I'm confused about the rotational work, defined as

$W=\int_{\theta_1}^{\theta_2} \tau_z d \theta $

Where $\tau_z$ is the component of the torque parallel to the axis of rotation $z$.

Consider a very common problem regarding pure rotational motion of a disk on a surface, under the effect of an external force. For istance if a disk is let free to roll on a incline it follows a pure rotational motion only if there is a force of static friction which exerts a torque on it, changing its angular speed $\omega$ (weight has zero torque on the body).

The force of static friction exerts a torque parallel to the axis of rotation which makes the disk rotate of a certain angle, so, from the expression above, I don't see the reason why it should not do work. On the other hand the force of static friction is an example of a force that does not work, since it does not cause displacement. How can that be?

The very same doubt is about ropes tensions in rigid bodies similar to yo-yos: these forces exert torques, but do they do rotational work?

  • $\begingroup$ Seems that there is some confusion about the role of static friction in your example. The force of static friction in your example doesn't do any work. If you just set your disk down on a flat table, there is static friction but nothing happens and no work is done. The work in your disk-on-incline example is due to the force of gravity, not the force of static friction. Static friction doesn't do the work; in effect, it just acts as an intermediary. $\endgroup$ – Samuel Weir Mar 24 '16 at 23:06
  • $\begingroup$ @SamuelWeir Ok but then what is the "role" of that formula (instead of $W=\int \vec{F} \cdot d\vec{s}$) in the case of the disk on the incline? Gravity does not exert torque (if calculated with respect to the CM of the disk) so how does it contribute to rotational work if friction does not? $\endgroup$ – Sørën Mar 24 '16 at 23:16
  • $\begingroup$ The force of gravity does exert a torque on the disk if the disk is put on an inclined plane (with friction). Gravitational potential energy is then converted into both rotational and translational kinetic energy of the disk. $\endgroup$ – Samuel Weir Mar 25 '16 at 0:18
  • $\begingroup$ @SamuelWeir Thanks for the answers, but I can't understand how gravity exerts this torque since it acts in the center of mass of the disk, even on a incline $\endgroup$ – Sørën Mar 25 '16 at 9:28

Seeing from the Centre of Mass (CoM) you are correct: it does indeed look like there is work done by static friction. But from this point of view it also seems like the incline is accelerating upwards... What force can do that?

The point is that the CoM is not an inertial frame of reference. It is an accelerating frame, and in accelerating frames we see fictious forces doing fictious work.

So let's stick to an inertial frame such as earth and pick the point of Centre of rotation there: For example at the contact point between wheel and ground. And actually it does make sense to consider a rotation about this point; it is like the wheel is pinpointed here and is about to turn about this point.

So, from here, what is going on? Gravity is now the only force causing a torque (about this new choice of Centre of rotation). Gravity pulls in the CoM which is moving constantly. So, work is being done by gravity.

Now, because the contact point is stationary now but moving in the next instant, this can be a bit tricky to intuitively understand. Because what is the angular displacement $\theta$ of the CoM, which you need for your work formula?

If you considered the wheel as rolling around a point in a circle, then the CoM will be having angular displacement. And gravity is the force that pulls it around. Back on the incline the same happens just in infinitely small bits infinitely many times because you always get a new point of contact and there start a new rotation about this point. If you in some way could sum these infinitely many infinitely small angular displacements together, then you would have the total $\theta$ for your formula. And the work formula will work again.

All in all, gravity is doing the work, not static friction.


If you study the problem in the frame of reference of the CM, the contact force of the plane or the rope does work, because the point of contact is moving in this frame.

However, note this frame is not galilean, so you may not blindly apply the work-energy theorem (in this exact case it is valid, but take the good habit not to apply it in non-galilean frames of reference).

If you study the problem in the lab frame of reference, you must express torque and rotation with respect to a fixed axis in that frame.


Friction force doesn’t do work because contact point of the disc with the ground changes. It is not same for a distance. At each moment, there is a new contact point. Hence, for all friction forces (for all contact points), $d=0$ and $W=Fd=0$.


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