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

Question

Quantrex Academy · Accepted Answer

Correct answer: C. (10)/(7) g h

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 ω^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) ω^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 ω] aligned =(1)/(5) M v^2+ (1)/(2) M v^2= (1)/(10) M v^2Now, gain in KE= loss in PE aligned (7)/(10) M v^2 & =M g h v & = (10)/(7) g h aligned

A solid sphere of mass M and radius 2 R rolls down an inclined...

Options

Correct answer

Step-by-step solution

More from Rotational Motion