Chapter 19.3, Problem 19.87P

To determine

The period of small oscillations of the system.

Answer to Problem 19.87P

The period of small oscillations of the system: ${\tau }_n=2\pi \sqrt{\frac{12r^2+2l^2}{3gl}}$

Explanation of Solution

Given information:

The mass of gear C is m, and the mass of gear A is 4m.

Calculations:

For the figures shown above:

$\begin{array}{l}Kinematics:\\ 2r{\theta }_A=r{\theta }_C\Rightarrow 2{\theta }_A={\theta }_C\\ 2{\dot{\theta }}_A={\dot{\theta }}_C\\ Let\\ {\theta }_A={\theta }_m\\ 2{\theta }_m={\left({\theta }_C\right)}_m\\ 2{\dot{\theta }}_m={\left({\dot{\theta }}_C\right)}_m\end{array}$

$\begin{array}{l}ForPosition1:\\ \text{Kinetic energy: }{T}_1=\frac{1}{2}{\overline{I}}_A{\stackrel{.}{\theta }}_m^2+\frac{1}{2}{\overline{I}}_C{\left(2{\stackrel{.}{\theta }}_m\right)}^2+\frac{1}{2}{\overline{I}}_{AB}{\stackrel{.}{\theta }}_m^2+\frac{1}{2}{\overline{I}}_{CD}{\left(2{\stackrel{.}{\theta }}_m\right)}^2+\frac{1}{2}{m}_{AB}{\left(\frac{l}{2}{\stackrel{.}{\theta }}_m\right)}^2+\frac{1}{2}{m}_{CD}{\left(\frac{l}{2}{\stackrel{.}{2\theta }}_m\right)}^2\\ where,\\ {\overline{I}}_A=\frac{1}{2}\left(4m\right){\left(2r\right)}^2=8m{r}^2\\ {\overline{I}}_C=\frac{1}{2}\left(m\right){\left(r\right)}^2=\frac{1}{2}m{r}^2\\ {\overline{I}}_{AB}=\frac{1}{12}m{l}^2\\ {\overline{I}}_{CD}=\frac{1}{12}m{l}^2\\ substituting,\\ T_1=\frac{1}{2}m\left[8r^2+\left(\frac{r^2}{2}\right)4+\frac{l^2}{12}+\frac{l^2}{3}+\frac{l^2}{4}+l^2\right]{\stackrel{.}{\theta }}_m^2\\ T_1=\frac{1}{2}m\left[10r^2+\frac{5}{3}{l}^2\right]{\stackrel{.}{\theta }}_m^2\\ \text{Potential energy: }{V}_1=0\end{array}$

$\begin{array}{l}ForPosition\text{ 2:}\\ \text{Kinetic energy: }{T}_2=0\\ \text{Potential energy: }{V}_2=mg\frac{l}{2}\left(1-\cos {\theta }_m\right)+\frac{mgl}{2}\left(1-\cos 2{\theta }_m\right)\\ Forsmallangles\left(1-\cos {\theta }_m=2{\sin }^2\frac{{\theta }_m}{2}\approx \frac{{\theta }_m^2}{2}\right)\\ \left(1-\cos 2{\theta }_m=2{\sin }^2{\theta }_m\approx 2{\theta }_m^2\right)\\ \therefore V_2=\frac{1}{2}mgl\left(\frac{{\theta }_m^2}{2}+2{\theta }_m^2\right)=\frac{1}{2}mgl\frac{5{\theta }_m^2}{2}\end{array}$

$\begin{array}{l}\text{Now, from the law of conservation of energy:}\\ T_1+V_1=T_2+V_2\\ with,\left({\dot{\theta }}_m^2={\omega }_n^2{\theta }_m^2\right)\\ \frac{1}{2}m\left[10r^2+\frac{5}{3}{l}^2\right]{\omega }_n^2{\theta }_m^2+0=0+\frac{1}{2}mgl\frac{5{\theta }_m^2}{2}\\ {\omega }_n^2=\frac{\frac{5}{2}gl}{10r^2+\frac{5}{3}{l}^2}\\ {\omega }_n^2=\frac{3gl}{12r^2+2l^2}\\ \text{Thus, the period of oscillation:}\\ {\tau }_n=\frac{2\pi }{\sqrt{{\omega }_n}}\\ \Rightarrow {\tau }_n=2\pi \sqrt{\frac{12r^2+2l^2}{3gl}}\end{array}$

Conclusion:

The period of small oscillations of the system is ${\tau }_n=2\pi \sqrt{\frac{12r^2+2l^2}{3gl}}$