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

Answer: (C)

$\sqrt{\frac{\mu g}{r}}$

Answer 2

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 $\omega$:

$\omega = \sqrt{\frac{\mu g}{r}}$