Uniform ring of radius R is moving on a horizontal surface with speed v and then climbs up a ramp of inclination `30^(@)` to a height h. There is no slipping in the entire motion. Then h is

A uniform solid sphere rolls on a horizontal surface with a linear speed of 10 m/s. It then rolls up a.plane inclined at 30° to horizontal. What is the height upto wlrich the sphere rises ? (See Fig.) Assume that the surface is frictionless. Also calculate the distance travelled by the Fig. sphere on the inclined plane ?

A uniform solid sphere rolls on a horizontal surface at 20 ms^(-1) . It then rolls up in incline having an angle of inclination at 30^(@) with the horizontal . If the friction losses are negligible , the value of height h above the ground where the ball stops is

A symmetrical body of mass M and radius R is rolling without slipping on a horizontal surface with linear speed v. Then its angular speed is

A ring of mass m and radius R rolls on a horizontal roudh surface without slipping due to an applied force 'F' . The frcition force acting on ring is :-

A ring of mass m and radius R rolls on a horizontal roudh surface without slipping due to an applied force 'F' . The frcition force acting on ring is :-

A uniform ring of mass m and radius R is in uniform pure rolling motion on a horizontal surface. The velocity of the centre of ring is V_(0) . The kinetic energy of the segment ABC is:

A ring, disc, spherical shell and solid sphere of same mass and radius are rolling on a horizontal surface without slipping with same velocity. If they move up an inclined plane, which can reach to a maximum height on the inclined plane?

A body of radius R and mass m is rolling smoothly with speed v on a horizontal surface. It then rolls up a hill to a maximum height h . If h = 3 v^2//4g . What might the body be ?