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Q14.23:
A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant $\alpha$ is defined by the relation $J = –\alpha \theta$, where $J$ is the restoring couple and $\theta$ the angle of twist).

Solution :

Step 1: Given Variables & Theoretical Foundation

To determine the torsional spring constant $\alpha$, we utilize the dynamics of a torsional pendulum. The system consists of a circular disc suspended by a wire, undergoing angular simple harmonic motion.

  • Mass of the disc ($M$): $10\text{ kg}$
  • Radius of the disc ($R$): $15\text{ cm} = 0.15\text{ m}$
  • Period of oscillation ($T$): $1.5\text{ s}$
  • Restoring torque relation: $J = -\alpha \theta$

[By the Principle of Angular Harmonic Motion, the period $T$ is given by $T = 2\pi \sqrt{\frac{I}{\alpha}}$, where $I$ is the Moment of Inertia of the disc about the axis of rotation.]

Step 2: Calculation of Moment of Inertia ($I$)

For a circular disc rotating about an axis passing through its centre and perpendicular to its plane, the moment of inertia is defined as:

$I = \frac{1}{2} M R^2$

Substituting the given values:

$I = \frac{1}{2} \times 10\text{ kg} \times (0.15\text{ m})^2$

$I = 5 \times 0.0225 = 0.1125\text{ kg}\cdot\text{m}^2$

Step 3: Deriving the Torsional Spring Constant ($\alpha$)

Rearranging the period formula $T = 2\pi \sqrt{\frac{I}{\alpha}}$ to solve for $\alpha$:

$\frac{T}{2\pi} = \sqrt{\frac{I}{\alpha}}$

$\frac{T^2}{4\pi^2} = \frac{I}{\alpha}$

$\alpha = \frac{4\pi^2 I}{T^2}$

Step 4: Final Substitution & Numerical Computation

Using $\pi^2 \approx 9.8696$:

$\alpha = \frac{4 \times 9.8696 \times 0.1125}{(1.5)^2}$

$\alpha = \frac{4.44132}{2.25}$

$\alpha \approx 1.9739\text{ N}\cdot\text{m/rad}$

R = 0.15m Wire (α) Mass (M) = 10kg

Final Solution: The torsional spring constant of the wire is approximately $1.97\text{ N}\cdot\text{m/rad}$.


More Questions from Class 11 Physics Oscillations Additional Exercises


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