A coin placed on a rotating table just slips when it is placed at a distance of $1 \mathrm{~cm}$ from the center. If the angular velocity of the table in halved, it will just slip when placed at a distance of _________ from the centre :

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 $\omega_2$. <br/><br/> Initially, when the coin is placed at a distance of 1 cm from the center, the centrifugal force acting on the coin is: <br/><br/> $F_1 = m r_1 \omega_1^2$ <br/><br/> where $r_1 = 1 \mathrm{~cm}$. <br/><br/> 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: <br/><br/> $F_2 = m r_2 \omega_2^2 = m r_2 \left(\frac{1}{2}\omega_1\right)^2$ <br/><br/> 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: <br/><br/> $m r_1 \omega_1^2 = m r_2 \left(\frac{1}{2}\omega_1\right)^2$ <br/><br/> Canceling the mass $m$ and the initial angular velocity $\omega_1^2$ from both sides, we get: <br/><br/> $r_1 = r_2 \left(\frac{1}{2}\right)^2$ <br/><br/> Now, we can solve for $r_2$: <br/><br/> $$r_2 = \frac{r_1}{\left(\frac{1}{2}\right)^2} = \frac{1 \mathrm{~cm}}{\frac{1}{4}} = 4 \mathrm{~cm}$$ <br/><br/> So, the coin will just slip when placed at a distance of 4 cm from the center when the angular velocity is halved.