# Circular Motion and Friction

1. Feb 2, 2007

Can anyone help me with this problem? I've tried to do part a, but I don't think I'm doing it right.

A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.

a.) The rate of rotation of the disk is gradually increased. The coefficient of static friction between the coin and the disk is 0.50. Determine the linear speed of the coin when it just begins to slip.

FN - Fg - Ff = ma
(FN = mg?)
mg - mg - u(mg) = (mv^2)/r
(.5)(9.8) = (v^2)/.14
v = .83 m/s

b.) If the experiment in part a were repeated with a second, identical coin glued to the top of the first coin, how would this affect the answer to part a? Explain your reasoning.

It would have no effect because the mass cancels out.

Thanks!!

2. Feb 2, 2007

### Staff: Mentor

Realize that Fn and Fg act vertically, while Ff acts horizontally. So you can't just add them all together! Treat vertical and horizontal components separately.

Luckily, Fn = Fg = mg, so your calculation works out OK. (But you'd better redo it so that you understand what you did.)

3. Feb 2, 2007

Oh, yes! How silly of me.
So it would be...
EFy = 0
mg = FN
and
EFx = ma
uFN = (mv^2)/r
umg
ug = (v^2)/r
(.5)(9.8) = (v^2)/.14
v = .83 m/s

Thank you!

4. Feb 2, 2007

### Staff: Mentor

That's more like it.

5. Feb 2, 2007

By the way, would the instantaneous acceleration be directed towards the center of the disk?

6. Feb 2, 2007

### Staff: Mentor

As long as the motion is uniformly circular, the acceleration is centripetal (which just means "towards the center").

But when the coin starts slipping, things get more complicated.