Question

A circular disc of radius \(15\,cm\) and mass \(10\,kg\) is suspended by a wire attached to its centre. When the wire is twisted by rotating the disc and released, the period of torsional oscillations of the disc is \(1.5\,s\). The torsional spring constant of the wire is nearly:

• A. \(2\,Nm\,rad^{-1}\)
• B. \(3\,Nm\,rad^{-1}\)
• C. \(4\,Nm\,rad^{-1}\)
• D. \(1\,Nm\,rad^{-1}\)

Hint:

For torsional oscillations: \[ T=2\pi\sqrt{\frac{I}{C}}. \] This is analogous to the spring formula \[ T=2\pi\sqrt{\frac{m}{k}}. \]

Answer 1

The Correct Option is A

Concept: The time period of torsional oscillation is \[ T = 2\pi \sqrt{\frac{I}{C}}, \] where \[ I = \text{moment of inertia}, \] and \[ C = \text{torsional constant}. \]

Step 1: Calculate moment of inertia of the disc.
Radius \[ R=0.15m. \] Mass \[ M=10kg. \] For a disc, \[ I=\frac12MR^2. \] \[ I = \frac12(10)(0.15)^2. \] \[ I = 0.1125\,kg\,m^2. \]

Step 2: Use the time period formula.
\[ 1.5 = 2\pi \sqrt{\frac{0.1125}{C}}. \] Squaring, \[ 2.25 = 4\pi^2 \frac{0.1125}{C}. \] \[ C = \frac{4\pi^2(0.1125)}{2.25}. \]

Step 3: Evaluate numerically.
\[ C = 0.2\pi^2. \] \[ C \approx 1.97. \] \[ C\approx2. \] Therefore \[ \boxed{2\,Nm\,rad^{-1}}. \]