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To mop-clean a floor, a cleaning machine presses a circular mop of radius R vertically downwards with a total force F and rotates with an angular speed about its axis. If the force F is distributed uniformly over the mop and the coefficient of friction of the floor is $\mu$ , the torque applied by the machine on the mop is? - JEE mains 2019

The point I am confused about here is how the friction is acting on the flat surface of the cleaning machine in contact with the ground. I can understand friction on a rotating object when it is the edges of the object which is translating against a wall (ref) but I can't understand it when the friction is caused by whole surface rotating.

So, in general, what is the correct mental model to understand the frictions caused when we have an area rotating about a surface?

Btw I am not asking solution to the quoted problem , it is available in many sites online already.

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I am not sure whether this is the most effective approach but here's how I would've deal with it.

It is stated in the question that the force is distributed uniformly. This means that any piece of the disk would experience a force proportional to its area. Keep this in mind.

Furthermore since the torque created by the friction at any point is proportional to this point's distance to the center of the disk, it makes sense to model the disk as many concentric rings. The reason this is sensible is that both the area of such a ring, thus the force on the ring, and the distance for all the points on it to the center can be written in terms of the radius of each ring. Thus it is easy to express the torque on each ring which is just the product of the distance and the force.

Then you can add all the disks back together to acquire the total torque on the disk. This simply means integrating with respect to r from 0 to R.

This is a sketch of the friction forces acting on different points. Note even they all have the same magnitude, the ones closer to the rim creates a larger torque. Also note all of them creates a torque in the same direction.

enter image description here

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  • $\begingroup$ Well nice, this is the sketch of the intended solution but doesn't explain my question of how to understand friction in case of relative rotating surfaces $\endgroup$
    – Brian
    Commented Apr 12, 2021 at 10:37
  • $\begingroup$ @Buraian The friction isn't different from the linear case. Remember the friction force is the normal force multiplied by the coefficient of friction. How you are moving on a surface is irrelevant. It only depends on two things: how rough the surface is, how much you are pushed onto it. So at any point on the disk the friction has the same magnitude, but a different direction. Doesn't matter how fast that point rubs against the floor. Since at any point the friction is anti-parallel to velocity, it's perpendicular to the radius at that point. Thus the torque for each point on the ring is equal $\endgroup$
    – GUNDOGAN
    Commented Apr 12, 2021 at 10:42
  • $\begingroup$ Can you show a picture for how exactly the 'force field' created by friction on the surface looks like? I think that's where I'm having the most difficulty understanding $\endgroup$
    – Brian
    Commented Apr 12, 2021 at 10:43
  • $\begingroup$ @Buraian I’ve edited my answer to include a sketch, hope it helped. $\endgroup$
    – GUNDOGAN
    Commented Apr 12, 2021 at 10:50
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    $\begingroup$ @Buraian Good luck on your JEE, heard it is super tough. $\endgroup$
    – GUNDOGAN
    Commented Apr 12, 2021 at 10:55
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The friction is due to the weight of the mop $\mu\times F$.

In a simple model split the circle in two along a diameter. On one side the mop moves one way, but on the other side the mop moves the other way, so the friction force acting is similar to a couple, two forces $\mu\times \frac{F}{2}$ each acting at a distance of $\frac{R}{2}$

That's the mental model to use, just like the torque applied to turn a tap, for example. The only difference is that the force is spread around the circle instead of along a line, but it works the same.

There is a torque on each circular element of the mop depending both on the area of the element and the distance from the centre, then you can find the total torque by integrating from $0$ to $R$.

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  • $\begingroup$ Shoot, that was a typo I meant friction. The idea was that like usually we say friction is caused b/w relative translaitonal motion between surface here it is relative rotational motion (which I can't understand) $\endgroup$
    – Brian
    Commented Apr 12, 2021 at 10:17
  • $\begingroup$ When the surfaces are sliding you can use maximum friction, perhaps you'll need to integrate from $0$ to $R$ to get the total moment, or torque. $\endgroup$
    – John Hunter
    Commented Apr 12, 2021 at 10:20
  • $\begingroup$ I still don't get how friction is acting in this case! And btw I am not looking for solution, simply how to think about friction being caused by relative rotation of two surfaces $\endgroup$
    – Brian
    Commented Apr 12, 2021 at 10:37
  • $\begingroup$ ok, the answer will be edited. $\endgroup$
    – John Hunter
    Commented Apr 12, 2021 at 10:38

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