Why Does the Penny Slide Off the Record at 0.080m?

[mizzy]

Homework Statement

A penny is placed on an LP record that is slowly accelerating up to 78 revolutions per minute. It is found that if the penny is placed at 0.080m or greater from the center, then the penny slides off the edge of the record. Find the coefficient of static friction if the mass of the penny is 0.0032kg.

Homework Equations

Ff = mu x n

a = v2/r

The Attempt at a Solution

I don't know how to start. Can someone guide me please??

 

[inutard]

Ok. Let us begin with this question. You are given the frequency of the LP (and thus the period as well), and you are given the mass of the penny. As you have written above,
Frictional Force = co-eff * Normal and centripetal acceleration = v^2/r = 4Pi^2r*frequency^2. For the penny to stay on the LP, friction has to provide enough centripetal force. When the radius is too great, the velocity of the penny is too large and it flys off the LP. Thus, we are looking for:
Force Friction = Force centripetal = mass of penny* centripetal acceleration.
The rest is plain math and some unit conversions.

 

[mizzy]

Ok. Let us begin with this question. You are given the frequency of the LP (and thus the period as well), and you are given the mass of the penny. As you have written above,
Frictional Force = co-eff * Normal and centripetal acceleration = v^2/r = 4Pi^2r*frequency^2. For the penny to stay on the LP, friction has to provide enough centripetal force. When the radius is too great, the velocity of the penny is too large and it flys off the LP. Thus, we are looking for:
Force Friction = Force centripetal = mass of penny* centripetal acceleration.
The rest is plain math and some unit conversions.

thanks.

Normal is just equal to mg, right?

 

[inutard]

yes. So youll notice that the mass does not actually matter in the question since it cancels out.

 

[rwisz]

I would suggest starting by breaking down the problem into smaller parts and identifying the relevant equations. In this case, we are dealing with rotational motion, so we can use the equation for centripetal acceleration (a = v^2/r) to find the acceleration of the penny on the record.

Next, we can use the coefficient of static friction equation (Ff = mu x n) to find the maximum frictional force that the penny can experience before sliding off the record. Here, n represents the normal force, which in this case is equal to the weight of the penny (mg).

Once we have these two equations, we can set them equal to each other and solve for the coefficient of static friction (mu). This will give us the minimum value of mu that is required for the penny to stay on the record at a given distance from the center.

It's also important to note that the acceleration of the record (78 revolutions per minute) will change the velocity of the penny, so we may need to use the equation for angular velocity (omega = 2pi/T) to find the velocity of the penny at a given distance from the center.

Overall, the key is to identify the relevant equations and use them to solve for the coefficient of static friction. I hope this helps guide you in the right direction.