# Coin on a turntable

1. Jul 30, 2013

### pluspolaritons

If I put a coin on a turntable at some distance away from the center and start turning the turntable eventually there will be a speed where the coin will fly off from the turntable.

If we put this in to calculation. We will equate the centripetal force and static friction to find out the point where the coin will fly off.

Now my question is this. The centripetal force points to the center and the static friction points the opposite direction. If centripetal force overcomes the friction, it seems to suggest the force will point towards the center instead of radially outward. But by common sense we know that the coin will fly off instead of going to the center when the revolution speed is high enough. Can someone explain why the coin fly off instead of going to the center of the turntable?

Thanks.

2. Jul 30, 2013

### pluspolaritons

OK here's an after thought, please let me know if I'm correct:

Maybe the coin does in fact go towards the center. But the point is it is no longer in a circular path, so it will eventually fly off the turntable once the static friction is overcome.

3. Jul 30, 2013

### HallsofIvy

The centripetal force is the friction force- which points toward the center. It has to overcome the "natural" motion in a straight line, tangent to the circular motion. Once that motion becomes sufficient to overcome the friction force, the coin moves "naturally", tangent to the circular motion.

4. Jul 30, 2013

### technician

When the friction force is not great enough to provide the centripetal force ( because speed is greater)
You should not describe the motion of the coin as " flying away from the centre"...... The coin moves in a straight line at a tangent to its circle....

Last edited: Jul 30, 2013
5. Jul 30, 2013

### 256bits

Not sure if the coin moves off with constant tangentially velocity. Dynamic friction continues to act on the coin causing acclerations.

6. Jul 30, 2013

### technician

I agree, if there was no further friction I think it would be simple tangential.
The coin may hit a part of the disc where the friction is great enough to make it move in circular motion again.

7. Jul 30, 2013

### rcgldr

As mentiond above, even when the coin slides, there is still kinetic (sliding) friction involved, so the coin will move in a curved path of increasing radius and increasing speed.

8. Jul 31, 2013

### pluspolaritons

Shouldn't the friction force always be in a direction opposite to the force applied? In this case the force is the centripetal force and the friction force should be in a direction opposite to the centripetal force?

9. Jul 31, 2013

### pluspolaritons

technician - i didn't mean that the coin will fly away from the center, but just simply it will fly away.

256bits/technician/rcgldr - It is possible that the coin will return to circular path if the turntable is big enough. But once the static friction is overcome it takes less force to keep the coin moving so once the energy is depleted enough by the kinetic friction, the coin should once again return to a circular path with wider radius.

10. Jul 31, 2013

### sophiecentaur

What is your point here? The centripetal force is towards the centre (which is what the term 'centripetal' means). If it is not great enough then the coin will no longer follow the same radius of circular motion. The path taken, when it slips, will be along a curve, with a radius determined by the force and the direction it acts. This curve will somewhere between a tangential line and the original circle. Actual details could be hard to work out but not impossible to do numerically, I think.
Are you confusing the reaction force on the turntable with the actual friction force on the coin?

11. Jul 31, 2013

### pluspolaritons

What I meant is shouldn't the friction force be in the opposite direction of the centripetal force (i.e pointing outward)? I thought that was the definition of friction.

12. Jul 31, 2013

### sophiecentaur

What force, other than friction, is available to keep the coin on its circular path? Sort out the answer to that and you will have answered your own question, I think.

13. Jul 31, 2013

### Tanya Sharma

Hi pluspolaritons

The basic flaw in your understanding is thinking centripetal force as a force just like gravity,tension,friction.It isn't.

The centripetal force is the net force acting on the rotating coin i.e mv2/r . It is same as ma in F = ma.

The actual force acting on the rotating coin is the static friction which provides the required centripetal acceleration .

