#### Question

A coin placed on a rotating turntable just slips. If it is placed at a distance of 4 cm from the centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of

1 cm

2 cm

4 cm

8 cm

#### Solution 1

8 cm

#### Solution 2

1 cm

Let the force of friction between the coin and the rotating turntable be *F*.

For the coin to just slip ,

we have : \[\text{ F = m }\omega^2 \text{r}\]

Here, \[\text{ m } \omega^2 \text{ r}\] is the centrifugal force acting on the coin.

For constant *F* and *m*, we have : \[\text{r} \propto \frac{1}{\omega^2}\]

Therefore,

\[\frac{\text{r}'}{\text{r}} = \left( \frac{\omega}{\omega'} \right)^2 \]

\[ \Rightarrow \text{ r' = 1 cm }\]

Is there an error in this question or solution?

Solution A Coin Placed on a Rotating Turntable Just Slips. If It is Placed at a Distance of 4 Cm from the Centre. Concept: Circular Motion.