enter image description here

If a force is applied tangentially to a rolling body then the equation relating the friction which comes into existence to prevent sliding and the force is:

$$ f_s = \frac{MR^2 - I}{MR^2 + I } F$$

This means that for a hollow cylinder/ring, the total friction force is zero (their $I= MR^2$).. but how? What makes their shape special such that they don't have any friction acting on them even while an external force is applied?


The condition being assumed here is rolling without slipping. This means that $v=\omega R$, where $v$ is the velocity of the center of mass, $\omega$ is the angular velocity of the rotation, and $R$ is the radius.

Now, by Newton's second law, the change in $v$ is influenced by the net force and mass, and the change in $\omega$ is influenced by the net torque and moment of inertia. Therefore, there are instances where the net force and net torque just "balance out" in terms of their influences so that $v=\omega R$ stays valid without the need of additional forces (e.g. friction$^*$) acting on the object. This is the case here. In other words, the geometry is such that the the velocity due to the net force evolves proportionally to the angular velocity due to the net torque without the need of additional forces like friction.

But the hoop isn't special, really. You can do this with a cylinder too, for example. You just need to apply the force half way between the edge and the center of the cylinder and its center. More generally, to not require friction we require $$\frac Fm=R\cdot\frac{\tau}{I}=\frac{\beta R^2 F}{I}\to\beta=\frac{I}{mR^2}$$ where $\beta R$ is the distance from the center to where the force is applied. In the special cases where $I=\gamma mR^2$ we end up with $\beta=\gamma$, consistent with what you have in your question for the hoop.

$^*$Specifically in the case of friction, remember that friction opposes relative motion between the two bodies. However, if rolling without slipping would already be happening in a frictionless environment, then there is no impending relative motion, and so friction will not act.

  • $\begingroup$ I do understand the equation but I'm trying to gain insights from the physical nature of the bodies. What particular properties from the geometries of the object causes this result to happen? I know that inertia is a function of geometry but I'm trying to figure out a explanation without usage of mathematical ideas. $\endgroup$ – Buraian Oct 18 '20 at 11:57
  • $\begingroup$ @Buraian My answer is physical, not mathematical. The geometry is such that the change in the velocity due to the net force evolves proportionally to the change in angular velocity due to the net torque without the need of additional forces like friction. $\endgroup$ – BioPhysicist Oct 18 '20 at 12:01
  • $\begingroup$ Yeah I understand that you are trying to say that it's would change the motion such that the rolling condition is met but I'm trying to figure out 'how' the 'such that' is achieved. $\endgroup$ – Buraian Oct 18 '20 at 12:03
  • $\begingroup$ @Buraian I don't think I'm fully following. Are you wondering how friction "knows" not to act? $\endgroup$ – BioPhysicist Oct 18 '20 at 12:06
  • $\begingroup$ Pretty much, yeah $\endgroup$ – Buraian Oct 18 '20 at 12:54

The equations of motion are $F+f_{\rm s} = Ma$ and $(F-f_{\rm s})R = I \alpha$.

$I=MR^2$ is a limiting condition and results in $f_{\rm s}=0$ for the no slip condition, $a_{\rm CoM} = \alpha R$, and the equations of motion are now $F = Ma$ and $FR = I \alpha$.

Now suppose you suddenly reduced the moment of inertia of the hoop, $I$, whilst keeping its mass, $M$, the same by moving some mass nearer the centre of the hoop but keeping the applied force, $F$, and the radius of the hoop, $R$, the same.

$F= Ma \Rightarrow$ the translational acceleration stays the same but $FR = I\downarrow \alpha\uparrow \Rightarrow $ the angular acceleration would now be larger than before the mass was moved.

This means that the no slip conditions is not satisfied and there is the potential of relative motion between the hoop and the ground.

If the no slip condition is to be maintained then the linear acceleration of the hoop must increase or the angular acceleration of the hoop must decrease or, both occur, which is in fact what happens, and produced by the frictional force $f_{\rm s}$.

The net force $F+f_{\rm s} \,\,(>F)$ increases thus increasing the linear acceleration and the net torque $(F-f_{\rm s})R \,\,(<FR)$ decreases this decreasing the angular acceleration.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.