4
$\begingroup$

There is a question in my physics book (by HC Verma) :

A small coin is placed on a record rotating at 33.33rev/minute. The coin does not slip on the record. Where does it get the required centripetal force from?

My teacher told me that its friction. But he didn't clarify it. I know friction opposes relative motion or if surfaces tend to have relative motions but there is no relative motion between the coin and the contact point on the disc in radially outward direction.

So how can friction act radially inward?

$\endgroup$
2
  • $\begingroup$ Maybe this helps: physics.stackexchange.com/q/35681 $\endgroup$ Feb 8, 2021 at 9:17
  • $\begingroup$ By definition, all uniform circular motion requires a center seeking (aka centripetal) force. Also note that static friction does not require relative motion. $\endgroup$ Feb 19, 2021 at 2:15

8 Answers 8

4
$\begingroup$

Static friction must point in the opposite direction to the relative motion that would occur in its absence. The confusion arises when trying to find the instantaneous relative velocity between the coin and the surface if friction were to disappear. Here's a quick mathematical proof that might help understand what's happening:

enter image description here

If friction were to suddenly disappear, the coin would move along the tangent, and the disk would rotate some angle $\Delta\theta$.

enter image description here

$$\vec{v}_A \text{ (velocity of coin)} = r\omega\hat{i}\\ \vec{v}_B \text{ (velocity of surface)} = r\omega\cos\Delta\theta \ \hat{i} - r\omega\sin\Delta\theta \ \hat{j}\\ \vec{v}_{AB} = r\omega\left[(1-\cos\Delta\theta)\hat{i} + \sin\Delta\theta\hat{j}\right]\\ $$

$$ \left|\vec{v}_{AB}\right| = r\omega\sqrt{1+\cos^2\Delta\theta - 2\cos\Delta\theta+\sin^2\Delta\theta}=r\omega\sqrt{2-2\cos\Delta\theta}\\ $$ $$ \tan\alpha = \frac{\sin\Delta\theta}{1-\cos\Delta\theta}=\frac{2\sin\frac{\Delta\theta}{2}\cos\frac{\Delta\theta}{2}}{1-1+2\sin^2\frac{\Delta\theta}{2}}=\cot\frac{\Delta\theta}{2} \\ \alpha = \tan^{-1}\left(\cot\frac{\Delta\theta}{2}\right)=\tan^{-1}\left(\tan\left(90-\frac{\Delta\theta}{2}\right)\right)=90-\frac{\Delta\theta}{2}\\ $$

$$ \text{magnitude of } \vec{v}_{AB} = r\omega\sqrt{2-2\cos\frac{\Delta\theta}{2}},\qquad \text{direction of } \vec{v}_{AB} = \alpha = 90 - \frac{\Delta\theta}{2} \\ \lim_{\Delta\theta\to 0} \left|\vec{v}_{AB}\right| = r\omega\sqrt{2-2\cos0}=0,\qquad \lim_{\Delta\theta\to 0} \alpha = 90 - \frac{0}{2} = 90 $$

enter image description here

Here, $\alpha$ is the angle between the relative velocity vector and the x-axis. Now, to find the instantaneous relative velocity, you take the limits of both the magnitude and direction as $\Delta \theta$ approaches $0$. As expected, the magnitude goes to $0$, however, surprisingly, the angle goes to $90$ i.e the vector points radially outwards. Ergo, static friction points radially inwards and opposes relative motion.

$\endgroup$
0
1
$\begingroup$

When you say this sentence you are very correct:

I know friction opposes relative motion

But remember that you can oppose relative motion in two ways:

  • You can oppose it by trying to stop it.
  • You can oppose it by trying to prevent it.

In the former case we call it kinetic friction. In the latter case we call it static friction.

It is the latter case with static friction you are experiencing in your scenario. Think of when you are pushing on a sofa but can't move it. It is static friction which is preventing you from initiating sliding (relative motion).

So, in conclusion, opposing relative motion does not require that a relative motion is already happening.

$\endgroup$
11
  • 2
    $\begingroup$ correct. But why should it tend to move radially outward ? $\endgroup$
    – Ankit
    Feb 8, 2021 at 9:49
  • $\begingroup$ @Astudent When you are sitting in your car, and the car turns, then the car is moving away from underneath you. You feel swung outwards because your surroundings (the car) are moving inwards. The coin in your scenario similarly is simply trying to keep moving straight (inertial motion) and the record is moving away underneath it because the record is turning. Moving away underneath it will cause sliding and relative motion. The coin must be "pulled along" with the record player in order to prevent relative motion. This "pulling along" is done by static friction which then must inwards. $\endgroup$
    – Steeven
    Feb 8, 2021 at 10:38
  • $\begingroup$ @Steeven, don't we have to specify a frame? If there were no friction, and we watched while standing on the floor, then we would see the coin move in an arc like this. It seems like the rule is "static friction on an object opposes the direction of sliding as viewed from the object's frame of reference." Is this correct? $\endgroup$
    – jdphys
    Feb 8, 2021 at 13:29
  • 1
    $\begingroup$ @Steeven, thanks for the clarification. How does the principle "static friction prevents relative motion" produce a radial force in the inertial frame? (That's what I see the question as asking.) If the coin's velocity is tangential, then how is the coin's relative motion purely radial and not at least somewhat tangential? And if there is some tangential relative movement in the inertial frame, then why don't we see a tangential component to static friction? $\endgroup$
    – jdphys
    Feb 8, 2021 at 14:43
  • 1
    $\begingroup$ @jdphysics The coin already has tangential velocity equal to the tangential velocity of the record, so there's no relative motion in that direction. If friction suddenly disappeared, the coin would fly off the record in a straight line, with both radial and tangential velocity. The tangential velocity component would remain unchanged, though, as friction is only acting radially. $\endgroup$ Feb 8, 2021 at 15:18
0
$\begingroup$

