Question

A thin metal disk with mass $\mathbf{2}\mathbf{.}\mathbf{00}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{k}\mathit{g}$and radius$\mathbf{2}\mathbf{.}\mathbf{20}\mathit{c}\mathit{m}$is attached at its centre to a long fibre. The disk, when twisted and released, oscillates with a period of$\mathbf{1}\mathbf{.}\mathbf{00}\mathit{s}$. Find the torsion constant of the fibre.

Short answer

$Thetorsionconstantofthefibreis1.9x{10}^{-6}$

Step-by-step solution

Definition of torsion constant

The ratio of the torque to the angle is called the torsion constant. It is denoted by c. For a rigid body formula for torsion constant is

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

Where,

l = Moment of inertia

T = Time period

calculation of torsion constant

Given,

$$\begin{gathered} M=2x{10}^{-3}kg \\ T=1s \\ R=2.2cm \end{gathered}$$

So, I of metal disc is localid="1668148513836" $=I=\frac{M{R}^2}{2}$

localid="1668148509505" $$\begin{gathered} PutvalueofMandRweget \\ I=4.84x{10}^{-7}kg-m^2 \\ Nowaccordingtoformula \\ c=\frac{4{\pi }^{2}I}{T^2} \end{gathered}$$

localid="1668148518008" $$\begin{gathered} putallthevaluesinformula \\ c=\frac{4x3.14x3.14x4.84x{10}^{-7}}{1} \\ C=1.9x{10}^{-6} \\ So,valueoftorsionconstantis1.9x{10}^{-6} \end{gathered}$$