Maximizing Rotation Rate for Block and Penny on Spinning Disk

[leena19]

Homework Statement

A penny of mass 3.1g rests on a small 20.0g block supported by a spinning disk of radius(r) 12cm.If the coefficients of friction between block and disk are 0.75(static) and 0.64(kinetic) while those for the penny and block are 0.45(kinetic) and 0.52(static).What is the maximum rate of rotation (in revolutions per minute) that the disk can have before either the block or the penny starts to slip?

Homework Equations

Fr=$$\mu$$R
F=m(omega)²r
where
omega=angular velocity

The Attempt at a Solution

Here's the force diagram I drew,
http://img24.imageshack.us/img24/8028/cicularmotion.png"

but I'm not very sure what I should do next.
I guess I should use Fr or Fr'=$$\mu$$R(or S as labelled in the diagram)
but I don't know which $$\mu$$ to use,the coefficient of static friction or kinetic friction ? and what is the difference between the 2 types of coefficients?
I know that static friction is the friction acting on the block when it is just about to move and is slightly more than the dynamic friction,but I'm not sure what to do with this info?

 

[Doc Al]

Apply Newton's 2nd law to the penny. What's the only centripetal force acting on it? What's the maximum value of that force before slipping occurs? Use that to find the maximum ω.

Since you are trying to avoid slipping, static friction is all you need. (Assuming I understand the problem correctly.)

 

[leena19]

Apply Newton's 2nd law to the penny.

Does this mean, the penny slips first?If so how do we decide it?

What's the only centripetal force acting on it?

The frictional force between the penny and the block,I guess?

What's the maximum value of that force before slipping occurs?

If my answer to the 1st question was correct,then it would be,
Fr' = $$\mu$$s * S which would be equal to mω²r
=$$\mu$$s*m'g = mω²r
ω² = (0.52*10)/(12*10^-2)
ω= 6.58rads^-1

=(6.58*60)/2pi = 62.83 rpm ?

 

[Doc Al]

Does this mean, the penny slips first?If so how do we decide it?

Assume the penny does not slip first, then do a similar analysis for the block. Compare the two speeds.

The frictional force between the penny and the block,I guess?

Sure.

If my answer to the 1st question was correct,then it would be,
Fr' = $$\mu$$s * S which would be equal to mω²r
=$$\mu$$s*m'g = mω²r
ω² = (0.52*10)/(12*10^-2)
ω= 6.58rads^-1

=(6.58*60)/2pi = 62.83 rpm ?

Looks good. (Are you supposed to use g = 10 m/s^2, instead of the usual 9.8?)

 

[leena19]

(Are you supposed to use g = 10 m/s^2, instead of the usual 9.8?)

We're asked to take g=10 if the question doesn't state g as equal to 9.8m/s^2

Assume the penny does not slip first, then do a similar analysis for the block.

OK.So
Fr= $$\mu$$s" * R = Mω²r
But now I'm a bit confused.Is the reaction force(R) on the block equal to the weight of the block alone or equal to the weights of both block and disk?
and is M= m+m' ?
where m' is the mass of the penny and m the mass of the block

 

[Doc Al]

But now I'm a bit confused.Is the reaction force(R) on the block equal to the weight of the block alone or equal to the weights of both block and disk?

I assume you mean "penny", not "disk". The normal force exerted by the disk on the block will equal the combined weight of block + penny.

and is M= m+m' ?
where m' is the mass of the penny and m the mass of the block

Yes, treat the "block+penny" as a single system. (Assume the penny doesn't slip.)

 

[leena19]

I assume you mean "penny", not "disk".

Yes,I meant disk.

Assuming the penny doesn't slip,I get ω= 25 rads^-1 which is more than the maximum angular velocity for the penny which would mean the penny slips first.

So in all such cases is it safe to assume the object on top slips first?

 

[Doc Al]

Yes,I meant disk.

Yes, you meant penny.

Assuming the penny doesn't slip,I get ω= 25 rads^-1 which is more than the maximum angular velocity for the penny which would mean the penny slips first.

Right.

So in all such cases is it safe to assume the object on top slips first?

Not at all. It depends on the coefficient of friction. (Swap the μs and see what happens.)

 

[leena19]

Yes, you meant penny.

Oops!I did it again.

It depends on the coefficient of friction. (Swap the μs and see what happens.)

Tried it.now I understand it.

THANK YOU VERY MUCH!