Question

A coin is placed on a rotating table. It is observed that the coin just slips when placed at a distance $4r$ from the centre. On doubling the angular velocity of the table, it will just slip when its distance from the centre is equal to:
A. 2r
B. 4r
C. r
D. $\dfrac{r}{4}$

Answer 1

Hint: Try and arrive at a relation between the angular velocity of the table and the distance of the coin from the centre of the table by identifying the forces that the coin is subjected to. In other words, we know that the centripetal and centrifugal forces acting on the coin become equal when the coin just begins to slide. Using this, arrive at an expression for the centripetal force acting on the coin in terms of the mass of the coin, angular velocity and its distance from centre and you should be able to deduce the relation between the two.

Formula used:
Frictional (centripetal) force for an angular motion $F_{centripetal} = m\omega^2 r$, where m is the mass of the body executing an angular motion, $\omega$ is the angular velocity experienced by it

Complete answer:
Let us try and understand what keeps the coin on the table and what makes it slip.

When a coin is placed on a rotating table, it stays on the table and rotates with it as long as the centripetal force of rotation is not less than the centrifugal force of rotation on the coin.
The centripetal force is brought about as a result of the static friction between the coin and the table which helps the coin hold its position on the table and is hence directed towards the centre of the table, whereas the centrifugal force is produced by the rotation of the table and is directed opposite to the centre of the table and strives to push any object placed on the table to its extremities and ultimately off the table.

The coin remains on the table as long as centripetal force acting on it is greater than the influence that centrifugal force has on it. This is possible when the coin possesses enough static friction.
The coin will begin to slip when the value of the static friction reaches a maximum value called the limiting friction. At this point the centripetal force experienced by the body will be equal to the centrifugal force, beyond which the coin is set in motion.

Now, centripetal force for a linear motion is given as
$F_{centripetal} = \dfrac{mv^2}{r}$, where m is the mass of the body, v is the velocity with which it rotates and r is its distance from the centre.
We know that angular velocity $\omega$ can be expressed in terms of linear velocity as $\omega = \dfrac{v}{r} \Rightarrow v = \omega r$
Substituting this in the centripetal force equation will give us the centripetal force for a circular motion:
$F_{centripetal} = \dfrac{m\omega^2 r^2}{r} = m\omega^2r$
For the initial case, we have:
$F_{centripetal} = m\omega^2(4r)$
For the final case we have:
$F_{centripetal} = m(2\omega)^2 r^{\prime}$
From the above two equations we get:
$\omega^2(4r) = (2\omega)^2 r^{\prime} $
$\Rightarrow 4r = 4r^{\prime} \Rightarrow r^{\prime} = r$
Therefore, on doubling the angular velocity of the table, the coin will just slip when its distance from the centre is equal to r.

So, the correct answer is “Option C”.

Note:
Though we’ve labelled the force that we have considered as $F_{centripetal}$ do not forget that this centripetal force is ultimately produced as a consequence of the frictional force between the coin and the table surface in this situation. We only used the centripetal form of this force to make it fit into the results that we needed to obtain, i.e., we only needed to arrive at the relation between the angular velocity and distance at which the coin was placed.

Note that it is always important to identify the forces that the components of a system are subjected to and the directions in which they are acting. This will help break down the physical situation into usable mathematical relations that can be employed to obtain the desired results.