach the ground first?

If a solid sphere, disc and cylinder are allowed to roll down an inclined plane from the same height

A solid sphere and a hollow sphere of the same mass and diameter, both initially at rest , roll down the same inclined plane. Which reaches the bottom first?

A body slides down a smooth inclined plane when released from the top , while another body falls freely from the same point. Which one will strike the ground earlier?

A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination, (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?

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)/ (i+k^2^l R^3 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.

A solid cylinder rolles down an inclined plane. Its mass is 2 kg and radius 0.1 m. If the height of the inclined plane is 4m, what is its rotational kinetic energy when it reaches the bottom.

One hollow and one solid cylinder of the same outer radius rolls down on a smooth inclined plane. The foot of the inclined plane is reached by :

A body of radius R and mass m is rolling horizontally without slipping with a unifrom speed v. It then rolls up in an inclined plane to a maximum height of h. If h = (3v^2)/(4g) , determine the MI of the body about its axis of rotation.