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

• A. $r=\mu g\omega^2$
• B. $r < \frac{\omega^2}{\mu g}$
• C. $r \le \frac{\mu g}{\omega^2}$
• D. $r \ge \frac{\mu g}{\omega^2}$

Answer 1

The Correct Option is C

To determine the condition under which the coin will revolve with the rotating gramophone record, we need to analyze the forces acting on the coin. When the coin is on the record, it tends to slide outward due to centrifugal force. The static friction between the coin and the record provides the necessary centripetal force to keep the coin in circular motion.

Let's consider the forces involved:

• The centrifugal force acting on the coin is given by F_{\text{centrifugal}} = m \omega^2 r, where m is the mass of the coin, \omega is the angular velocity of the record, and r is the distance of the coin from the center of the record.
• The frictional force, which acts as the centripetal force, is given by F_{\text{friction}} = \mu m g, where \mu is the static coefficient of friction and g is the acceleration due to gravity.

For the coin to revolve with the record without slipping, the frictional force must be greater than or equal to the centrifugal force:

F_{\text{friction}} \ge F_{\text{centrifugal}}

Substituting the expressions for these forces, we get:

\mu m g \ge m \omega^2 r

Canceling out m from both sides (assuming m \neq 0), we arrive at:

\mu g \ge \omega^2 r

Rearranging for r, we have:

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

Thus, the coin will revolve with the record if the distance from the center, r, is less than or equal to \frac{\mu g}{\omega^2}. This is the condition needed to prevent the coin from sliding off the record.

The correct answer is: r \le \frac{\mu g}{\omega^2}