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NEET PHYSICSLAWS OF MOTIONMedium

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:

A

27 cm

B

9 cm

C

3 cm

D

1 cm

Step-by-Step Solution

  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: mω2r=μmgm \omega^2 r = \mu m g [Source 70] Here, mm is the mass, ω\omega is the angular velocity, rr is the distance from the center, μ\mu is the coefficient of friction, and gg is gravity.
  3. Relation: Since mm, μ\mu, and gg are constants for the same coin and surface, we have: ω2r=constant\omega^2 r = \text{constant} r∝1ω2r \propto \frac{1}{\omega^2}
  4. Calculation:
  • Initial state: r1=9 cmr_1 = 9 \text{ cm}, ω1=ω\omega_1 = \omega
  • Final state: ω2=3ω\omega_2 = 3\omega Using the inverse proportionality: r2r1=(ω1ω2)2\frac{r_2}{r_1} = \left( \frac{\omega_1}{\omega_2} \right)^2 r29=(ω3ω)2\frac{r_2}{9} = \left( \frac{\omega}{3\omega} \right)^2 r2=9×(13)2=9×19=1 cmr_2 = 9 \times \left( \frac{1}{3} \right)^2 = 9 \times \frac{1}{9} = 1 \text{ cm}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSLAWS OF MOTIONplacedrotatingturntableplaceddistance

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