Chapter 19.3, Problem 19.88P

To determine

The period $\left({\tau }_n\right)$ of small oscillation of the system.

Answer to Problem 19.88P

The period $\left({\tau }_n\right)$ of small oscillation of the system is $\underset{\_}{2\pi \sqrt{\frac{60r^2+10l^2}{9gl}}}$.

Explanation of Solution

Given information:

The length of rod AB and CD is l.

The mass of the gear C is m.

The mass of the gear A is 4m.

The mass of rod AB $\left(m_{AB}\right)$ and CD $\left(m_{AB}\right)$ is m.

Calculation:

Show the position 1 and position 2 of the system with velocity as in Figure (1).

Write the kinematics relations of gear:

$\begin{array}{l}2r{\theta }_A=r{\theta }_C\\ 2{\theta }_A={\theta }_C\\ 2{\dot{\theta }}_A={\dot{\theta }}_C\end{array}$

Let, ${\theta }_A$ is equal to ${\theta }_m$.

Substitute ${\theta }_m$ for ${\theta }_A$ in above relations.

$\begin{array}{c}2{\theta }_m={\left({\theta }_C\right)}_m\\ 2{\dot{\theta }}_m={\left({\dot{\theta }}_C\right)}_m\end{array}$

For position 1:

Write the expression for mass moment of inertia gear C$\left(I_C\right)$:

$I_C=\frac{1}{2}m{r}^2$

Write the expression for mass moment of inertia of gear A $\left(I_A\right)$:

$\begin{array}{c}{I}_A=\frac{1}{2}\left(4m\right){\left(2r\right)}^2\\ =8m{r}^2\end{array}$

Write the expression for mass moment of inertia of rod AB $\left(I_{AB}\right)$:

$I_{AB}=\frac{1}{12}m{l}^2$

Write the expression for mass moment of inertia of rod CD $\left(I_{CD}\right)$:

$I_{CD}=\frac{1}{12}m{l}^2$

Write the expression the velocity of rod AB$\left(v_{AB}\right)$:

$v_{AB}=\frac{l}{2}{\dot{\theta }}_m$

Write the expression the velocity of rod CD $\left(v_{CD}\right)$:

$v_{CD}=\frac{l}{2}2{\dot{\theta }}_m$

Write the expression for the kinetic energy $\left(T_1\right)$:

$T_1=\left\{\begin{array}{l}\frac{1}{2}\left(I_A\right){\dot{\theta }}_m^2+\frac{1}{2}\left(I_C\right){\left(2{\dot{\theta }}_m\right)}^2+\frac{1}{2}\left(I_{AB}\right){\dot{\theta }}_m^2+\frac{1}{2}\left(I_{CD}\right){\left(2{\dot{\theta }}_m\right)}^2\\ +\frac{1}{2}{m}_{AB}{\left(v_{AB}\right)}^2++\frac{1}{2}{m}_{CD}{\left(v_{CD}\right)}^2\end{array}\right\}$

Substitute  m for $m_{AB}$, m for $m_{CD}$,$\frac{1}{2}m{r}^2$ for $I_C$, $8m{r}^2$ for $I_A$, $\frac{1}{12}m{l}^2$ for $I_{AB}$, $\frac{1}{12}m{l}^2$ for $I_{CD}$, $\frac{l}{2}{\dot{\theta }}_m$ for $v_{AB}$, and $\frac{l}{2}2{\dot{\theta }}_m$ for $v_{CD}$.

$\begin{array}{c}{T}_1=\left\{\begin{array}{l}\frac{1}{2}\times 8m{r}^2\times {\dot{\theta }}_m^2+\frac{1}{2}\times \left(\frac{1}{2}m{r}^2\right)\times {\left(2{\dot{\theta }}_m\right)}^2+\frac{1}{2}\left(\frac{1}{12}m{l}^2\right){\dot{\theta }}_m^2+\\ \frac{1}{2}\left(\frac{1}{12}m{l}^2\right){\left(2{\dot{\theta }}_m\right)}^2+\frac{1}{2}\times m\times {\left(\frac{l}{2}{\dot{\theta }}_m\right)}^2+\frac{1}{2}\times m\times {\left(\frac{l}{2}2{\dot{\theta }}_m\right)}^2\end{array}\right\}\\ =\frac{1}{2}\left[8m{r}^2+2m{r}^2+\frac{1}{12}m{l}^2+\frac{1}{3}m{l}^2+\frac{1}{4}m{l}^2+m{l}^2\right]{\dot{\theta }}_m^2\\ =\frac{1}{2}m\left[10r^2+\frac{5}{3}{l}^2\right]{\dot{\theta }}_m^2\end{array}$

Write the expression for the potential energy $\left(V_1\right)$:

