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. 2 cm
• C. 4 cm
• D. 1 cm

Answer 1

The Correct Option is C

To solve the problem of the coin slipping on the rotating table, we need to understand the relationship between the angular velocity, the radius at which the coin is placed, and the centripetal force required to keep the coin moving in a circle.

The centripetal force required to keep the coin from slipping is given by the formula:

\(F_c = m \cdot r \cdot \omega^2\)

where \(F_c\) is the centripetal force, \(m\) is the mass of the coin, \(r\) is the radius from the center, and \(\omega\) is the angular velocity.

Initially, the coin just slips at a distance of 1 cm. Thus, the equation becomes:

\(F_c = m \cdot 1 \cdot \omega^2\)

Now, if the angular velocity of the table is halved, the new angular velocity is \(\frac{\omega}{2}\). Let the new distance where the coin just slips be \(r'\). The new force equation will be:

\(F_c = m \cdot r' \cdot \left(\frac{\omega}{2}\right)^2\)

Simplifying gives:

\(F_c = m \cdot r' \cdot \frac{\omega^2}{4}\)

Since the coin just slips, the centripetal force required should be the same in both situations. Therefore, we equate the two expressions for \(F_c\):

\(m \cdot 1 \cdot \omega^2 = m \cdot r' \cdot \frac{\omega^2}{4}\)

Cancelling \(m \cdot \omega^2\) from both sides, we get:

\(1 = \frac{r'}{4}\)

Solving for \(r'\) gives:

\(r' = 4\)

Therefore, when the angular velocity is halved, the coin will just slip when placed at a distance of 4 cm from the center.

Hence, the correct answer is 4 cm.