A solid sphere of mass M and radius 2 R rolls down an inclined plane of height h without slipping. The speed of its centre of mass when it reaches the bottom is

Step-by-step solution

When solid sphere rolls down on an inclined plane, then it has both rotational and translational kinetic energy or array ll & K=K_ rot +K_ trans & K= 1 2 l ^2+ 1 2 M v^2 array where, I= moment of inertia of solid sphere = 2 5 M R^2 aligned K & = 1 2 ( 2 5 M(2 R)^2 ) ^2+ 1 2 M v^2 & [ R=2 R] & = 4 5 M R^2 ( v 2 R )^2+ 7 2 M v^2 & [ v=r ] aligned = 1 5 M v^2+ 1 2 M v^2= 1 10 M v^2 Now, gain in KE = loss in PE aligned 7 10 M v^2 & =M g h v & = 10 7 g h aligned