Question

A coin is placed on a disc. The coefficient of friction between the coin and the disc is \( \mu \). If the distance of the coin from the center of the disc is \( r \), the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is:

• A. \( \frac{\mu g}{r} \)
• B. \( \sqrt{\frac{r}{\mu g}} \)
• C. \( \sqrt{\frac{\mu g}{r}} \)
• D. \( \frac{\mu}{\sqrt{r g}} \)

Answer 1

The Correct Option is C

For a coin placed on a rotating disc, the forces acting on it are the normal force \( N = mg \) and the frictional force \( f \) that provides the centripetal force:

\[ f = m \omega^2 r \]

Since the frictional force is given by:

\[ f = \mu N = \mu mg \]

Equating the centripetal force and the frictional force:

\[ \mu mg = m \omega^2 r \]

Simplifying for \( \omega \):

\[ \omega = \sqrt{\frac{\mu g}{r}} \]