Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

$r = \frac{\mu g}{\omega^2}$

$r < \frac{\omega^2}{\mu g}$

$r \le \frac{\mu g}{\omega^2}$

$r \ge \frac{\mu g}{\omega^2}$

Step-by-Step Solution

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.
Centripetal Force: The required centripetal force is $F_c = mr\omega^2$ , where $m$ is the mass of the coin.
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, $f_{s,max} = \mu mg$ .
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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