Question

A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a thin rim with radius 0.55 cm, connected to the balance shaft by thin spokes of negligible mass. The total mass of the balance wheel is 0.90 g. (a) What is the moment of inertia of the balance wheel about its shaft? (b) What is the torsion constant of the coil spring (Fig. 14.19)?

Short answer

1. The moment of inertia of the balance wheel about its shaft is$\mathbf{I}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{77}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{\hspace{0.33em}}\mathbf{kg}\mathbf{\hspace{0.33em}}{\mathbf{m}}^{\mathbf{2}}\mathbf{.}$
2. The torsion constant of the coil spring isrole="math" localid="1668147826890" $\mathit{K}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{37}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathit{N}\mathit{m}$.

Step-by-step solution

Determine the formula of moment of inertia of a rotating body

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,

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

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

Write the given information and calculate the moment of inertia

Given,

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

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

Now, from equation (1), you get,

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

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

Determine the equation of the torsion constant calculate

Period of a tortional pendulum is,

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

Here, K is torsion constant, then;

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

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