Physics: Circular motion, acceleration?

Question

I was working on my work for Physics and I just absolutely can't figure out this problem even though I've spent such a long time on it:

A coin is placed on a stereo record, which takes 3 seconds to make a complete revolution. Assume that the coin has a velocity of 5 m/s.

a. What is the acceleration of the coin?

b. What is the distance from the center of the stereo record to the coin?

c. How big is the coefficient of the force of friction between the stereo and the coin?

Could you explain how to figure out the answers? Thanks!

Answer 1

A] The accel would be "0" the coin is in constant velocity - not acceleration.

B] the coin is traveling at 15 m per revolution - this is the circumference - Pi *r squared = circumference     SOLVE THE EQUATION FOR "r"

C] the force of friction is greater than the mass of the coin * the velocity

since the coin is staying in place on the record, the friction ( holding it down) must be higher than the forces driving the coin off the edge of the rotating record...

Answer 2

Let r = distance from centre of the stereo record to the coin.

Distance travelled in one revolution = 2*pi*r

Velocity of the coin = 5 m/s

Therefore, time taken in making one revolution = 2*pi*r/5

This is given as 3 seconds

Therefore, 2*pi*r/5 = 3

Or, r = (3*5)/(2*pi) = 15/(2*3.14) = 15/6.28 = 2.39 m

Acceleration a = v^2/r = 5^2/2.39 = 25/2.39 = 10.46 m/s^2

Force of friction provides centripetal force.

Therefore, force of friction = ma, where m = mass of coin

Normal force on the coin by the stereo record balances weight of coin.

Therefore, normal force = mg

Coefficient of friction = force of friction/normal force = (ma)/(mg) = a/g = 10.46/9.8 = 1.07

Ans:

a) 10.46 m/s^2

b) 2.39 m

c) 1.07

Note: I have made an assumption that the coin does not skid on the stereo record.