# Two different works done by friction to roll a sphere

Consider a ring with string wound upon it, on a rough surface with $$\mu$$ enough for pure rolling. If we pull the string tangentially with force $$F$$, in which direction would the frictional force be?

Let $$f$$ be the frictional force taken along the positive $$x$$ direction. Let the mass be $$m$$, and the inertia around its center of mass be $$I$$. Then if $$a, \alpha$$ are the translational and angular accelerations respectively, then we should have $$ma = F + f$$ $$I\alpha = Fr - fr$$ $$a = \alpha R$$ where the last equation comes from the fact that the point of contact must be at rest.

From these equations I am getting $$f=0$$. So I think there should be no frictional force. Am I correct?

• hi - looks correct to me, why do you think otherwise? Commented Jun 1, 2020 at 15:58
• @aman_cc the official answer was coming otherwise, it said that friction would be backwards Commented Jun 2, 2020 at 4:31

Your result is a correct one, and it is valid only when the radius of gyration $$R_g$$ of the body is $$r$$, as for a ring. This can be checked by looking at the first two equations for $$f=0$$. They become
$$ma = F$$ $$I\alpha = F r$$
$$a=\frac{I}{mr} \alpha$$
This is the pure rolling condition only if $$I=mr^2$$. So for a ring there is no need of a frictional force to enforce the pure rolling. In the general case $$f = F \left(\frac{1-\frac{I}{mr^2}}{1+\frac{I}{mr^2}}\right) = F \left(\frac{1-\frac{R_g^2}{r^2}}{1+\frac{R_g^2}{r^2}}\right)$$
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