0
$\begingroup$

Suppose a coin is placed on a turn table which is then rotated. The coin initially rotates along with the disk and may fly off eventually.

My question is, why is static friction acting radially inward and thus providing centripetal acceleration? I know it acts opposite to the potential direction of relative motion, but why is that radially inward? I have seen this asked several times but is not answered satisfactorily and I cannot get the intuition behind it.

Secondly, if the turn table is rotated from rest, what causes the the coin to obtain a velocity in the first place (i.e. what causes tangential acceleration if the static friction acts radially inward)?

Lastly, why may the coin eventually fly off; what is causing the centrifugal force required to increase making the static friction insufficient?

I would be extremely grateful if someone could intuitively explain this to me.

$\endgroup$
0
$\begingroup$

Imagine that a coin is on a turntable at 12 O'clock. When you start turning counterclockwise, the coin first is pushed by surface horizontally. The coin "wants" to stay in place, so it "resists" being moved, but the surface pushes it (by friction).

Next, once coin started moving a little bit, it "wants" to keep moving in that initial direction but then a turntable makes it change the direction towards 11 O'clock. That's how the surface friction pushes the coin toward the center.

The surface friction has a maximum limit, that is defined by a weight of a coin and friction coefficient. So, while the turntable speeds up at some point the surface friction is not enough to keep the coin in place, so it flies off

$\endgroup$
4
  • $\begingroup$ Would it be correct to say that the direcrion of the relarive velocity vector, in the limit as the time interval tends to zero, is toward the center?0 $\endgroup$ – OVERWOOTCH Jun 23 '20 at 3:52
  • $\begingroup$ watch this first, then ask more questions youtube.com/watch?v=WdcfWf07UIg $\endgroup$ – Aksakal almost surely binary Jun 23 '20 at 3:55
  • $\begingroup$ I have and it has only confused me more. If you are slowly increasing the angular velocity of the disk and thus the coin, surely there must be some tangential acceleration. Sholdnt friction have 2 components? With only Fsinx providing the centripetal force. $\endgroup$ – OVERWOOTCH Jun 23 '20 at 6:04
  • $\begingroup$ Of, course there is angular acceleration. If your turntable was super quick, the coin would immediately start sliding outward, because it would't be able to gain speed $\endgroup$ – Aksakal almost surely binary Jun 23 '20 at 13:13
0
$\begingroup$

why is static friction acting radially inward and thus providing centripetal acceleration?

Because the disk underneath the coin is being accelerated radially inward.

The linear case may be easier to understand. We place a package on a conveyor belt. The belt accelerates to the right. The frictional forces between the package and the belt accelerate the package to the right.

The turntable is a rigid object. Whatever is applying torque to the table causes the entire object to spin. The rigid cohesive forces mean that parts of the turntable at some distance from the axis are accelerating toward the center. Just like the conveyor belt, friction attempts to accelerate the coin to the same extent. Either the coin rotates with the turntable, or friction is insufficient and the coin moves in some other path.

what causes the the coin to obtain a velocity in the first place (i.e. what causes tangential acceleration if the static friction acts radially inward)?

The radial acceleration is for the constant rotational speed case. If the turntable is changing speeds (starting from rest), then there will be tangential acceleration as well.

what is causing the centrifugal force required to increase making the static friction insufficient?

Required centrifugal force depends on the mass of the coin, the distance from the axis, and the rotational speed of the table. If all of them are constant, you would expect the required force to remain constant and the coin to never move (relative to the turntable).

In practice, the table might not be perfectly flat. A bump might disturb the coin causing friction to reduce. Or maybe the motor changed speed for a second. The conditions of the experiment would determine if the coin were to stay or not.

i thought it was the relative velocity and not the acceleration that determined friction.

That's correct (at least for kinetic friction). But assuming static friction in this case, the relative velocity is always zero. So the force of friction is sufficient to accelerate the coin exactly as much as the turntable underneath is accelerating (up until it begins slipping). But if the turnable underneath were not accelerating, then the coin would not require friction.

The coin would still speed up if i placrd it on a moving disk; whats providing the tangential acceleration then?

Kinetic friction. It is only in the case where the coin has no relative velocity, and that the turntable is not changing the rotational speed that friction is purely radial. If you are in a slipping configuration, then friction will act in other directions as well.

$\endgroup$
1
  • $\begingroup$ Im sorry but i thought it was the relative velocity and not the acceleration that determined friction. A mass moving up an inclined slop without an applied force, the Kinetic friction acts oppsoite to the acceleration. For static friction, the tendendcy of motion is always in the firrctoon of accelerstion, despite the instantaneous direction of velocity.? Lastly? The coin would still speed up if i placrd it on a moving disk; whats providing the tangential acceleration then? $\endgroup$ – OVERWOOTCH Jun 23 '20 at 3:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.