Question

A coin rests 11.0 cm from the center of a turntable. The coefficient of static friction between the coin and turntable surface is 0.340. The turntable starts from rest at t = 0 and rotates with a constant angular acceleration of 0.820 rad/s2. (b) At what angular speed (ω) will the coin start to slip?

Answer 1

Answer:

5.5 rad/s

Explanation:

The friction between the coin and the turntable provides the centripetal force that keeps the coin in circular motion. Therefore, we can write:

$\mu mg = m\omega^2 r$

where

$\mu = 0.340$ is the coefficient of friction

m is the mass of the coin

$g=9.8 m/s^2$ is the acceleration of gravity

$\omega$ is the angular speed

r is the distance of the coin from the centre of rotation

In this problem,

r = 11.0 cm = 0.11 m

The coin starts to slip when the centripetal force becomes larger than the maximum frictional force:

$m\omega^2 r > \mu m g$

Solving for $\omega$, we find the angular speed at which this happens:

$\omega = \sqrt{\frac{\mu g}{r}}=\sqrt{\frac{(0.340)(9.8)}{0.11}}=5.5 rad/s$