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JEE Main 202431 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Morning Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2024 · 31 January · Morning Shift · Physics · Laws of Motion

A coin is placed on a disc. The coefficient of friction between the...

A coin is placed on a disc. The coefficient of friction between the coin and the disc is μ\mu. If the distance of the coin from the center of the disc is rr, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is :A coin is placed on a disc. The coefficient of friction between the coin and the disc is μ . If the distance of the

Options

  1. A\mugr\dfrac{\mug}{r}
  2. Br\mug\sqrt{\dfrac{r}{\mug}}
  3. C\mugr\sqrt{\dfrac{\mug}{r}}
  4. Dμrg\dfrac{\mu}{\sqrt{rg}}

Correct answer

C. \mugr\sqrt{\dfrac{\mug}{r}}

Step-by-step solution

A coin is placed on a disc. The coefficient of friction between the coin and the disc is μ . If the distance of the

Normal force acting on the coin is, N=mgN = mg.

Required friction force to provide centripetal force for the coin is, f=mω2rf = m{\omega}^{2}r and maximum value of static friction force available is, f=\muN=\mumgf = \muN = \mumg.

For limiting condition, we can write

⇒\mumg=mrω2⇒ \mumg = mr{\omega}^{2}

⇒ω=\mugr⇒ \omega = \sqrt{\dfrac{\mug}{r}}

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