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

To find the maximum angular velocity \( \omega \) that can be given to a disc without a coin slipping off, we need to consider the forces acting on the coin due to the rotation of the disc. The key force here is the centripetal force required to keep the coin in circular motion.

The centripetal force needed to keep the coin moving in a circle of radius \( r \) with angular velocity \( \omega \) is given by:

\(F_c = m \omega^2 r\) 

where:

• \(m\) is the mass of the coin,
• \(\omega\) is the angular velocity,
• \(r\) is the distance from the center of the disc to the coin.

The frictional force \( F_f \) between the coin and the disc must be equal to or greater than the centripetal force to prevent slipping. This frictional force is given by:

\(F_f = \mu m g\)

where:

• \(\mu\) is the coefficient of friction,
• \(g\) is the acceleration due to gravity.

For the coin not to slip, the frictional force must be equal to or greater than the centripetal force:

\(m \omega^2 r \leq \mu m g\)

Dividing throughout by \(m\) (assuming \(m \neq 0\)) and solving for \( \omega \), we get:

\(\omega^2 \leq \frac{\mu g}{r}\)

\(\omega \leq \sqrt{\frac{\mu g}{r}}\)

Thus, the maximum angular velocity \( \omega \) that can be given to the disc without the coin slipping is:

\(\omega = \sqrt{\frac{\mu g}{r}}\)

Therefore, the correct answer is: \( \sqrt{\frac{\mu g}{r}} \).

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 \( \omega \):

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