Question

A DVD drive is spinning at 100.0 rpm. A dime (2.00 gm) is placed 3.00 cm from the center of the DVD. What must the coefficient of friction be to keep the dime on the disk?

Answer 1

Answer:

0.3375

Explanation:

w = angular speed of the DVD drive = 100.0 rpm = $100.0 \frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}$ = $10.5\frac{rad}{sec}$

m = mass of the dime = 2 g = 0.002 kg

r = radius = 3 cm = 0.03 m

μ = Coefficient of friction

The frictional force provides the necessary centripetal force to move in circle. hence

frictional force = centripetal force

μ mg = m r w²

μ g =  r w²

μ (9.8) =  (0.03) (10.5)²

μ = 0.3375