A coin, placed on a rotating turn-table slips, when it is placed at a distance of 9 cm from the centre. If the

Question

A coin, placed on a rotating turn-table slips, when it is placed at a distance of 9 cm from the centre. If the angular velocity of the turn-table is tripled, it will just slip, if its distance from the centre is:

Accepted Answer

1. **Principle:** For a coin placed on a rotating turntable, the necessary centripetal force is provided by the static friction between the coin and the surface. The coin slips when the required centripetal force exceeds the maximum limiting friction.
2. **Formula:** The condition for the coin to just slip is when the centripetal force equals the limiting friction:
   $$ m \omega^2 r = \mu m g $$
   [Source 70]
   Here, $m$ is the mass, $\omega$ is the angular velocity, $r$ is the distance from the center, $\mu$ is the coefficient of friction, and $g$ is gravity.
3. **Relation:** Since $m$, $\mu$, and $g$ are constants for the same coin and surface, we have:
   $$ \omega^2 r = 	ext{constant} $$
   $$ r \propto \frac{1}{\omega^2} $$
4. **Calculation:**
   - Initial state: $r_1 = 9 	ext{ cm}$, $\omega_1 = \omega$
   - Final state: $\omega_2 = 3\omega$
   Using the inverse proportionality:
   $$ \frac{r_2}{r_1} = \left( \frac{\omega_1}{\omega_2} ight)^2 $$
   $$ \frac{r_2}{9} = \left( \frac{\omega}{3\omega} ight)^2 $$
   $$ r_2 = 9 	imes \left( \frac{1}{3} ight)^2 = 9 	imes \frac{1}{9} = 1 	ext{ cm} $$

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