I know friction opposes relative motion or if surfaces tend to have relative motions but there is no relative motion between the coin and the contact point on the disc in radially outward direction.

The reason there's no relative motion is because of static friction!

I'd first suggest a little experiment.

Take an empty transparent bottle (plastic water/soda bottle will do). Take a coin or a small object like a marble and put it in the bottle. Put the cap on. Holding the bottle by the cap, spin it around (you can spin yourself around). If you do this right, you will notice that the small object migrates outwards. Even if you slant the bottle such that the object has to move upwards, it will still work.

Here is the experiment setup:

enter image description here

Next, check out this animation I made: https://www.desmos.com/calculator/s8roerbvub. Slide/play the value of $t_0$ to analyze the motion of a particle moving at $1\ \rm m/s$ in a straight line and in circular motion.

Hopefully this builds your intuition that without some center seeking force, the object will go outwards.

Now, time for some reasoning.

From the animation, you can see that the particle "wants" to go outwards. At $t_0=6.28$, the particle is instantaneously following the circular trajectory of a point on the rotating disk, and so there is no relative motion. But the particle will try to push out, so friction will push inwards to resist the relative motion.

Why is friction pointing directly inwards? Because there is no angular acceleration. If the disk accelerates, then the friction force will not be directed towards the center -- a component of it will speed the particle up.

Nonetheless, in the case where the angular speed is constant, why does friction point directly inward?

First, consider the following sketch:

enter image description here

It shows a particle moving in uniform circular motion with constant speed, and geometrically you can see that on some "small" time interval $\delta t$, friction will always point inwards.

Second, we know that if the speed of the particle is to remain constant, force must always be perpendicular to trajectory. Therefore, there is nowhere else for friction to point except inwards.

Friction pointing directly inwards will occur if the particle and disk are moving at the same uniform angular speed. The particle will want to go directly outwards, so friction will want to resist that motion. The only geometric way to resist that motion is for the friction force to point directly inwards.

$\endgroup$
0
$\begingroup$

The statement that static friction acts in a direction opposite to that of impending motion is a fact and further explanation of such requires the development of a theoretical model.

The model that explains this phenomenon requires two assumptions:

  1. The instant two objects' surface come in close contact, chemical bonds start forming and complete bonding is attained in finite amount of time.

The strength of these bonds depends on the nature of the surface. The total bond strength is hence proportional to the surface area of contact.This total bond strength adds to the friction between the surfaces.

  1. All the objects have surfaces such that there are microscopic projections protruding outward. These structures reduce the effective area of contact.

Eventhough these structures reduce the effective area of contact, they do provide their own addition to the friction by applying force force when there is strain due to a relative or impending motion between the surfaces.

(Further read: If you need less sliding friction, should you make the surface rougher or smoother?)

Both of these together give a force known as friction.

When there is some impeding motion between two surfaces, what really happens is that, the bonds start to feel strain and in response it applies a restoring force which adds up to give static friction, which opposes the impending motion.

(Note that friction depends on the effective area of contact which in turn depends on normal reaction, hence friction depends on normal reaction)

$\endgroup$
0
$\begingroup$

Here is an explanation that provides a physical discussion of what is happening.

First consider the motion of the coin as viewed by an observer rotating with the record. The coin is fixed to the rotating record; not moving as seen by the observer. That is, the coin is fixed (not moving) in a reference frame rotating with the record. In this rotating reference frame the coin experiences a centrifugal force in the radially outward direction, because this is a non-inertial reference frame. To keep the coin fixed in this frame there must be a counter-force in the radially inward direction, and that is the force of friction. A force of friction counters the actual or impending motion, and in this case it counters the impending motion from the centrifugal force that is radially outward. (An object does not have to be moving relative to the surface to experience a frictional force. Consider a box at rest on an inclined plane; the force of friction up the plane counters the component of the force of gravity down the plane, but the object is not moving.) For the coin to remain fixed to the record, the angular velocity of the record (speed of rotation, 33.33 rpm) must be sufficiently low such the force of friction can counter the centrifugal force which increases with angular velocity. If the angular velocity is too great or the coefficient of static friction too small, the centrifugal force is sufficiently large such that the coefficient of static friction cannot provide enough frictional force and the coin will not remain fixed but move radially outward; as the coin moves outward it will also experience the Coriolis force side-ways.

