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

1. Understanding the Centripetal Formula
The centripetal force required to keep the coin moving in a circle is provided by the force of friction. At the point of slipping, the centripetal force is equal to the maximum static friction force:

$ F_{\text{centripetal}} = m \cdot a_{\text{centripetal}} = m \cdot r \cdot \omega^2 $,
$ F_{\text{friction}} = \mu \cdot N = \mu \cdot m \cdot g $.

Where:

$ m $: Mass of the coin,
$ r $: Distance from the center,
$ \omega $: Angular velocity,
$ \mu $: Coefficient of static friction,
$ N $: Normal force,
$ g $: Acceleration due to gravity.

At the point of slipping: $$ m \cdot r \cdot \omega^2 = \mu \cdot m \cdot g, $$ which simplifies to: $$ r \cdot \omega^2 = \mu \cdot g. $$

2. Constant Friction: Since the coin and turntable remain the same, $ \mu $ and $ g $ are constants. Let $ \mu \cdot g = C $ (a constant). Therefore: $$ r \cdot \omega^2 = C. $$

3. Applying the Condition:

Initially: $ r_1 \cdot \omega_1^2 = C $,
Finally: $ r_2 \cdot \omega_2^2 = C $.
Since both are equal to $ C $: $$ r_1 \cdot \omega_1^2 = r_2 \cdot \omega_2^2. $$

4. Given Values:

$ r_1 = 4 \, \text{cm} $,
$ \omega_2 = 2 \cdot \omega_1 $.

5. Solving for $ r_2 $:

$$ 4 \cdot \omega_1^2 = r_2 \cdot (2 \cdot \omega_1)^2, $$ $$ 4 \cdot \omega_1^2 = r_2 \cdot 4 \cdot \omega_1^2, $$ $$ r_2 = \frac{4 \cdot \omega_1^2}{4 \cdot \omega_1^2}, $$ $$ r_2 = 1 \, \text{cm}. $$

Final Answer: The correct answer is (A) 1 cm.

Answer 2

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.