A round uniform body of radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is

Question

Accepted Answer

(2) (g sin θ)/(1+I/(MR²)). Take torques about the contact point, where the normal force and friction have no moment. Only the weight contributes, with moment arm R sin θ: τ=MgR sin θ. Moment of inertia about the contact point, by the parallel-axis theorem: I_c=I+MR². α=(MgR sin θ)/(I+MR²), and rolling gives a=α R. a=(MgR² sin θ)/(I+MR²). Dividing top and bottom by MR²: a=(g sin θ)/(1+I/(MR²)). Two checks: a body with I=0 slides at g sin θ, and a ring (I=MR²) manages only half that. Both come out right.

A round uniform body of radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is

Step-by-step solution

Why the other options are wrong