Question

A coin placed on a rotating table just slips when it is placed at a distance of 1 cm from the center. If the angular velocity of the table is halved, it will just slip when placed at a distance of ----- from the centre:

• A. 6 cm
• B. 4 cm
• C. 2 cm
• D. 1 cm

Hint:

When the angular velocity is reduced, the distance at which the coin slips increases according to the square of the ratio of the angular velocities.

Answer 1

The Correct Option is B

The frictional force is responsible for causing the coin to slip. This force is given by \( f_r = m \cdot \omega^2 r \), where \( \omega \) is the angular velocity and \( r \) is the distance from the center.
Given that the angular velocity is halved, we use the equation \( \omega^2 \cdot r^2 = \text{constant} \). When the angular velocity is halved, the distance \( r \) from the center at which the coin slips will be: \[ r_2 = 4 \, \text{cm} \] Thus, when the angular velocity is halved, the coin will slip at a distance of 4 cm from the center.