Stuck on concept of a particle moving on horizontal circle (m3) :help:

1)On the point of slipping, its currently in equilibrium with respect to its radial distance from the centre of the disk.
2)I dont think its important what force is moving the disk. It does not however rotate due to the presence of the particle.

"Here, the centripetal force is provided by the frictional force F N between the particle and the disc."
This means the centripetal force on the particle is provided by the frictional force between the particle and the disk. $F_(centripetal) = \frac{mv^2}{r} = F_(friction) = \mu R$

3)It is the fact that the disk is not smooth that is causing the particle to move in a circle. If the disk were smooth there would be no centripetal force acting on the particle. If there were no other forces acting the particle would remain stationary.

1)On the point of slipping, its currently in equilibrium with respect to its radial distance from the centre of the disk.
2)I dont think its important what force is moving the disk. It does not however rotate due to the presence of the particle.

"Here, the centripetal force is provided by the frictional force F N between the particle and the disc."
This means the centripetal force on the particle is provided by the frictional force between the particle and the disk. $F_(centripetal) = \frac{mv^2}{r} = F_(friction) = \mu R$

3)It is the fact that the disk is not smooth that is causing the particle to move in a circle. If the disk were smooth there would be no centripetal force acting on the particle. If there were no other forces acting the particle would remain stationary.

Thanks, one thing though, if it slips, in which direction does the particle move? Towards the centre? I just cannot bring myself imagine a particle moving on rotating cd, it just seems to fall off!
Oh got it ,the particle moves in circle.

The radial distance of the particle would increase.

The centrifugal force, due to the rotation, acts outwards.
The centripetal force which counteracts this is provided by the friction between the mass and the disk and this acts in the opposite direction to the centrifugal force, i.e. towards the centre.
On the point of slipping means that the frictional force is only just strong enough to counteract the centrifugal forces; if the centrifugal force were any stronger (i.e. the disk was spinning faster) then the mass would fly off away from the centre.
When a body is on the point of slipping then the formula F="mu" R (where "mu" is the coefficient of static friction) can be used. If it wasn't slipping then F would be less than "mu" R.

If the disk was smooth then there is no interaction between the disk and the mass, so nothing would happen.

Whoa. Let's just calm down.

I am not really accusing you of grossly not understanding this problem, and I am sure that you are a very able persson when it comes to the core modules.


Unfortunately, after reading the first post, it is clear you have not grasped Newton's Laws. You may be able to remeber them and Believe in them, but you do not understand them.

I really think if you stopped your M£ work now, and really tried to secure what is going on in the way forces are concerned, etc, you would understand how the forces come into play for these types of questions.

Sorry, can't explain. Ignore the comment.

But they say centrifugal force is fictious

Centrifugal force is only fictitious in a non-rotating referernce frame (i.e. one in which newton's laws apply). If you want to look at the problem from the frame of reference of the particle you can do, but it may also be solved from the frame of reference of a stationary observer.

In this case I would simply use the non-rotating refernce frame, $\frac{mv^2}{r} = \mu R$ If the frictional force is constant, what happens if the disk spins faster given that $v = r\omega$ ? Play around with these and you should find r must increase. These equations apply when looking as an outside observer, and do not require cetrifugal force to be invoked.