# Centripetal acceleration and coefficient of static friction

• JSapit
In summary, to find the maximum distance from the center of the rotating record that the coin can be placed without slipping off, we can use the formula a = v^2/r, where v is the velocity of the record and r is the distance from the center. We can also use the formula Ff = Fn*.3, where Ff is the force of friction, Fn is the normal force, and .3 is the coefficient of static friction. By equating these two equations, we can solve for r and find the maximum distance.
JSapit

## 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?

a=v^2/r

## The Attempt at a Solution

I have no idea where to start.

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.

## 1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and its magnitude can be calculated using the formula a = v^2/r, where v is the velocity of the object and r is the radius of the circular path.

## 2. How does centripetal acceleration relate to coefficient of static friction?

Centripetal acceleration is dependent on the coefficient of static friction in the case of an object moving in a circular path on a horizontal surface. The coefficient of static friction is the maximum frictional force that can be exerted on an object before it starts to slide. It is this force that provides the centripetal acceleration necessary to keep the object in its circular path.

## 3. What is the significance of the coefficient of static friction in centripetal acceleration?

The coefficient of static friction is important in centripetal acceleration because it determines the maximum speed at which an object can move in a circular path on a given surface without slipping. If the speed of the object exceeds this maximum value, it will start to slip and lose its circular motion.

## 4. How does changing the coefficient of static friction affect centripetal acceleration?

If the coefficient of static friction is increased, the maximum speed at which an object can move in a circular path without slipping also increases. This means that the centripetal acceleration required to keep the object in its circular path also increases. Similarly, if the coefficient of static friction decreases, the maximum speed and the required centripetal acceleration decrease as well.

## 5. How can I calculate the coefficient of static friction needed for a given centripetal acceleration?

You can use the formula F = μs * m * a to calculate the coefficient of static friction required for a given centripetal acceleration, where F is the maximum frictional force, μs is the coefficient of static friction, m is the mass of the object, and a is the centripetal acceleration. Rearranging this formula, we get μs = F/(m*a). So, if you know the maximum frictional force and the mass of the object, you can calculate the coefficient of static friction needed for a specific centripetal acceleration.

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