question_answer

A force F is applied at the top of a ring of mass M and radius R placed on a rough horizontal surface as shown in the figure. Friction is sufficient to prevent slipping. The frictional force acting on the ring is

A)  \[\frac{F}{2}\]towards right       

B) \[\frac{F}{3}\]towards left

C)  \[\frac{2F}{3}\] towards right   

D)  0

Correct Answer: D

Solution :

Let f be the force of friction on the ring which is directed to the right, a be the acceleration of its center of mass and\[\alpha \]the angular acceleration as shown in the figure. The point of contact P is at instantaneous rest so the ring rotates about P. \[\therefore \]\[\alpha =\frac{{{\tau }_{P}}}{{{I}_{P}}}=\frac{F(2R)}{2M{{R}^{2}}}=\frac{F}{MR}\] As \[F+f=Ma=M(R\alpha )=F\]\[\therefore \]\[f=0\]