A circular disc of mass 10kg is suspended by a wire attached to i… - Vidyadip

Question

A circular disc of mass $10kg$ 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 $1.5s$. The radius of the disc is $15cm$. Determine the torsional spring constant of the wire. (Torsional spring constant $\alpha$ is defined by the relation $\text{J}=-\alpha\theta,$ where J is the restoring couple and $\theta$ the angle of twist).

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Answer

Mass of the circular disc, m = 10kg Radius of the disc, r = 15cm = 0.15m The torsional oscillations of the disc has a time period, T = 1.5s The moment of inertia of the disc is:$\text{l}=\frac{1}{2}\text{mr}^2$
$=\frac{1}{2}\times(10)\times(0.15)^2$
$= 0.1125kg m^2$ Time period, $\text{T}=2\pi\sqrt{\frac{\text{l}}{\alpha}}$$\alpha$ is the torsional constant.
$\alpha=\frac{4\pi^2\text{l}}{\text{T}^2}$
$=\frac{4\times(\pi)^2\times0.1125}{(1.5)^2}$
= 1.972Nm/rad Hence, the torsional spring constant of the wire is $1.972Nm\ rad^{-1}$.

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