Why Does a Coin Fly Off a Turntable?

  • Thread starter pluspolaritons
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In summary: the coin will fly off if the speed is too great. and the coin will eventually return to the center if the speed is low enough.
  • #1
pluspolaritons
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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 into 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.
 
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  • #2
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
pluspolaritons said:
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 into 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.
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.

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.
 
  • #4
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...
 
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  • #5
Not sure if the coin moves off with constant tangentially velocity. Dynamic friction continues to act on the coin causing acclerations.
 
  • #6
256bits said:
Not sure if the coin moves off with constant tangentially velocity. Dynamic friction continues to act on the coin causing acclerations.

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
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
HallsofIvy said:
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.

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
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
pluspolaritons said:
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?

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
sophiecentaur said:
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?

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
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
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 .
 
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  • #14
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
pluspolaritons said:
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.
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
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
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
WannabeNewton said:
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.

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 acquires 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)
 
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  • #19
sophiecentaur said:
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.
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
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.
 
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  • #21
technician said:
My post 14 and sophiecentres contributions are good for me.
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:

attachment.php?attachmentid=60618.jpg
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:

attachment.php?attachmentid=60619.jpg
I generated the above using a python-based simulation and gnuplot. Here's the python script:
Code:
# 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)# Advance position and velocity.
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()
 

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  • #22
D H said:
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:

attachment.php?attachmentid=60618.jpg



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:

attachment.php?attachmentid=60619.jpg



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


# Advance position and velocity.
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()


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
technician said:
PS your diagrams show it perfectly
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
D H said:
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.

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 definately have a component of acceleration towards the centre
as well as a component 'tangential?'

Diagrams look ok for me.
 
  • #25
D H said:
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.

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.
 
  • #26
sophiecentaur said:
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.
I think we all agree! the only issue is about 'language' I write in the way I would talk to students watching me with the model...nothing wrong with that.
'all over the table'...OK...wherever the coin is on the table there is friction.
I (in my class) would like to 'remove'/'ignore' the 'tangential' friction which keeps the coin traveling along the circle... I want to try to show/talk about the friction responsible for keeping the coin in circular motion... I think these are 2 separate aspects of the friction...ω2r for the centripetal force... no idea what the other one is.

So... I think, regardless of language, we agree that
1) there is a force of friction between the coin and the table
2) the radial component of this friction force is the centripetal force
3)the 'tangential' component stops the table slipping under the coin

DH's diagrams illustrate this nicely...the green arrow is not at 90o to the red arrow, except near the start so the green arrow is the resultant of 2 friction forces.

My ball bearing demo on a rotating table effectively 'removes' the tangential component of friction. I also have a rotating railway track that does the same thing.

I am perfectly content with everything here.

PS...DH's diagrams are the closest anyone has come to showing what could be called 'free body diagrams' the diagrams I draw go well with these diagrams...it is a great pity more free body diagrams are not drawn to illustrate these problems.
 
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  • #27
The centripetal force is not "radial" unless the motion is in a circle about the turntable spindle (no slipping). The centre about which it is orbiting is not the spindle.
The tangential component speeds the coin up and the coin is slipping.
I hope you sort this out before confusing your students.
 
  • #28
sophiecentaur said:
The centripetal force is not "radial" unless the motion is in a circle about the turntable spindle (no slipping). The centre about which it is orbiting is not the spindle.
The tangential component speeds the coin up and the coin is slipping.
I hope you sort this out before confusing your students.

I read your comments.
No need for you to worry about my students.
thanks for your concern for them
 
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  • #29
technician said:
I think we all agree!
I think not.


the only issue is about 'language' I write in the way I would talk to students watching me with the model...nothing wrong with that.
There is a whole lot wrong with you teaching your students non-standard and arguably wrong nomenclature.


So... I think, regardless of language, we agree that
1) there is a force of friction between the coin and the table
2) the radial component of this friction force is the centripetal force
3)the 'tangential' component stops the table slipping under the coin
1. Correct. However, if the only thing we can agree on is the fact that a frictional force exists, we might as well hang it up and call it a day.

2. Very incorrect. By this definition, even a centrifugal force such as that between two charged particles with the same sign charge qualifies as centripetal. The term you should be using is transverse, not centripetal.

3. Also incorrect. Now you are talking about angular momentum. In general, both the transverse and tangential components of the frictional force result in a torque on the turntable.


DH's diagrams illustrate this nicely...the green arrow is not at 90o to the red arrow, except near the start so the green arrow is the resultant of 2 friction forces.
More nonsense. There is only one frictional force. Yes, you can split that into components in an arbitrary number of ways, but this partitioning is arbitrary. You are mistaking the map for the territory here.


My ball bearing demo on a rotating table effectively 'removes' the tangential component of friction. I also have a rotating railway track that does the same thing.
There are two things wrong here. One is that you are wrong. I suggest you do some reading on the problem of a ball rolling on a turntable. The other is that this is off-topic. Create your own thread on this topic if you wish. Do not bring it up again in this thread.


sophiecentaur said:
hope you sort this out before confusing your students.
I hope so, too.
 
  • #30
D H said:
I think not.
There is a whole lot wrong with you teaching your students non-standard and arguably wrong nomenclature.
1. Correct. However, if the only thing we can agree on is the fact that a frictional force exists, we might as well hang it up and call it a day.

2. Very incorrect. By this definition, even a centrifugal force such as that between two charged particles with the same sign charge qualifies as centripetal. The term you should be using is transverse, not centripetal.

3. Also incorrect. Now you are talking about angular momentum. In general, both the transverse and tangential components of the frictional force result in a torque on the turntable.
More nonsense. There is only one frictional force. Yes, you can split that into components in an arbitrary number of ways, but this partitioning is arbitrary. You are mistaking the map for the territory here.
There are two things wrong here. One is that you are wrong. I suggest you do some reading on the problem of a ball rolling on a turntable. The other is that this is off-topic. Create your own thread on this topic if you wish. Do not bring it up again in this thread.
I hope so, too.

Thank you for your comments...duly noted

Time to close this thread then?
 
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1. Why does a coin fly off a turntable?

A coin flies off a turntable due to a combination of centripetal force and inertia. As the turntable spins, the coin is subject to centripetal force, which pulls it towards the center of rotation. However, the coin also has inertia, which causes it to resist changes in its motion. When the centripetal force becomes too weak to keep the coin on the turntable, the coin will fly off in a straight line tangent to the edge of the turntable.

2. Does the weight of the coin affect its flight off the turntable?

Yes, the weight of the coin does affect its flight off the turntable. Heavier coins have more inertia, making it more difficult for the centripetal force to keep them on the turntable. As a result, heavier coins are more likely to fly off the turntable compared to lighter coins.

3. Can the speed of the turntable affect the flight of the coin?

Yes, the speed of the turntable can affect the flight of the coin. The faster the turntable spins, the stronger the centripetal force acting on the coin. This means that the coin will require more inertia to stay on the turntable, making it more likely to fly off at higher speeds.

4. Is the shape of the turntable important in determining the flight of the coin?

Yes, the shape of the turntable can play a role in determining the flight of the coin. A flat turntable will provide a more stable surface for the coin to spin on, making it less likely to fly off. However, a curved or sloped turntable may cause the coin to roll or slide off more easily due to the change in direction of the centripetal force.

5. Can other factors, such as air resistance, affect the flight of the coin off the turntable?

Yes, other factors such as air resistance can affect the flight of the coin off the turntable. Air resistance can create a force that opposes the motion of the coin, making it more difficult for the coin to stay on the turntable. This can be especially noticeable at higher speeds or with lighter coins that are more affected by air resistance.

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