JEE Main 2023 — Circular Motion Question with Solution

When a coin is placed on a rotating table and is just about to slip, the centrifugal force acting on the coin equals the maximum static friction force. Let's denote the mass of the coin as $$m$$, the initial angular velocity as $${\omega }_1$$, and the final angular velocity as $${\omega }_2$$.

Initially, when the coin is placed at a distance of 1 cm from the center, the centrifugal force acting on the coin is:

$$F_1=m{r}_1{\omega }_1^2$$

where $$r_1=1\mathrm{cm}$$.

When the angular velocity is halved ($${\omega }_2=\frac{1}{2}{\omega }_1$$), the centrifugal force acting on the coin when it just slips is:

$$F_2=m{r}_2{\omega }_2^2=m{r}_2{\left(\frac{1}{2}{\omega }_1\right)}^2$$

Since the coin is just about to slip in both cases, the maximum static friction force remains the same. Therefore, we can equate the centrifugal forces:

$$m{r}_1{\omega }_1^2=m{r}_2{\left(\frac{1}{2}{\omega }_1\right)}^2$$

Canceling the mass $$m$$ and the initial angular velocity $${\omega }_1^2$$ from both sides, we get:

$$r_1=r_2{\left(\frac{1}{2}\right)}^2$$

Now, we can solve for $$r_2$$:

$$r_2=\frac{r_1}{{\left(\frac{1}{2}\right)}^2}=\frac{1\mathrm{cm}}{\frac{1}{4}}=4\mathrm{cm}$$

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