• # 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 leftC)  $\frac{2F}{3}$ towards right   D)  0

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$