Ncert Exercises & Solutions

14.23 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 α is defined by the relation J= –αθ, where J is the restoring couple and θ the angle of twist).

Given:

Mass of the circular disc, m = 10 kg
Radius of the disc, r = 15 cm = 0.15 m
The torsional oscillations of the disc have a time period, T = 1.5 s
The moment of inertia of the disc is:

$I = \frac{1}{2} {mr}^{2}$
$= \frac{1}{2} \times (10) \times {(0.15)}^{2}$
$= 0.1125 kg m^{2}$
$Time period, T = 2 π \sqrt{\frac{I}{α}}$
$\Rightarrow α = \frac{4 π^{2} I}{T^{2}}$
$= \frac{4 \times {(π)}^{2} \times 0.1125}{{(1.5)}^{2}}$
$= 1.972 Nm / rad$
$Hence, the torsional spring constant of the wire is 1.972 Nm {rad}^{- 1} .$

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