How Much Static Friction is Needed for a Penny to Stay on a Rotating Record?

[DoctorB2B]

Homework Statement

What is the minimum coefficient of static friction necessary to allow a penny to rotate without sliding off a 33.5rpm record (diameter of these old-fashioned records is 0.3), when the penny is placed at the edge of the record. Ignore the size of the penny.

The Attempt at a Solution

I know the equation to determine the static friction coefficient. However, I am clueless as to how to proceed from here.

 

[rl.bhat]

Which force slides off the penny from the record? What is its magnitude?
What is the direction and magnitude of the frictional force?

 

[nlightner]

I would approach this problem by first understanding the concept of static friction and its relationship to the coefficient of friction. The coefficient of static friction is a measure of the force required to initiate movement between two surfaces in contact with each other. In this case, the two surfaces would be the record and the penny.

To find the minimum coefficient of static friction necessary for the penny to rotate without sliding off the record, we can use the following equation:

μs = F/N

Where μs is the coefficient of static friction, F is the force required to initiate movement, and N is the normal force between the two surfaces. In this case, the normal force would be the weight of the penny acting downward on the record.

To find the force required to initiate movement, we can use the equation:

F = ma

Where F is the force, m is the mass of the penny, and a is the acceleration of the penny. In this case, the acceleration of the penny would be the centripetal acceleration required for it to rotate at 33.5 revolutions per minute (rpm). We can find this acceleration using the equation:

a = ω^2r

Where ω is the angular velocity, which is equal to 33.5 revolutions per minute (rpm) converted to radians per second (rad/s), and r is the radius of the record, which is equal to half of its diameter, or 0.15 meters.

Now, we can plug in the values into the equation for the force required to initiate movement:

F = ma = m(ω^2r)

Next, we can substitute this value for F into the equation for the coefficient of static friction:

μs = F/N = m(ω^2r)/mg

Where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.

Finally, we can plug in the mass of the penny (which we are ignoring the size of), the converted angular velocity, and the radius of the record into the equation to solve for the coefficient of static friction:

μs = (0.01 kg)(2π(33.5 rpm/60 s)^2(0.15 m))/(0.01 kg)(9.8 m/s^2) = 0.16

Therefore, the minimum coefficient of static friction necessary for the penny to rotate without sliding off the record is approximately 0.16. This means that the