Centripetal acceleration and coefficient of static friction

AI Thread Summary
The discussion revolves around calculating how far a coin can be placed from the center of a rotating record without slipping, given a coefficient of static friction of 0.3 and a rotation speed of 33.3 rpm. Participants clarify the relationship between centripetal acceleration, angular velocity, and the forces acting on the coin. The key equations involve expressing acceleration in terms of radius and angular speed, ultimately leading to the frictional force equating to the centripetal force. The process involves converting rpm to angular velocity and using it to derive the necessary radius. The final goal is to solve for the maximum radius where the coin remains stationary on the record.
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Homework Statement



A coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the record is 0.3, how far from the center of the record can the coin be placed without having it slip off?

Homework Equations



a=v^2/r

The Attempt at a Solution



I have no idea where to start.
 
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Take the mass of the coin to be m. What's the maximum frictional force acting on the coin? You've got a formula for a in terms of r, equate ma to the previous force and solve for r.
 
By saying that I have a formula for m in terms of r, are you saying talking about the formula I mentioned in my post, or a different formula? Also, when you say take the mass of the coin as m, how which formula would I use for m? I converted the revolutions per minute to per second, thinking that I could use that for my angular velocity, but I'm not sure where to go from there.

Thanks for the quick reply.
 
JSapit said:
By saying that I have a formula for m in terms of r, are you saying talking about the formula I mentioned in my post, or a different formula? Also, when you say take the mass of the coin as m, how which formula would I use for m? I converted the revolutions per minute to per second, thinking that I could use that for my angular velocity, but I'm not sure where to go from there.

Thanks for the quick reply.

I said a formula for 'a' in terms of 'r'. Yes, the one in your post. You will also want to express 'v' in terms of r and your angular speed.
 
Sorry, I'm still a little confused. I'm not sure what you mean by, in your first post, "equate ma to the previous force and solve for r." Also, in your last post, how would I express v in terms of r and angular speed?

I THINK my angular speed is 1.11*Pi rad/sec. Is this right?

Again, thank you for the replies, and being patient with me. Physics is not my strong point.
 
Your angular speed is about right. Call it w. Then v=w*r, right? So a=v^2/r=w^2*r^2/r=w^2*r. Equate m*a=m*w^2*r to your frictional force and solve for r.
 
How do I find the acceleration then?
 
JSapit said:
How do I find the acceleration then?

a=w^2*r. You know w, you don't know r. You can still write the equation even if you don't know r. You are going to solve for it.
 
From v^2/r: V = velocity = m/s we can get meters from the equation for circumference
2*pi*r so v = 2*pi*r/s, once you convert the rpm to rps you multiply the velocity by that, so you should have V= 1.11*pi*r/s

Go back to the equation plugging in our velocity you get (1.11^2*pi^2*r^2)/(s^2*r), which simplifies to 1.2321*pi^2*r (seconds is a unit so I just got rid of it, it isn't used)

Now for the coin not to move the Ff (force of friction) = Fc (Centripetal Force), so Ff = 1.2321*pi^2*r*m, Ff = Fn*.3 = m*g*.3

So m*g*.3 = 1.2321*pi^2*r*m, the m's cancel out and if you plug in and do some rearranging you get r.
 
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