A gramophone record is revolving with an angular velocity $\omega$. A coin is placed at a distance $r$ from th

Question

A gramophone record is revolving with an angular velocity $\omega$. A coin is placed at a distance $r$ from the centre of the record. The static coefficient of friction is $\mu$. The coin will revolve with the record if:

Accepted Answer

1. **Identify Forces:** For the coin to revolve with the record without slipping, it requires a centripetal force directed towards the centre. The only force capable of providing this in the horizontal plane is the static friction ($f_s$) between the coin and the record.
2. **Centripetal Force:** The required centripetal force is $F_c = mr\omega^2$, where $m$ is the mass of the coin.
3. **Friction Limit:** The maximum value of static friction is limiting friction, $f_{s,max} = \mu N$. Since the coin is on a horizontal surface, the normal reaction $N = mg$. Thus, $f_{s,max} = \mu mg$.
4. **Condition for No Slipping:** The coin will revolve with the record only if the required centripetal force is less than or equal to the maximum available static friction.
   $$F_c \le f_{s,max}$$
   $$mr\omega^2 \le \mu mg$$
   Canceling mass $m$ (since $m > 0$):
   $$r\omega^2 \le \mu g$$
   $$r \le \frac{\mu g}{\omega^2}$$
   (Reference: NCERT Class 11, Physics Part I, Chapter 5, Section 5.10, Motion of a car on a level road).

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