Question

A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of μs = 0.70 and μk= 0.50. The turntable very slowly speeds up.
What is the angular speed in rpm when the coin slides off?

Answer 1

Answer:

$\omega=64.55rpm$

Explanation:

The coin will slide when the centripetal force is too much for the friction to hold, that is, when it is equal to the maximum static friction, so:

$F_{cp}=F_{s}^{max}$

Or:

$ma_{cp}=\mu_sN$

Which means (since on the vertical direction only the normal and weight are acting):

$mr\omega^2=\mu_smg$

So for our values we have:

$\omega=\sqrt{\frac{\mu_sg}{r}}=\sqrt{\frac{(0.7)(9.8m/s^2)}{0.15m}}=6.76rad/s$

Since $1rpm=2\pi rad/60s$, we convert our angular speed using this conversion factor (which is equal to 1 thus not altering our result):

$\omega=6.76rad/s=6.76rad/s(\frac{1rpm}{2\pi rad/60s})=64.55rpm$