Last edited: Jul 31, 2013
14. Jul 31, 2013

### technician

There are 2 aspects to friction in this discussion.
If you picture an object on a linear conveyor belt then there is a friction force keeeping the object in contact.
With a rotating table this friction force is still present but some of the friction force now acts towards the centre....to provide the centripetal force. It means that the resultant friction force is at some angle between the radial direction and the tangential direction.
A good analogy is a car turning in a circle, the centripetal force arises from some of the friction force being 'directed' towards the centre by turning the steering wheel to turn the front wheels.
If I had to demonstrate the effect I can imagine a turntable with a ball bearing on the table. As the table rotates there is effectively no (very little) friction between the ball and the table so the table will rotate underneath the ball. I could arrange a radial, vertical arm fixed to the table so that the arm will make contact with the ball, if the vertical wall of the arm was covered in some friction material (sandpaper?) so that the force of friction between the ball and the arm was great enough to prevent sliding along the wall then the ball would rotate in circular motion on the table.
There is then only 1 effective friction force...radial and it would be the centripetal force.

I have made similar demonstrations to show some aspects of circular motion but not exactly as I have described here.....must now do it !!

PS... agree with Tanya...'centripetal' means resultant (net) force in relation to circular motion. In the same way 'restoring' means resultant in relation to SHM

15. Jul 31, 2013

### WannabeNewton

No that is not the definition of friction. Friction simply opposes relative motion that would occur in its absence. In the absence of a radially inward friction force on the puck from the turntable, the puck will acquire a non-vanishing outward radial velocity across the turntable when you turn the turntable so if you want the puck to have vanishing radial velocity across the turntable (i.e. travel in a circle) then the turntable needs to have a sufficient radially inward friction force acting on the object. This radially inward friction force is the centripetal force.

16. Jul 31, 2013

### WannabeNewton

There is clearly a radial component to the velocity of the puck if it is not traveling in a circle; it's distance from the origin is not constant in time. This is standard terminology.

17. Jul 31, 2013

### sophiecentaur

As the coin moves over the surface, friction will have two components, one which is normal to the direction of motion (providing some centipetal force to give it a curved path) and one which is in a direction opposite to its velocity.

18. Jul 31, 2013

### 256bits

As a sidenote, that leads me to address that the viewpoint of the observer or setup of the axis is important, and if not chosen correctly can lead to some bizarre intepretations.

for example:
One could argue that a rotating ball moving off tangentially when the string breaks has the same radial and tangential velocities along the path as the ball did when the string broke. Since there is no force acting on the ball to change either velocity that must be the case, and vice versu.

On the otherhand, one could also argue that since the radial distance from ball to the centre is changing, there MUST be a radial velocity. And on further analysis, as time progresses and the ball gets farther and farther away, the tangential velocity becomes zero and the ball aquires only radial velocity. The obvious conclusion here would be that there must be some unkown force acting upon the ball to change the tangential velocity into a radial velocity.

The two scenarios appear to be in contradiction. In one case there is obviously no forces acting upon the ball and in the second there obviously must be.

Well, note that both have different chosen axis. If we take the z-axis as the axis of the centre.

In the first case the x and y axis are chosen to be parallel and perpendicular to the tangent of the circle of the initial velocity of the ball when the string breaks.

In the second case, the axis are chosen to be centred upon the moving ball and ROTATING. The ROTATION of the radial axis ( the line connecting the ball and the z-axis ) and tangential axis gives rise to the ficticious force acting upon the ball.

Wasn't there some kind of problem with the motion of celestial objects years ago due to an ill-chosen, but understandable, viewpoint from an earth-center, that baffled thinkers for centuries. ( ie Copernicus and Aristotle )

PS. was technician trying to impress the same idea ( some posts are deleted and I had no chance to read them)

19. Jul 31, 2013

### D H

Staff Emeritus
That's not right, no matter how you determine velocity. In a frame rotating with the turntable, the frictional force is directed against the velocity vector, period. However, this frame also has centrifugal and coriolis accelerations that contribute to the acceleration (rotating frame).

From the perspective of an inertial frame, the frictional force is the only horizontal force present. I'll use a frame in which the center of the turntable is fixed. Once again, the frictional force is directed against the relative velocity between the coin's velocity and the point on the turntable directly underneath the coin. That point on the turntable beneath the coin will be moving faster than the coin as the coin migrates outward. Because that point is moving faster than the coin, friction with the turntable will act to speed the coin up a bit (but never to a speed as fast as that point on the turntable beneath the coin).

20. Aug 1, 2013

### technician

256: I did make some posts along those lines but they were considered to be 'irrelevant' so I deleted them rather than get involved in such discussion.
My post 14 and sophiecentres contributions are good for me.

