Question

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

• A. 1 cm
• B. 2 cm
• C. 4 cm
• D. 8 cm

Answer 1

The Correct Option is A

The centripetal force is given by $F_c = m\omega^2r$, where:

• $m$ is the mass of the coin,
• $\omega$ is the angular velocity of the turntable, and
• $r$ is the distance of the coin from the center.

The maximum static friction is given by $F_f = \mu_s mg$, where:

• $\mu_s$ is the coefficient of static friction,
• $m$ is the mass of the coin, and
• $g$ is the acceleration due to gravity.

For the coin to just slip, the centripetal force must equal the maximum static friction:

$m\omega^2r = \mu_s mg$

$\omega^2r = \mu_s g$

Since $\mu_s$ and $g$ are constant, we can say that $\omega^2r$ is constant.

Initially, the coin slips at $r_1 = 4$ cm with angular velocity $\omega_1$. When the angular velocity is doubled ($\omega_2 = 2\omega_1$), let the distance at which it slips be $r_2$. Since $\omega^2r$ is constant, we have:

$\omega_1^2 r_1 = \omega_2^2 r_2$

$\omega_1^2 (4\,\text{cm}) = (2\omega_1)^2 r_2$

$4\omega_1^2 = 4\omega_1^2 r_2$

$r_2 = 1$ cm

The correct answer is (A) 1 cm.