Chapter 19.3, Problem 19.85P

To determine

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

Answer to Problem 19.85P

The period $\left({\tau }_n\right)$ of small oscillation of the rod is $\underset{\_}{1.834\text{s}}$.

Explanation of Solution

Given information:

The weight $\left(W_A\right)$ of the sphere A is 14 oz.

The weight $\left(W_C\right)$ of the sphere C is 10 oz.

The weight $\left(W_{AC}\right)$ of the rod AC is 20 oz.

Assume the acceleration due to gravity (g) as $32.2{\text{ft/s}}^{\text{2}}$.

Calculation:

Show the free body diagram of the system for position 1 and position 2 as in Figure (1).

For position 1:

Write the expression for the velocity of sphere C ${\left(v_C\right)}_m$ at position 2:

$\begin{array}{c}{\left(v_C\right)}_m=8\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}{\dot{\theta }}_m\\ =\frac{2}{3}{\dot{\theta }}_m\end{array}$

Write the expression for the velocity of sphere A ${\left(v_A\right)}_m$ at position 2:

$\begin{array}{c}{\left(v_A\right)}_m=5\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}{\dot{\theta }}_m\\ =\frac{5}{12}{\dot{\theta }}_m\end{array}$

Write the expression for the velocity of sphere A ${\left(v_{AC}\right)}_m$ at position 2:

${\left(v_{AC}\right)}_m=\frac{1}{8}{\dot{\theta }}_m$

Calculate the mass $\left(m_C\right)$ of the sphere C:

$m_C=\frac{W_C}{g}$

Substitute 10 oz for $W_C$ and $32.2{\text{ft/s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_C=\frac{10\text{oz}\times \frac{1\text{lb}}{16\text{oz}}}{32.2}\\ =0.01941\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\end{array}$

Calculate the mass $\left(m_A\right)$ of the sphere A:

$m_A=\frac{W_A}{g}$

Substitute 14 oz for $W_A$ and $32.2{\text{ft/s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_C=\frac{14\text{oz}\times \frac{1\text{lb}}{16\text{oz}}}{32.2}\\ =0.027174\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\end{array}$

Calculate the mass $\left(m_{AC}\right)$ of the sphere A:

$m_{AC}=\frac{W_{AC}}{g}$

Substitute 20 oz for $W_A$ and $32.2{\text{ft/s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_{AC}=\frac{20\text{oz}\times \frac{1\text{lb}}{16\text{oz}}}{32.2}\\ =0.03882\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}\end{array}$

Write the expression for moment of inertia of road AC:

${\overline{I}}_{AC}=\frac{1}{12}{m}_{AC}{\left(\left(5+8\right)\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2$

Substitute $0.03882\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for $m_{AC}$.

$\begin{array}{c}{\overline{I}}_{AC}=\frac{1}{12}\left(0.03882\right){\left(\left(5+8\right)\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ =3.797\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

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

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

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

$T_1=\frac{1}{2}{m}_C{\left(v_C^{}\right)}_m^2+\frac{1}{2}{m}_A{\left(v_A^{}\right)}_m^2+\frac{1}{2}{m}_{AC}{\left(v_{AC}^{}\right)}_m^2+\frac{1}{2}{\overline{I}}_{AC}{\dot{\theta }}_m^2$

Substitute $0.01941\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for $m_A$, $0.027174\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for $m_C$,$0.03882\text{lb}\cdot {\text{s}}^{\text{2}}\text{/ft}$ for $m_{AC}$,  $\frac{2}{3}{\dot{\theta }}_m$ for ${\left(v_C\right)}_m$, $\frac{5}{12}{\dot{\theta }}_m$ for ${\left(v_A\right)}_m$, $\frac{1}{8}{\dot{\theta }}_m$ for ${\left(v_{AC}\right)}_m$ and $3.797\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for ${\overline{I}}_{AC}$

$\begin{array}{c}{T}_1=\frac{1}{2}\left(0.01941\right){\left(\frac{2}{3}{\dot{\theta }}_m\right)}^2+\frac{1}{2}\left(0.027174\right){\left(\frac{5}{12}{\dot{\theta }}_m\right)}^2+\frac{1}{2}\left(0.03882\right){\left(\frac{1}{8}{\dot{\theta }}_m\right)}^2+\frac{1}{2}\left(3.797\times {10}^{-3}\right){\dot{\theta }}_m^2\\ =\frac{1}{2}{\dot{\theta }}_m^2\left(8.6267\times {10}^{-3}+4.7177\times {10}^{-3}+6.0656\times {10}^{-4}+3.797\times {10}^{-3}\right)\\ =0.01775\frac{{\dot{\theta }}_m^2}{2}\end{array}$

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

$V_1=0$

For position 2:

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

$T_2=0$

Write the expression for the displacement $\left(h_C\right)$ of sphere C:

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

For small oscillation $\sin \theta \approx \theta$.

