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

For a coin on a rotating table, the coin just slips when the centripetal force equals the maximum static friction force:

$$f_s = m\omega^2 r$$

At the point of slipping:

$$\mu m g = m\omega^2 r$$

$$\mu g = \omega^2 r$$

We are given that The coin just slips at $$r_1 = 1$$ cm when angular velocity is $$\omega$$.

So: $$\mu g = \omega^2 \times 1$$ ... (i)

When the angular velocity is halved to $$\frac{\omega}{2}$$, let the coin just slip at distance $$r_2$$:

$$\mu g = \left(\frac{\omega}{2}\right)^2 \times r_2 = \frac{\omega^2}{4} \times r_2$$ ... (ii)

From equations (i) and (ii):

$$\omega^2 \times 1 = \frac{\omega^2}{4} \times r_2$$

$$r_2 = 4 \text{ cm}$$

The coin will just slip when placed at a distance of 4 cm from the centre.