$V_1=0$

For position 2:

Write the expression for the kinetic energy $\left(T_2\right)$:

$T_2=0$

Write the expression for the displacement $\left(h_1\right)$:

$\begin{array}{c}{h}_1=\left(\frac{l}{2}-\frac{l}{2}\cos {\theta }_m\right)\\ =\frac{l}{2}\left(1-\cos {\theta }_m\right)\left(\because \left(1-\cos {\theta }_m=2{\sin }^2\frac{{\theta }_m}{2}\right)\right)\\ =\frac{l}{2}2{\sin }^2\frac{{\theta }_m}{2}\end{array}$

For small oscillation $2{\sin }^2\frac{{\theta }_m}{2}\approx \frac{{\theta }_m^2}{2}$.

$h_1=\frac{l}{2}\frac{{\theta }_m^2}{2}$

Write the expression for the displacement $\left(h_2\right)$:

$\begin{array}{c}{h}_2=\left(\frac{l}{2}-\frac{l}{2}\cos 2{\theta }_m\right)\\ =\frac{l}{2}\left(1-\cos {\theta }_m\right)\left(\because \left(1-\cos 2{\theta }_m=2{\sin }^2{\theta }_m\right)\right)\\ =\frac{l}{2}2{\sin }^2{\theta }_m\end{array}$

For small oscillation $2{\sin }^2{\theta }_m\approx 2{\theta }_m^2$.

$h_2=\frac{l}{2}2{\theta }_m^2$

Write the expression for the potential energy $\left(V_2\right)$:

$V_2=-mg{h}_1+mg{h}_2$

Substitute $\frac{l}{2}\frac{{\theta }_m^2}{2}$ for $h_1$ and $\frac{l}{2}2{\theta }_m^2$ for $h_2$.

$\begin{array}{c}{V}_2=-mg\left(\frac{l}{2}\frac{{\theta }_m^2}{2}\right)+mg\left(\frac{l}{2}2{\theta }_m^2\right)\\ =\frac{1}{2}\left(-mg\left(\frac{l}{2}{\theta }_m^2\right)+mg\left(l2{\theta }_m^2\right)\right)\\ =\frac{1}{2}mgl\left(-\frac{{\theta }_m^2}{2}+2{\theta }_m^2\right)\\ =\frac{1}{2}mgl\left(\frac{3{\theta }_m^2}{2}\right)\end{array}$

Express the term $\left({\dot{\theta }}_m\right)$:

${\dot{\theta }}_m={\omega }_n{\theta }_m$

Write the expression for the conservation of energy:

$T_1+V_1=T_2+V_2$

Substitute $\frac{1}{2}m\left[10r^2+\frac{5}{3}{l}^2\right]{\dot{\theta }}_m^2$ for $T_1$, 0 for $V_1$, 0 for $T_2$ and $\frac{1}{2}mgl\left(\frac{3{\theta }_m^2}{2}\right)$ for $V_2$.

$\frac{1}{2}m\left[10r^2+\frac{5}{3}{l}^2\right]{\dot{\theta }}_m^2+0=0+\frac{1}{2}mgl\left(\frac{3{\theta }_m^2}{2}\right)$

Substitute ${\omega }_n{\theta }_m$ for ${\dot{\theta }}_m$.

$\begin{array}{l}\frac{1}{2}m\left[10r^2+\frac{5}{3}{l}^2\right]{\omega }_n^2{\theta }_m^2+0=0+\frac{1}{2}mgl\left(\frac{3{\theta }_m^2}{2}\right)\\ \frac{1}{2}m\left[10r^2+\frac{5}{3}{l}^2\right]{\omega }_n^2{\theta }_m^2=\frac{1}{2}\left(\frac{3}{2}mgl\right){\theta }_m^2\\ {\omega }_n^2=\frac{\frac{3}{2}mgl}{m\left[10r^2+\frac{5}{3}{l}^2\right]}\\ {\omega }_n=\sqrt{\frac{9gl}{60r^2+10l^2}}\end{array}$

Calculate the period $\left({\tau }_n\right)$ of small vibration of the system using the relation:

${\tau }_n=\frac{2\pi }{{\omega }_n}$

Substitute $\sqrt{\frac{9gl}{60r^2+10l^2}}$ for ${\omega }_n$.

$\begin{array}{c}{\tau }_n=\frac{2\pi }{\sqrt{\frac{9gl}{60r^2+10l^2}}}\\ =2\pi \sqrt{\frac{60r^2+10l^2}{9gl}}\end{array}$

Therefore, the period $\left({\tau }_n\right)$ of small oscillation of the system is $\underset{\_}{2\pi \sqrt{\frac{60r^2+10l^2}{9gl}}}$.