This is a solved example in my text book"Engineering mechanics dynamics. R.C.Hibbeler" :

The 3 kg disk is attached to the end of a cord. The other end of the cord is attached to a ball and socket joint located at the center of the platform. If the platform rotates rapidly, and the disk is placed on it and released from rest, determine the time it takes for the disk to reach a spead great enough to break the cord. The maximum tension the cord can sustain is 100 N.. And the coefficient of the kinetic friction between the disk and the platform is 0.1 enter image description here

when the free body diagram of the particle was drawn, the friction force was considered in the tangential direction,, but it wasn't considered in the normal direction(tension was only considered)why? enter image description here

Also, it is written "the frictional force has a sense of direction that opposes the relative motion of the disk with respect to the platform" I don't know how is this thing is true .I was expecting that the friction force acts in the same direction this of the relative motion of the disk with respect to the platform , since this force causes the speed of the disk to increase .


Circular motion requires a centripetal force pointing to the center of the circle that the disk is moving around, and this is provided by the tension in the cable. In addition, the disk is slipping on the turntable, and the kinetic friction force is trying to stop this slippage, so the disk is getting accelerated in a tangential direction, as shown in the diagram. Your problem requires that you use Newton's 2nd law to determine the tangential acceleration of the disk. This will allow you to calculate the tangential velocity, and angular velocity, of the disk with respect to time. Once you know this, you can determine how long the disk will remain on the turntable, because the string breaks when the centripetal force requirement exceeds the breaking point of the string.

  • $\begingroup$ Why isn't there friction force in the radial direction ? $\endgroup$ – Eman.suradi Jul 7 '17 at 18:13
  • $\begingroup$ If the kinetic friction force is trying to stop the slippage, then this will decrease the speed of the disk , but the speed is increasing $\endgroup$ – Eman.suradi Jul 7 '17 at 18:15
  • $\begingroup$ @Eman.suradi, the disk is dropped onto the rotating table, and it is initially at rest. From the disk's perspective, the rotating table is slipping under it. Friction from the rotating table is trying to prevent this slippage, so it is accelerating the disk, in the same direction that the rotating table is turning. This means that indeed, the friction force is trying to stop the slippage, but it can only do this if the disk rotates at the same speed as the rotating table. This necessarily requires rotational acceleration from the disk. $\endgroup$ – David White Jul 8 '17 at 0:33
  • $\begingroup$ There is no friction force in the radial direction because the cord will not let the disk move in that direction. Kinetic friction opposes motion, so where no motion exists, no kinetic friction exists. $\endgroup$ – David White Jul 8 '17 at 0:36

Everything that David White said is helpful. I would also like to point out one of your main misconceptions, which might help you understand the textbook's solution and similar examples.

Also, it is written "the frictional force has a sense of direction that opposes the relative motion of the disk with respect to the platform" I don't know how is this thing is true.

The general rule is that friction opposes the direction of relative motion between two surfaces at contact. If friction were to act in the same direction of relative motion, as you suggested, then anything could move without an applied force.

  • $\begingroup$ If the speed of the disk is increasing , then there must be a force in the direction of motion not in the opposite direction ,, the platform is rotating counterclockwise with some angular speed and the disk is also moving counterclockwise with greater angular speed , and this speed is increasing , so friction must act in the counterclockwise direction , is it true ? $\endgroup$ – Eman.suradi Jul 7 '17 at 18:26
  • $\begingroup$ @Eman.suradi Where do you see that the disk has greater angular speed? The question does not suggest that. $\endgroup$ – JMac Jul 7 '17 at 18:36
  • $\begingroup$ @JMac kinetic friction is given . This means the disk is moving relative to the platform $\endgroup$ – Eman.suradi Jul 7 '17 at 18:37
  • $\begingroup$ @Eman.suradi That just means they are moving relative to each other. The disk starts from rest and is accelerated by the platform. It can never exceed the angular velocity of the platform. If the rope was strong enough to allow the disk to move the same speed as the platform; the disk would stop once it reached that speed; as there would be no relative motion for the dynamic friction; only a static friction keeping it in place. $\endgroup$ – JMac Jul 7 '17 at 18:39
  • $\begingroup$ @JMac I think I misunderstood the scenario of the problem , and you explain it to me .. So this problem is different from putting a disk at a stationary platform and then rotating this platform ,, In this problem ,, static friction will act first and once the disk starts to slip dynamic friction will act on this disk ,, is it true ? $\endgroup$ – Eman.suradi Jul 7 '17 at 18:58

opposes the relative motion of the disk with respect to the platform

If the platform is spinning clockwise, then the relative motion of the disk relative to the platform is counterclockwise... while the force on the disk is clockwise. So yes, that is true.

And as long as the string is attached at the center, it is always pointing radially outward - so the direction of the friction is always perpendicular to the direction the string is pointing. As a result there is no component of the friction along the string.

This means that the only force that is capable of breaking the string is the centripetal force due to the disk. This means you have to solve the question "at what rate of rotation is the centripetal force on a 3 kg object at the end of a 1 m string equal to 100 N?". This happens when

$$\rm F = m\omega^2 r = 100~N$$

With a coefficient of kinetic friction of 0.1, the (linear) acceleration of the mass will be 0.1 g (where g is the acceleration due to gravity, 9.81 m/s$^2$).

Finally, you know that $v = \omega r$. Now you have everything you need to solve this.


This diagram helps to explain why the force on the disk due to friction is perpendicular to the string:

enter image description here

The disk is rotating (slower than the platform). If we look at things in a rotating frame of reference (moving so the disk is stationary), we see the platform is still rotating underneath the disk. The motion of the platform right below the disk is perpendicular to the radial vector (from the center to the disk) and thus the force of friction on the disk will be perpendicular as well.

If you consider the disk is not an infinitesimal point, then there will be points to the left and right where the force is not quite perpendicular to the string - but to the extent there is a force on one side that points out, there is an equal force on the other side that points in. This does leave a small residual torque on the disk: I'm pretty sure that the person asking the question is not asking you to think about that (although it does matter - as the disk starts to rotate about the center of the platform, it also ends up with a rotation about its own center... otherwise the attachment point could not continue to point inwards).

  • $\begingroup$ "so the direction of the friction is always perpendicular to the direction the string is pointing." .Can you help me understand how did you reach to this result , please ? I didn't understand this point $\endgroup$ – Eman.suradi Jul 7 '17 at 22:01
  • $\begingroup$ I have tried to update the answer to address the point you raised in your comment. $\endgroup$ – Floris Jul 7 '17 at 22:23
  • $\begingroup$ @Floris What software do you use to draw these diagrams? They are extremely neat! $\endgroup$ – user139621 Jul 8 '17 at 2:07
  • $\begingroup$ @Blue it's just PowerPoint followed by screen capture... a rare thing that Microsoft did well. $\endgroup$ – Floris Jul 8 '17 at 3:06

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