Question

a solid cylinder of mass m and radius r rolls without slipping down an inclined plane of length l and height h the speed of its centre of mass when reach the bottom of inclined plane

Answer 1

The speed of its center of mass when reach the bottom of inclined plane is r(4gh/3r²)^1/2

Given-

• Mass of solid cylinder = M
• Radius of cylinder = r
• Length of the inclined plane = l
• Height of the plane = h

Let v be the velocity of center of mass. Now according to the conservation of energy.

Potential energy at the top = Kinetic energy of the rotation + Kinetic energy of the translational motion at the bottom.

By substituting the value we get

mgh = 1/2 mv² + 1/2 Iω² where

v = rω and I = 1/2 mr²

So,

ω² = 4gh/3r²

Kinetic energy = 1/2 mv² = 1/2 mr²ω²

v = rω = r (4gh/3r²)^1/2