Imagine a rough turn table with a wooden block (let it have negligible dimensions for the sake of simplicity) on it. The turn table is perfectly horizontal and is rotating with a constant angular speed in such a way that the block doesn’t slip on it.

It is a well established fact that in the case of kinetic friction, friction opposes relative motion and for static friction, it will oppose the “tendency” of a body to move. By “tendency”, I’m talking about the direction of relative velocity that the body would have if the friction were to be absent.

Now, in this scenario, friction will be the only force that will provide centripetal force. But as per the above point discussed, there has to be a tendency for the block to move radially outwards for the friction to act inwards. Where is the tendency?

  • $\begingroup$ This question is very similar to these questions: physics.stackexchange.com/q/353191 physics.stackexchange.com/q/495120 $\endgroup$ Commented Mar 13, 2023 at 7:04
  • $\begingroup$ But since the concept was discussed with the help of a car, it was hard to understand with the complex wheel turning happening. This question eliminates the said complexity and focuses solely on the concept of friction and centripetal force. $\endgroup$ Commented Mar 13, 2023 at 7:07
  • $\begingroup$ How fast do you need to turn the table to overcome friction? What happens after that? $\endgroup$
    – MS-SPO
    Commented Mar 14, 2023 at 6:07
  • $\begingroup$ the maximum angular speed for rotation without slipping can be found by equating the limiting friction to centripetal force. And any angular speed greater than that will result in slipping. $\endgroup$ Commented Mar 15, 2023 at 0:58
  • $\begingroup$ Ok, so you are aware of it. What‘s your problem, then? $\endgroup$
    – MS-SPO
    Commented Mar 15, 2023 at 4:30

3 Answers 3


If you have a body undergoing circular motion is has a centripetal acceleration towards the centre of rotation and hence there must be a net force acting in that direction.
In your example there is only one force and that is the force due to friction.

Frictional forces do one of two things, try and reduced the relative motion between two bodies (kinetic friction) or try and prevent the relative motion between two bodies (static friction).

For a given situation the magnitude of the static friction adjusts itself from $0^+$ to $\mu_{\rm static}\, N$ so that there is no relative movement between the two bodies, where $\mu_{\rm static}$ is the coefficient of static friction and $N$ the normal force.

If the frictional force was suddenly switched off the body would move at a tangent to its motion at a constant velocity.
That is what the body would "like" to do, the body has a tendency to travel at constant velocity, but the static frictional force prevents it from doing so.


Imagine the friction to disappear suddenly. The block and the point on the turn table with which it is in contact will have a tendency to move tangentially. But after an infinitesimally small interval $dt$, the block will still have the tendency to move in the same direction but the point of contact will have its velocity changed. enter image description here

$$\vec{v}{_{_B}} =v\hat{j}$$ $$\vec{v}_{_{POC}} =-v\sin{d\theta}\hat{i}+v\cos{d\theta}\hat{j}$$ The relative velocity of the block with respect to the point of contact is going to be $$\vec{v}{_{_{BPOC}}}= \vec{v}{_{_B}}-\vec{v}_{_{POC}} $$ $$=v\hat{j}-[-v\sin{d\theta}\hat{i}+v\cos{d\theta}\hat{j}] $$ $$=(v-v\cos{d\theta})\hat{j}+v\sin{d\theta}\hat{i}$$ $$\lim_{d\theta\rightarrow0} \vec{v}{_{_{BPOC}}}= v {d\theta}\hat{i} $$

This shows that with respect to the point of contact, the block will have a tendency to move outwards. Therefore the frictional force has to act radially inwards.

  • 1
    $\begingroup$ is this tendency related to centrifugal force by any means ? because i would like to think that the tendency of the friction acting on the center comes from the static friction from centrifugal force of the object $\endgroup$
    – Naveen V
    Commented Mar 13, 2023 at 7:37
  • $\begingroup$ Tendency has nothing to do with the centripetal force. Tendency is decided just by looking at the relative velocity $\endgroup$ Commented Mar 13, 2023 at 7:43

If there's no static friction, the block cannot maintain uniform circular motion. It would definitely drop from the table because no force supports centripetal force. Now you see the tendency. The static friction just prevent this tendency from being away from the position. Plus, the centripetal force is provided by static friction.


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