Chapter 19.3, Problem 19.71P

To determine

Period of small oscillation of rod.

Answer to Problem 19.71P

The period of small oscillation of rod is $3.18\text{}\text{ s}$.

Explanation of Solution

Given:

Weight of sphere A is $14\text{-oz}$.

Weight of sphere B is $10\text{-oz}$.

Concept used:

Draw the diagram for the system considers position 1 as datum.

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Write the expression for height $h_C$.

$h_C=BC\left(1-\cos {\theta }_m\right)$

Write the expression for height $h_A$.

$h_A=BA\left(1-\cos {\theta }_m\right)$

Write the identity as $1-\cos {\theta }_m=2{\sin }^2\frac{{\theta }_m}{2}$

As the value of ${\theta }_m$ is very small therefore, $\sin {\theta }_m\approx {\theta }_m$

Substitute $\frac{{\theta }_m^2}{2}$ for $1-\cos {\theta }_m$ in the value of $h_A$ and $h_C$.

$\begin{array}{l}{h}_A=BA\frac{{\theta }_m^2}{2}\\ h_C=BC\frac{{\theta }_m^2}{2}\end{array}$

Write the expression for velocity of point C.

${\left(v_C\right)}_m=\left(BC\right){\dot{\theta }}_m^2$ ...... (1)

Here, ${\left(v_C\right)}_m$ is the velocity of point C, $BC$ is the length of bar between the points B and C and ${\dot{\theta }}_m$ is the angular velocity of rod.

Write the expression for velocity of point A.

${\left(v_A\right)}_m=BA{\dot{\theta }}_m^2$ ...... (2)

Here, ${\left(v_A\right)}_m$ is the velocity of point A and $BA$ is the length of bar between the points B and A.

Write the expression for the conservation of energy.

$T_1+V_1=T_2+V_2$ ...... (3)

Here, $T_1$ is the kinetic energy system at position (1), $V_1$ is the potential energy system at position (1), $T_2$ is the kinetic energy system at position (2) and $V_2$ is the potential energy system at position (2).

The kinetic energy of the system at position one is zero as the system is in rest.

Write the expression for the potential energy at point (1).

$V_1=w_C{h}_C-w_A{h}_A$

Substitute $BC\frac{{\dot{\theta }}_m^2}{2}$ for $h_C$ and $BA\frac{{\theta }_m^2}{2}$ for $h_A$ in the above expression.

$V_1=\left[w_{C}BC-w_{A}BA\right]\frac{{\theta }_m^2}{2}$ ...... (4)

Here, $w_A$ is the weight of sphere A and $w_C$ is the weight of sphere C.

Write the expression for the kinetic energy at position (2).

$T_2=\frac{1}{2}{m}_C{\left(v_C\right)}_m^2+\frac{1}{2}{m}_A{\left(v_A\right)}_m^2$

Substitute $\frac{w_C}{g}$ for $m_C$, $\frac{w_A}{g}$ for $m_A$, ${\omega }_n{\theta }_m$ for ${\dot{\theta }}_m^2$, $BA{\dot{\theta }}_m^2$ for ${\left(v_A\right)}_m$ and $BC{\dot{\theta }}_m^2$ for ${\left(v_C\right)}_m$ in above expression.

$T_2=\frac{\left({\omega }_n^2{\theta }_m^2\right)}{2g}\left[w_C{\left(BC\right)}^2+w_A{\left(BA\right)}^2\right]$ ...... (5)

Here, $g$ is the acceleration due to gravity and ${\omega }_n$ is the natural frequency of oscillation.

Potential energy at position (2) is zero.

Write the expression for the time period of oscillation.

$\tau =\frac{2\pi }{{\omega }_n}$ ...... (6)

Here, $\tau$ is the time period of oscillation.

Calculation:

Substitute $10\text{-oz}$ for $w_C$, $14\text{-oz}$ for $w_A$, $8\text{ in}$ for $BC$ and $5\text{ in}$ for $BA$ in equation (4).

$\begin{array}{c}{V}_1=\left[10\text{-oz}\left(\frac{1\text{ lb}}{16\text{-oz}}\right)\left(8\text{ in}\right)\left(\frac{1\text{ ft}}{12\text{ in}}\right)-14\text{-oz}\left(\frac{1\text{ lb}}{16\text{-oz}}\right)\left(\text{5 in}\right)\left(\frac{1\text{ ft}}{12\text{ in}}\right)\right]\frac{{\theta }_m^2}{2}\\ =0.05208\left(\frac{{\theta }_m^2}{2}\right)\end{array}$

Substitute $32.2{\text{ ft/s}}^2$ for $g$, $10\text{-oz}$ for $w_C$, $14\text{-oz}$ for $w_A$, $8\text{ in}$ for $BC$ and $5\text{ in}$ for $BA$ in equation (5).

$\begin{array}{c}{T}_2=\frac{{\left({\omega }_n^2{\theta }_m^2\right)}^2}{2\left(32.2\right)}\left[\begin{array}{l}10\text{-oz}\left(\frac{1\text{ lb}}{16\text{-oz}}\right){\left[\left(8\text{ in}\right)\left(\frac{1\text{ ft}}{12\text{ in}}\right)\right]}^2\\ +14\text{-oz}\left(\frac{1\text{ lb}}{16\text{-oz}}\right){\left[\left(\text{5 in}\right)\left(\frac{1\text{ ft}}{12\text{ in}}\right)\right]}^2\end{array}\right]\\ =\left(6.672\times {10}^{-3}\right)\left({\omega }_n^2{\theta }_m^2\right)\end{array}$

Substitute $0$ for $T_1$, $\left(6.672\times {10}^{-3}\right)\left({\omega }_n^2{\theta }_m^2\right)$ for $T_2$, $0$ for $V_2$ and $0.05208\left(\frac{{\theta }_m^2}{2}\right)$ for $V_1$ in equation (3).

$\begin{array}{l}0+0.05208\left(\frac{{\theta }_m^2}{2}\right)=\left(6.672\times {10}^{-3}\right)\left({\omega }_n^2{\theta }_m^2\right)+0\\ 0.02604\left({\theta }_m^2\right)=\left(6.672\times {10}^{-3}\right)\left({\omega }_n^2{\theta }_m^2\right)\end{array}$

Simplify the above expression for ${\omega }_n$.

$\begin{array}{c}{\omega }_n^2=3.903\\ {\omega }_n=\sqrt{3.903}\\ =1.976\text{rad/s}\end{array}$

Substitute $1.976\text{ rad/s}$ for ${\omega }_n$ in equation (6).

$\begin{array}{c}\tau =\frac{2\pi }{1.976}\\ =3.18\text{ s}\end{array}$

The period of small oscillation of rod is $3.18\text{}\text{ s}$.

Conclusion:

Thus, the period of small oscillation of rod is $3.18\text{}\text{ s}$.