Now consider the motion of the coin as viewed by an observer on the ground watching the rotating record. In this inertial reference frame there is no centrifugal force, because the centrifugal force only appears in a non-inertial frame. To the observer in the inertial frame the coin is moving in a circle and must experience a force that provides the required centripetal force for circular motion. That centripetal force is the force of friction directed radially inward.

$\endgroup$
0
$\begingroup$

First, let's imagine a disk with no friction. What would happen if we place a coin on it and rotate it? The coin will have the same angular velocity as the disk. Because of this angular velocity, the coin also possesses a tangential velocity of $v_{T}=\omega r$.

By Newton's first law, we know that the coin will tend to move in the tangential direction because no force is acting on the coin rather than gravity and normal force. If we want to stop the coin from this movement, we need to pull it towards the center with some sort of force. But this force must continuously be applied if we want to keep the coin in the circular motion. That is where the friction comes into play. Friction force must always pull the coin radially inwards to keep it from tangential movement. That is an intuitive way to this about it.

Because the frame of reference is accelerating, we must introduce a pseudo force for the newton laws to be valid. And this pseudo force is called the Centrifugal Force. This fictitious force acts radially outwards and has the same magnitude as the Centripetal Force.

$F_{centrifugal}=-F_{centripetal}=m{\omega}^2 r$

Thus, for the coin to balance in the radial direction another force must balance out the Centrifugal Force. And this force is the Friction. As of now, it is trivial that Centripetal force is provided by the Friction force in order to balance out the Centrifugal force.

$\endgroup$
0
$\begingroup$

As Many People here have given the Mathematical Explanation to this problem, I will try my best to stick to the basic concepts of physics only. The answer can be long so please try to bear it with utmost patience

So lets consider it from start suppose there is a coin on the table and the record suddenly gets turned on so if we look closely at the surface the whole record is moving at an angular velocity of 33.33 rad/s so relative to the surface there will be a tendency of slipping and hence kinetic friction will be there in opposite direction where there is tendency of slipping. this concludes the 1st part of the tangential frictional acceleration of the coin which will make it stable and after some time the relative velocity between them will be zero.

Now Coming back to your problem. Now lets take an example from our surroundings... Suppose you go to an amusement park and there you saw a ride like merry go round which spins at very high speed.. now as soon as you will get into it you will feel a force pushing you away from the ride and you might pick up the safety handle to prevent yourself from falling... what exact force is this.. this is regarded as pseudo force and this is caused due to the fact that as the whole ride is performing a circular motion, centripetal acceleration will be there but as you are sitting on the ride, the ride will be an non inertial reference and you will experience a force away from the ride

A simple FBD of the whole situation

Now Coming to the conclusion.. as the record is performing circular motion centripetal acceleration will be there. so relative to record (or we can say the point just bottom of the coin) there will be a tendency of slipping away from the record.. and using the same argument I used in the beginning of the question.. friction will be there which will oppose the slipping between record and coin and will prevent it from getting away from the disk and as some user mentioned it it will go tangentially which you can observe by rotating a stone with a rope and then letting go of rope. you will find that the stone will go tangentially... so it will be friction only that will stabilize the coin and will help it to perform circular motion with disk.. Hope it helps...

$\endgroup$
-1
$\begingroup$

Dynamic friction opposes relative motion. Static friction doesn't oppose motion itself, it opposes motion the coin would experience if there was no friction.

Static friction is the opposite to the net force the coin would experience without friction, up to some maximum value. If the net force (without friction) reaches that value, the static friction won't be able to counteract other forces, the coin will start moving and the friction will be dynamic friction.

At least in 101 physics courses, we often model dynamic friction by $$\mathbf{F}_{\text{dynamic}} = \mu_{\text{D}} ||\mathbf{N}||$$ where $\mathbf{N}$ is the normal force and $\mu_{D}$ is the dynamic friction coefficient (determined experimentally). Note that dynamic friction does not depend on the speed. The maximum for static friction has an analogous expression $$\mathbf{F}_{\text{static (max.)}} = \mu_{\text{S}} ||\mathbf{N}||\text{.}$$

$\endgroup$
3
  • $\begingroup$ Mind explaining downvote? $\endgroup$
    – Gilgamesh
    Feb 8, 2021 at 17:03
  • $\begingroup$ I did not downvote. But you do seem to be missing the point: there are no horizontal forces acting on the coin aside from friction. Saying the static friction is opposite the net force isn’t helpful without a discussion of pseudo forces. $\endgroup$
    – Ben51
    Feb 16, 2021 at 0:39
  • $\begingroup$ Ok, I agree that sentence is pretty pointless in this scenario if I don't explain pseudo-forces. The answer should be in the first sentence thoguht. Thanks for the comment. $\endgroup$
    – Gilgamesh
    Feb 16, 2021 at 0:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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