QJEEM23Phb307: JEE Main 2023 Physics - Class 11 JEEM23Phb307 Solution: A coin placed on a rotating table just slips when it is placed…

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.