COMEDK · Physics · 8. Rotational Motion
A circular disc of mass \(20 \mathrm{~kg}\), having radius \(10 \mathrm{~cm}\) 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 \(1 \mathrm{~s}\). The torsional spring constant of the wire is
- A \(6 \mathrm{~N~m} \mathrm{~rad}^{-1}\)
- B \(9.86 \mathrm{~N~m} \mathrm{~rad}^{-1}\)
- C \(3.94 \mathrm{~N~m} \mathrm{~rad}^{-1}\)
- D \(1.264 \mathrm{~N~m} \mathrm{~rad}^{-1}\)
Answer & Solution
Correct Answer
(C) \(3.94 \mathrm{~N~m} \mathrm{~rad}^{-1}\)
Step-by-step Solution
Detailed explanation
The moment of inertia \(I\) of a circular disc about its central axis is given by \(I = \dfrac{1}{2}MR^2\). Given \(M = 20 \text{ kg}\) and \(R = 10 \text{ cm} = 0.1 \text{ m}\), we have \(I = \dfrac{1}{2} \times 20 \times (0.1)^2 = 10 \times 0.01 = 0.1 \text{ kg m}^2\). The…
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