I have read various answers , on PSE and elsewhere , and most of them explain that the point of contact of the rolling object undergoes 0 instantaneous displacement in the direction of friction, I agree , but then there is a feeling that the friction does provide a torque , so how can it do 0 work . ( The pure rolling here is under a constant external force on the object so friction does act ).

  • $\begingroup$ Note that there is a difference between work and torque, even though both variables can have the same units associated with them (e.g., N-m). When changing a tire on your automobile, you can put a lot of torque on your lug wrench but still not be able to loosen the lug nut because that lug nut was initially put on too tightly. Thus, even though you are applying a lot of torque, you are not doing any work on the lug nut because it is not moving. $\endgroup$ – David White Mar 7 at 17:19

In a scenario of pure rolling of a rigid wheel on a flat plane, you don't need any friction. Once the wheel is rolling, it will continue to do so, even if the friction coefficient becomes zero. If this were not the case, you'd be violating conservation of angular momentum. There is no force, no torque and therefore no work done.

The more interesting case is that where the object is rolling under an external force, say down an inclined plane. See the diagram below

enter image description here

Now, you can analyze it in two ways. One is similar to Farcher's answer where the point of contact moves perpendicular to the frictional force and hence, there is no work done. But you were interested in it from the point of view of torques (where we consider the whole wheel, not just the point of contact) so let's do that.

Friction does two things to the wheel as a whole. It does negative work when you look at the linear motion of the wheel. Indeed,

$$W_1 = -f.dS,$$

where $f$ is the force of friction and the wheel has moved a linear distance $dS$. Next, the friction provides a torque about the center of the wheel and the wheel has angular displacement. Hence, it does positive rotational work i.e.

$$W_2 = \tau. d\alpha,$$

where $\tau$ is the torque and $d\alpha$ is the angular displacement of the wheel. But note that $\tau = fR$ and $R d\alpha = dS$. Hence $W_2 = f.dS$ and you get

$$W_{tot} = W_1 + W_2 = 0$$

  • $\begingroup$ Just wondering what would happen if the wheel,originally stationary, was on a flat smooth plane and I gave it a push right at the top of the wheel. If i took moments about the CoM, there would be a net torque. Doesn't mean that the wheel will start to roll? $\endgroup$ – Vishal Jain Mar 7 at 18:12
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    $\begingroup$ @VishalJain It will "roll" but not in the sense where the point of contact does not slide. You will give it some angular momentum and some linear momentum. $\endgroup$ – nr2618 Mar 7 at 19:01
  • $\begingroup$ How you make this diagram. Looks nice, want tool. ;) $\endgroup$ – Almo Mar 7 at 19:51
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    $\begingroup$ @Almo, I didn't make it - I searched for rolling motion on inclined plane on Google images. $\endgroup$ – nr2618 Mar 7 at 19:54
  • $\begingroup$ @nr2618 also a good solution! :) $\endgroup$ – Almo Mar 8 at 3:44

The pure rolling here is under a constant external force on the object so friction does act.

As an example consider an object rolling, without slipping, down an inclined plane.
There will be a frictional force $F$ acting, at the point of contact between the object and the inclined plane, upwards and parallel to the slope as an object is rolling down an inclined plane as well as the weight of the object $mg$ and the normal reaction $N$ both of which act through the centre of mass of the rolling object.

If you consider the motion of a point on the rolling object its path is a cycloid.
The image below is taken from the Wikipedia article on the cycloid and I need you to imagine that the black line at the bottom is inclined to the horizontal and thus representing the inclined plane.

enter image description here

At the point of contact the cycloid is perpendicular to the inclined plane ie along the line $XC$ where $X$ is the point of contact and $C$ the centre of mass of the rolling object.

enter image description here

This means that the direction of motion of the point of contact is perpendicular to the frictional force and so the frictional force does no work on the object.

It terms of the torque about the centre of mass $C$ because the motion of the point of contact is towards the centre of mass of the object there is no rotation of the torque about the centre of mass and so no work is done by the frictional torque.

  • $\begingroup$ @AaronStevens If you consider the whole wheel instead of the point of contact like that, you also need to take into account the (negative) $\vec{F}_\mu \cdot \Delta \vec{s}$ work done by static friction, whose effect you can see in the fact that a rolling non-sliding wheel accelerates slower than a sliding object. $\endgroup$ – JiK Mar 7 at 14:24
  • $\begingroup$ @AaronStevens Well, I'd say it's a big difference to say "I disagree. In the case of an object rolling down an incline the static friction force does do work." and to say that and acknowledge that the total work done by the force is still $0$. $\endgroup$ – JiK Mar 7 at 14:29
  • $\begingroup$ @AaronStevens But the total work is $0$. No matter how you treat the situation, the work done by gravity in the lab coordinate system is $m\vec{g} \cdot \Delta \vec{s}$, where $\Delta \vec{s}$ is the displacement of the centre of mass of the object, and it is equal to the linear + rotational kinetic energy gained by the object. Therefore the work done by other forces must be $0$, and since the support force does no work, the friction force can't either. $\endgroup$ – JiK Mar 7 at 14:38
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    $\begingroup$ @AaronStevens You can find the total change in linear + rotational kinetic energy just by considering what forces act on the object, no need to assume a priori anything about the work done by friction. Everyone in this page agrees what forces act on the object, so it's an easy exercise. The question here was to which force we should attribute what amount of work. $\endgroup$ – JiK Mar 7 at 14:59
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    $\begingroup$ @JiK I think we are just talking past each other. It's irrelevant. I agree the work done by friction here is in fact $0$. Thanks for the discussion. $\endgroup$ – Aaron Stevens Mar 7 at 15:00

Yep you're right. Consider the case of a cylinder rolling down a hill, starting from rest. Friction does provide a torque and definitely does work on the cylinder. However, in the case of pure rolling motion, i.e no slipping, the work done by friction = gain in rotational KE. So we can say that the change in gpe = change in total KE.

It therefore appears based on that conservation statement that the frictional force does no work. Derivation

Edit: Note delta KE means the change in linear KE, i.e $mv^2/2$


If you have a solid wheel rolling on a perfectly smooth plane and the material of the wheel and the surface have zero elasticity, no work is done .


If you have a perfectly spherical cow of uniform density and zero elasticity rolling on a smooth, flat, perfectly rigid plane in a vacuum, and no external force, then no work is done.

In your question, adding an external force changes the equation, such that the linear acceleration applied by that external force is partially (but negligibly) countered by the change in rotational speed, due to friction applying torque.

As with most spherical cow analogies, the calculations are good enough in most circumstances. A baseball's flight can be predicted fairly accurately by taking just a point-object ballistic trajectory through a vacuum most of the time. Throwing a crumpled ball of paper, though, you'll need to take drag into account.

The same with rolling objects. For most calculations, you can ignore friction, because it will be negligible compared to other forces, such as slope. If you reduce the rigidity of either the object or the plane it's rolling in, or add other external forces though, then friction starts to matter.


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