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COMEDK2024PhysicsRotational MotionActual

A circular disc of mass 20 ~kg , having radius 10 ~cm is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The time period of torsional oscillations is found to be 1 ~s . The torsional spring constant of the wire is

Options

  1. A6 ~N~m ~rad ⁻¹
  2. B9.86 ~N~m ~rad ⁻¹
  3. C3.94 ~N~m ~rad ⁻¹
  4. D1.264 ~N~m ~rad ⁻¹

Correct answer

C. 3.94 ~N~m ~rad ⁻¹

Step-by-step solution

The moment of inertia I of a circular disc about its central axis is given by I = 1 2 MR^2 . Given M = 20 kg and R = 10 cm = 0.1 m , we have I = 1 2 20 (0.1)^2 = 10 0.01 = 0.1 kg m ^2 . The time period T of torsional oscillations is given by T = 2 I C , where C is the torsional spring constant. Squaring both sides, T^2 = 4 ^2 I C , which implies C = 4 ^2 I T^2 . Substituting the given values T = 1 s and I = 0.1 kg m ^2 : C = 4 (3.14)^2 0.1 = 4 9.8596 0.1 = 3.94384 N m rad ⁻¹ . Rounding to two decimal places, C 3.94

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