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:
[tex]\mu mg = m\omega^2 r[/tex]
where
[tex]\mu = 0.340[/tex] is the coefficient of friction
m is the mass of the coin
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]\omega[/tex] 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:
[tex]m\omega^2 r > \mu m g[/tex]
Solving for [tex]\omega[/tex], we find the angular speed at which this happens:
[tex]\omega = \sqrt{\frac{\mu g}{r}}=\sqrt{\frac{(0.340)(9.8)}{0.11}}=5.5 rad/s[/tex]