$h_C=BC\frac{{\theta }_m^2}{2}$

Write the expression for the displacement $\left(h_A\right)$ of sphere A:

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

For small oscillation $\sin \theta \approx \theta$.

$h_A=BA\frac{{\theta }_m^2}{2}$

Write the expression for the displacement $\left(h_{AC}\right)$ of sphere A:

$\begin{array}{c}{h}_{AC}=\frac{1}{8}\left(1-\cos {\theta }_m\right)\left(1-\cos \theta =2{\sin }^2\frac{\theta }{2}\right)\\ =\frac{1}{8}{\sin }^2\frac{{\theta }_m}{2}\end{array}$

For small oscillation $\sin \theta \approx \theta$.

$h_{AC}=\frac{1}{8}\frac{{\theta }_m^2}{2}$

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

$V_2=-W_A{h}_A+W_C{h}_C+W_{AC}{h}_{AC}$

Substitute $BC\frac{{\theta }_m^2}{2}$ for $h_C$, $BA\frac{{\theta }_m^2}{2}$ for $h_A$ and $\frac{1}{8}\frac{{\theta }_m^2}{2}$ for $h_{AC}$.

$\begin{array}{c}{V}_2=-W_A\left(BA\frac{{\theta }_m^2}{2}\right)+W_C\left(BC\frac{{\theta }_m^2}{2}\right)+W_{AC}\frac{1}{8}\frac{{\theta }_m^2}{2}\\ =\left(-W_A\left(BA\right)+W_C\left(BC\right)+W_{AC}\frac{1}{8}\right)\frac{{\theta }_m^2}{2}\end{array}$

Substitute 10 oz for $W_C$, 14 oz for $W_A$, 20 oz for $W_{AC}$, $8\text{in}\text{.}$ for $BC$, and $5\text{in}\text{.}$ for $BA$.

$\begin{array}{c}{V}_2=\left(-\left(14\text{oz}\times \frac{1\text{lb}}{16\text{oz}}\right)\left(5\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)+\left(10\text{oz}\times \frac{1\text{lb}}{16\text{oz}}\right)\left(8\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)+\left(20\text{oz}\times \frac{1\text{lb}}{16\text{oz}}\right)\frac{1}{8}\right)\frac{{\theta }_m^2}{2}\\ =\left(-0.3646+0.4167+0.1563\right)\frac{{\theta }_m^2}{2}\\ =0.2084\frac{{\theta }_m^2}{2}\end{array}$

Write the expression for conservation of energy:

$T_1+V_1=T_2+V_2$

Substitute $0.01775\frac{{\dot{\theta }}_m^2}{2}$ for $T_1$, 0 for $V_1$, 0 for $T_2$, and $0.2084\frac{{\theta }_m^2}{2}$ for $V_2$.

$\begin{array}{l}0.01775\frac{{\dot{\theta }}_m^2}{2}+0=0+0.2084\frac{{\theta }_m^2}{2}\\ 0.01775\frac{{\dot{\theta }}_m^2}{2}=0.2084\frac{{\theta }_m^2}{2}\end{array}$

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

$\begin{array}{l}0.01775\frac{{\left({\omega }_n{\theta }_m\right)}^2}{2}=0.2084\frac{{\theta }_m^2}{2}\\ 0.01775\frac{{\omega }_n^2{\theta }_m^2}{2}=0.2084\frac{{\theta }_m^2}{2}\\ {\omega }_n^2=\frac{0.2084}{0.01775}\\ {\omega }_n=\sqrt{11.741}\\ {\omega }_n=3.426\text{rad/s}\end{array}$

Calculate the period $\left({\tau }_n\right)$ of small oscillation of the rod:

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

Substitute $3.426\text{rad/s}$ for ${\omega }_n$.

$\begin{array}{c}{\tau }_n=\frac{2\pi }{3.426}\\ =1.834\sec \end{array}$

Therefore, the period $\left({\tau }_n\right)$ of small oscillation of the rod is $\underset{\_}{1.834\text{s}}$.