Last edited: Aug 1, 2013
21. Aug 1, 2013

### D H

Staff Emeritus
technician, your post #14 is wrong, for (at least) two reasons. First off, a ball rolling on a turntable and a coin sliding on a turntable are very different problems. Secondly, in neither case is the force centripetal.

This thread is about a coin on a turntable, so it's best to keep the discussion to this topic.

Suppose we place a coin on a non-rotating turntable and slowly increase the turntable's rotation rate. We'll stop increasing the rotation rate at the instant the coin starts slipping. I'll define μ (mu) as the ratio of kinetic friction to static friction, and set the axes such that at the instant the coin starts slipping, xhat points from the center of the turntable to the coin and yhat along the coin's inertial velocity vector. Below is a family of lab frame trajectories on the turntable that result from this:

Focusing on the mu=0.9 curve, here's a plot of the lab frame velocity and acceleration vectors at various points along the trajectory:

I generated the above using a python-based simulation and gnuplot. Here's the python script:
Code (Text):

# Coin sliding off a turntable simulation.
#
# Usage:
#   python <script_name> [mu_value [print_option]]
#
# Arguments:
#   mu_value - The fractional part of the ratio of kinetic to static friction.
#   print_option - 0 to print summary only, 1 to print data for plotting.
#
# Example:
#   The following simulates a coin sliding off a turntable with mu=0.8.
#   Data suitable for use with gnuplot are saved to the files r8.dat (position),
#   v8.dat (velocity vectors), and a8.dat (acceleration vectors).
#
#   python <script_name> 8 1

from __future__ import print_function
import copy
import math
import sys

# Compute the inertial acceleration vector due to kinetic friction.
# The magnitude of the acceleration is constant but the direction is
# against the relative velocity between the coin and the point on the
# turntable directly beneath the coin.
def accel (x, v, omega, amag) :
rsq = x[0]*x[0] + x[1]*x[1]
rmag = math.sqrt(rsq)
rhat = [x[0]/rmag, x[1]/rmag]
rw = rmag*omega
vp = [-rw*rhat[1], rw*rhat[0]]
vrel = [v[0]-vp[0], v[1]-vp[1]]
vrelsq = vrel[0]*vrel[0] + vrel[1]*vrel[1]

if (vrelsq < 1e-15) :
a = [-amag*rhat[0], -amag*rhat[1]]
else :
vrelmag = math.sqrt(vrelsq)
vrelhat = [vrel[0]/vrelmag, vrel[1]/vrelmag]
a = [-amag*vrelhat[0], -amag*vrelhat[1]]

return a

# Advance state by timestep=dt using RK4.
def step (x, v, omega, amag, dt) :
x0 = copy.copy(x)
v0 = copy.copy(v)
a0 = accel (x, v, omega, amag)
(x, v) = step_pos_vel (x0, v0, v0, a0, 0.5*dt)

v1 = copy.copy(v)
a1 = accel (x, v, omega, amag)
(x, v) = step_pos_vel (x0, v1, v0, a1, 0.5*dt)

v2 = copy.copy(v)
a2 = accel (x, v, omega, amag)
(x, v) = step_pos_vel (x0, v2, v0, a2, dt)

a = accel (x, v, omega, amag)
v3 = [(v0[0] + 2.0*(v1[0]+v2[0]) + v[0]) / 6.0,
(v0[1] + 2.0*(v1[1]+v2[1]) + v[1]) / 6.0]
a3 = [(a0[0] + 2.0*(a1[0]+a2[0]) + a[0]) / 6.0,
(a0[1] + 2.0*(a1[1]+a2[1]) + a[1]) / 6.0]

(x, v) = step_pos_vel (x0, v3, v0, a3, dt)

return (x, v)

def step_pos_vel (x, xdot, v, vdot, dt) :
return ([x[0]+xdot[0]*dt, x[1]+xdot[1]*dt],
[v[0]+vdot[0]*dt, v[1]+vdot[1]*dt])

