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?

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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?

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

1 Answer 1


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

$$ma = F$$ $$I\alpha = F r$$

which gives immediately

$$a=\frac{I}{mr} \alpha$$

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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