Chapter 19.3, Problem 19.89P

(a)

To determine

The frequency $\left(f_n\right)$ of small oscillation for a is equal to 0.5 m.

Answer to Problem 19.89P

The frequency $\left(f_n\right)$ of small oscillation for a is equal to 0.5 m is $\underset{\_}{0.802\text{Hz}}$.

Explanation of Solution

Given information:

The mass $\left(m_S\right)$ of the sphere is 2 kg.

The mass $\left(m_{ABC}\right)$ of the bar ABC is 1.5 kg.

The length (l) of the bar is 0.7 m.

The spring constant (k) is $200\text{N/m}$.

The value of a is 0.5 m.

Assume the acceleration due to gravity (g) as $9.81{\text{m/s}}^{\text{2}}$.

Calculation:

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

For position 1(Maximum velocity):

Calculate the mass moment of inertia bar ABC $\left(I_{ABC}\right)$ using the relation:

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

Substitute 1.5 kg for $m_{ABC}$ and 0.7 m for l.

$\begin{array}{c}{I}_{ABC}=\frac{1}{12}\times 1.5\times {\left(0.7\right)}^2\\ =0.06125\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Write the expression the maximum velocity of sphere ${\left(v_S\right)}_m$:

${\left(v_S\right)}_m=l{\dot{\theta }}_m$

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

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

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

$T_1=\frac{1}{2}{m}_S{\left(v_S\right)}_m^2+\frac{1}{2}\left(I_{ABC}\right){\dot{\theta }}_m^2+\frac{1}{2}{m}_{ABC}{\left(v_{ABC}\right)}_m^2$

Substitute 2 kg for $m_S$, $l{\dot{\theta }}_m$ for ${\left(v_S\right)}_m$, $0.06125\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I_{ABC}$, 1.5 kg for $m_{ABC}$, $\frac{l}{2}{\dot{\theta }}_m$ for $v_{ABC}$, and 0.7 m for l.

$\begin{array}{c}{T}_1=\frac{1}{2}\times 2\times \left({0.7}^2{\dot{\theta }}_m^2\right)+\frac{1}{2}\left(0.06125\right){\dot{\theta }}_m^2+\frac{1}{2}\times 1.5\times {\left(\frac{0.7}{2}\right)}^2{\dot{\theta }}_m^2\\ =\left[0.49+0.030625+0.091875\right]{\dot{\theta }}_m^2\\ =0.6125{\dot{\theta }}_m^2\end{array}$

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 in sphere $\left(h_S\right)$:

$\begin{array}{c}{h}_S=\left(l-l\cos {\theta }_m\right)\\ =l\left(1-\cos {\theta }_m\right)\left(\because \left(1-\cos {\theta }_m=2{\sin }^2\frac{{\theta }_m}{2}\right)\right)\\ =l2{\sin }^2\frac{{\theta }_m}{2}\end{array}$

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

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

Write the expression for the displacement of rod ABC $\left(h_{ABC}\right)$:

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

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

$h_{ABC}=\frac{l}{2}\frac{{\theta }_m^2}{2}$

Write the expression for the displacement of spring $\left(x_m\right)$:

$x_m=a\sin {\theta }_m$

For small angle $\sin {\theta }_m\approx {\theta }_m$.

$x_m=a{\theta }_m$

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

$V_2=\frac{1}{2}k{x}_m^2-m_{S}g{h}_S-m_{ABC}g{h}_{ABC}$

Substitute $a{\theta }_m$ for $x_m$, $\frac{l}{2}\frac{{\theta }_m^2}{2}$ for $h_{ABC}$, and $l\frac{{\theta }_m^2}{2}$ for $h_S$.

$V_2=\frac{1}{2}k{\left(a{\theta }_m\right)}^2-m_{S}g\left(l\frac{{\theta }_m^2}{2}\right)-m_{ABC}g\left(\frac{l}{2}\frac{{\theta }_m^2}{2}\right)$

Substitute $200\text{N/m}$ for k, 0.7 m for l, 2 kg for $m_S$, 1.5 kg for $m_{ABC}$, and $9.81{\text{m/s}}^{\text{2}}$ for g.

$\begin{array}{c}{V}_2=\frac{1}{2}\times 200\times {\left(a{\theta }_m\right)}^2-2\times 9.81\times \left(0.7\times \frac{{\theta }_m^2}{2}\right)-1.5\times 9.81\times \left(\frac{0.7}{2}\frac{{\theta }_m^2}{2}\right)\\ =100a^2{\theta }_m^2-6.867{\theta }_m^2-2.575{\theta }_m^2\\ =\left(100a^2-9.442\right){\theta }_m^2\end{array}$

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 $0.6125{\dot{\theta }}_m^2$ for $T_1$, 0 for $V_1$, 0 for $T_2$ and $\left(100a^2-9.442\right){\theta }_m^2$ for $V_2$.

$\begin{array}{l}0.6125{\dot{\theta }}_m^2+0=0+\left(100a^2-9.442\right){\theta }_m^2\\ 0.6125{\dot{\theta }}_m^2=\left(100a^2-9.442\right){\theta }_m^2\end{array}$

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

$\begin{array}{l}0.6125{\omega }_n^2{\theta }_m^2=\left(100a^2-9.442\right){\theta }_m^2\\ {\omega }_n^2=\frac{\left(100a^2-9.442\right)}{0.6125}\end{array}$ (1)

Substitute 0.5 m for a in Equation (1).

$\begin{array}{l}{\omega }_n^2=\frac{\left(100{\left(0.5\right)}^2-9.442\right)}{0.6125}\\ {\omega }_n=\sqrt{25.401}\\ {\omega }_n=5.0399\text{rad/s}\end{array}$

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

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

Substitute $5.0399\text{rad/s}$ for ${\omega }_n$.

$\begin{array}{c}{f}_n=\frac{5.0399}{2\pi }\\ =0.802\text{Hz}\end{array}$

Therefore, the frequency $\left(f_n\right)$ of small oscillation for a is equal to 0.5 m is $\underset{\_}{0.802\text{Hz}}$.

(b)

To determine

The smallest value of a for which oscillation will occur.

Answer to Problem 19.89P

The smallest value of a for which oscillation will occur is $\underset{\_}{0.307\text{m}}$.

Explanation of Solution

Given information:

The mass $\left(m_S\right)$ of the sphere is 2 kg.

The mass $\left(m_{ABC}\right)$ of the bar ABC is 1.5 kg.

The length (l) of the bar is 0.7 m.

The spring constant (k) is $200\text{N/m}$.

The value of a is 0.5 m.

Assuming the acceleration due to gravity (g) as $9.81{\text{m/s}}^{\text{2}}$.

Calculation:

Calculate the value of a:

Substitute 0 for ${\omega }_n$ in Equation (1).

$\begin{array}{l}0=\frac{\left(100a^2-9.442\right)}{0.6125}\\ 0=100a^2-9.442\\ a^2=\frac{9.442}{100}\\ a=\sqrt{0.09442}\\ a=0.307\text{m}\end{array}$

Therefore, the smallest value of a for which oscillation will occur is $\underset{\_}{0.307\text{m}}$.