# Integrate state and record data until the coin flies off the turntable.
def integ (x, v, omega, amag, dt, rfinal, mode, smu) :
t = 0

if (mode > 0) :
fd_r = open ("r" + smu + ".dat", 'w')
fd_v = open ("v" + smu + ".dat", 'w')
fd_a = open ("a" + smu + ".dat", 'w')

while (vector_mag(x) < rfinal) :
if (mode > 0) :
print_xva (t, x, v, omega, amag, fd_r, fd_v, fd_a)
(x, v) = step (x, v, omega, amag, dt)
t = t + dt

if (mode > 0) :
print_xva (t, x, v, omega, amag, fd_r, fd_v, fd_a)

print ("t=", t)
print ("x=", x[0], ', ', x[1], ', magnitude = ', vector_mag(x))
print ("v=", v[0], ', ', v[1], ', magnitude = ', vector_mag(v))

def print_xva (t, x, v, omega, amag, fd_r, fd_v, fd_a) :
print (x[0], x[1], file=fd_r)
if (abs(int(t*5+0.5) - t*5) < 1e-12) :
a = accel (x, v, omega, amag)
print (x[0], x[1], v[0], v[1], file=fd_v)
print (x[0], x[1], a[0], a[1], file=fd_a)

def vector_mag (vec) :
return math.sqrt (vec[0]*vec[0] + vec[1]*vec[1])

def main () :

# Get the fractional mu value and output option from the argument list.
# mu is the ratio of kinetic friction to static friction;
# the value must be in the range [0.0, 1.0). This is enforced
# by prefixing "0." to the input value.
if (len(sys.argv) > 1) :
smu = sys.argv[1]
mu = float("0." + smu)
else :
smu = "8"
mu = 0.8

# mode = 0 means print just summary data to stdout.
# A non-zero value prints position, velocity, and acceleration
# data to files in a format suitable for use with gnuplot.
if (len(sys.argv) > 2) :
mode = int(sys.argv[2])
else :
mode = 0

# Turntable angular velocity, in radians per time unit.
omega = 1

# The integration time step.
# This is close to optimal for RK4 for this value of omega.
dt = 0.005

# The initial radius of the coin and and the radius of the turntable.
rinit = 1
rfinal = 4

# The magnitude of the acceleration due to kinetic friction.
amag = mu*rinit*omega*omega

# Initial state; coin just started slipping (so non-slip velocity).
x = [rinit, 0.0]
v = [0.0, rinit*omega]

# Integrate state and record data.
integ (x, v, omega, amag, dt, rfinal, mode, smu)

main()

#### Attached Files:

File size:
26.5 KB
Views:
487
• ###### coin_turntable_vec.jpg
File size:
17.1 KB
Views:
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22. Aug 1, 2013

### technician

I know they are different!
My post 14 is to highlight a demonstration I would do to show the friction force on the 'object'
Towards the centre,,,,,the centripetal force. I referred to a ball so that no 'linear/tangential' (my terms) friction would complicate matters. I suggested that sandpaper stuck to the vertical, radial arm in my demonstration could provide the friction. The same effect could be demonstrated with a small magnet on the arm to hold the ball bearing in place. As long as a clear demonstration of how the centripetal force is provided.

PS your diagrams show it perfectly

23. Aug 1, 2013

### D H

Staff Emeritus
Shows what perfectly? I don't know what you're seeing, but what I see is that the frictional force is not centripetal. The force vectors don't point to a common center and they are not normal to the velocity vector.

24. Aug 1, 2013

### technician

because you have friction acting over the whole table not just in a radial directio....that is what my ball bearing on a rotating table is to illustrate.
It looks to me as though you definetly have a component of acceleration towards the centre
as well as a component 'tangential?'

Diagrams look ok for me.

25. Aug 1, 2013

### sophiecentaur

Those two diagrams are very useful - especially the red / green vector one. The fact that the velocity vectors are increasing in length must imply that there is some tangential force component and that implies that the acceleration is not quite towards the centre of the curve at each point (i.e not quite normal).
@techncian: your comments still seem to imply that you are still seeing this from a bit 'off beam'. How can a force act "all over the table"? It can only act in one direction at a time and only at the point of contact with the coin. It would be easier if you tried to use less 'personalised' terms if you want to converse usefully on a topic. That's the way Science works best.
@DH It would be interesting to see how the rate of energy supplied (work done) by the motor increases as the path of the coin progresses. The forces increase and so does the speed.