# Rotational work done by static friction?

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1. Apr 2, 2016

### Soren4

1. The problem statement, all variables and given/known data
A disk roll without slipping down a incline plane. Identify the forces acting on the disk, explain qualitatively which of these forces do work.

2. Relevant equations
The rotational work is given by $W=\int \tau_z d\theta$ (1) , where $\tau_z$ is the component of the torque parallel to the axis of rotation.

3. The attempt at a solution
I'm ok with the first part of the exercise, my main doubt is about works. In particular I know that only gravity do work, while static friction force does not (the contact point does not move in every time istant considered). Nevertheless, looking at the definition of (rotational) work (1) I came up with a big doubt. Consider the motion of the disk as a combination of a traslational motion of the CM and a rotation about the CM, take as pivot point for the calculation of torques the CM. Then static friction force exerts a torque, which causes the rotation, while gravity does not. Hence, from (1), static friction force should do (rotational) work. But that's not possible, because static friction does no work. How can that be?

2. Apr 2, 2016

### Simon Bridge

You can check by using the work-energy relation.

When a wheel rolls without slipping, the physical pivot is the point of contact ... which is where the static friction acts. Thus the torque provided by static friction is zero. Of course you could calculate torques about any other point - like the com - but then, are you working in an inertial frame?

Consider opening a door - there is a reaction force at the hinges right?
This is the physical pivot that you would calculate the torques about ... but you could choose to calculate the torques about the centre of mass of the door couldn't you?

Back to an object rolling without slipping:
Following the energy - some of the energy from gravity goes to the translational motion, and some of it gets stored in rotation: telling you that, however you did the maths, it is the gravity that did the work.

3. Apr 2, 2016

### PeroK

Your analysis is correct. If there were no friction, all the GPE would become linear KE. Static friction translates some of that linear KE into rotational KE. From that point of view, friction only translates one form of KE to another.

But, you can also look at it from just the rotational point of view. As far as the disk is concerned, friction is working its way round the circumference and doing work to rotate it.

It depends to some extent on your frame of reference.

PS as Simon rightly points out above, that second reference frame is non-inertial: it's a frame accelerating down the incline with the disk's centre of mass. In that frame, the incline is moving and doing work rotating the disc.

Last edited: Apr 2, 2016
4. Apr 2, 2016

### Soren4

Thanks for your answer @Simon Bridge ! I think that is not a problem to use the CM frame, even if not inertial. The net torque calculated with respect to CM equals the net torque only of real forces (fictitious forces do not influence the calculation of torque w.r.t. CM).

I'm ok with the use of the work-energy relation, but I still do not see the correct application of (1), which should hold anyway, if the CM is taken as pivot to calculate torques. Who is in that case $\vec{\tau}$?

Thanks for the reply @PeroK ! Rotational point of view is what I'm trying to understand and, as I said above, I'm having troubles to get what is the role of friction from that point of view. You mentioned that it does work to rotate the disk, could you specify a little deeper this concept? Because that's exactly what I do not get here

5. Apr 2, 2016

### PeroK

In the acclerating reference frame, you have the real tangential force of gravity, $F_g$ acting down the slope, the real friction force $f$ acting up the slope and a fictitious force $F_f$ acting up the slope, such that:

$F_g = f + F_f$ as there is no linear acceleration in this frame.

This leaves the frictional force as an unbalanced torque acting on the disk, as per your analysis.

6. Apr 2, 2016

### PeroK

Here's another way to look at it:

Gravity does positive work down the slope: $W_g$

Friction does negative work up the slope: $-W_f$ and equal positive work rotating the disk: $W_f$

So:

Linear KE = $W_g - W_f$

Rotational KE = $W_f$

Total KE = $W_g - W_f + W_f = W_g$

So, all the nett work is done by gravity and friction does no nett work.