91Ó°ÊÓ

Moment of inertia defined with respect to rotation axis, as a quantity that decides the amount of torque required for a desired angular acceleration, expressed as,

$\mathrm{l}=\mathrm{m}×{\mathrm{r}}^2$ (1)

m is sum of the product of the masses; r is the distance from the axis of the rotation

Given,

total mass of the balance wheel, m = 0.95 g = $9.5×{10}^{-4}\mathrm{kg}$.

radius of the wheel, r = 0.54 cm = $5.4×{10}^{-3}m$.

Now, from equation (1), you get,

$\begin{array}{l}l=9.5×{10}^{-4}\begin{array}{ll}kg×(5.4x{10}^{-3}& m{)}^2\end{array}\\ =2.77x{10}^{-8}kg{m}^2\end{array}$

Therefore, the moment of inertia of the balance wheel about its shaft is$\mathit{l}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{77}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathit{k}\mathit{g}\mathbf{}{\mathit{m}}^{\mathbf{2}}$

Period of a tortional pendulum is,

$T=2\mathrm{Ï€}\sqrt{\frac{\mathrm{I}}{\mathrm{K}}}$

Here, K is torsion constant, then;

$\begin{array}{r}K=\frac{4{\mathrm{π}}^2×1}{T^2}=\frac{4{\mathrm{π}}^2×2.77×{10}^{−8}{\mathrm{kgmm}}^2}{(0.5\mathrm{s}{)}^2}\\ K=4.375×{10}^{−6}\mathrm{Nm}\end{array}$

Hence, the torsion constant of the coil spring is$\mathit{K}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{37}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathit{N}\mathit{m}$.