Question

A coin is placed on a disc rotating with an angular velocity \( \omega \). The co-efficient of friction between the disc and the coin is \( \mu \). The maximum distance of the coin from the centre of the disc up to which it will rotate with the disc is

• A. \( \sqrt{\frac{\mu}{\omega^2}} \)
• B. \( \frac{\mu g}{\omega^2} \)
• C. \( \sqrt{\frac{\mu g}{\omega^2}} \)
• D. \( \frac{\mu g}{\omega} \)

Hint:

For circular motion on a rotating disc, static friction provides the centripetal force until its limiting value is reached.

Answer 1

The Correct Option is B

Step 1: Identify the required force.
For the coin to rotate along with the disc, it must get centripetal force towards the centre.

Step 2: Write centripetal force.
If the coin is at distance \( r \) from the centre, then centripetal force is:
\[ F_c = m\omega^2 r \]

Step 3: Identify the source of centripetal force.
The required centripetal force is provided by friction between the coin and the disc.

Step 4: Write maximum frictional force.
Maximum static friction is:
\[ f_{\max} = \mu mg \]

Step 5: Apply limiting condition.
At maximum distance, friction just provides centripetal force:
\[ m\omega^2 r = \mu mg \]

Step 6: Solve for maximum distance.
Cancel \( m \) from both sides:
\[ \omega^2 r = \mu g \] \[ r = \frac{\mu g}{\omega^2} \]

Step 7: Final Answer.
Therefore, the maximum distance is:
\[ \boxed{\frac{\mu g}{\omega^2}} \]