Chapter 19.2, Problem 19.50P

A small collar of mass 1 kg is rigidly attached to a 3-kg uniform rod of length L = 750 mm. Determine (a) the distance d to maximize the frequency of oscillation when the rod is given a small initial displacement, (b) the corresponding period of oscillation.

Fig. P19.50

To determine

The distance d to maximize the frequency of oscillation when the rod is given a small initial displacement.

Answer to Problem 19.50P

The distance d to maximize the frequency of oscillation when the rod is given a small initial displacement is $\underset{\_}{227\text{mm}}$.

Explanation of Solution

Given information:

The mass $\left(m_C\right)$ of the collar is 1 kg.

The mass $\left(m_R\right)$ of the rod AB is 3 kg.

The length (L) of the rod AB is 750 mm.

The acceleration due to gravity (g) is $\text{9}\text{.81}{\text{m/s}}^{\text{2}}$.

Calculation:

Show the free-body-diagram equation as in Figure (1).

The external forces in the system are forces due mass of the collar and the rod. The effective force in the system is $m_c{\overline{a}}_t$ and the effective restoring couple is $\overline{I}\ddot{\theta }$.

Take moment about A in the system for external forces.

$\begin{array}{c}{\left(\sum M_A\right)}_{external}=\left(m_{r}g\times \frac{L}{2}\sin \theta \right)+\left(m_{c}g\times d\sin \theta \right)\\ =\left(\frac{m_{r}L}{2}+m_{c}d\right)g\sin \theta \end{array}$

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

Take moment about A in the system for effective forces.

$\begin{array}{c}{\left(\sum M_A\right)}_{effective}=-\left(m_r{\left({\overline{a}}_t\right)}_r\times \frac{L}{2}\right)-\left(m_c{\left({\overline{a}}_t\right)}_c\times d\right)-\left(\overline{I}\ddot{\theta }\right)\\ =-m_r\frac{L}{2}{\left({\overline{a}}_t\right)}_r-m_{c}d{\left({\overline{a}}_t\right)}_c-\overline{I}\ddot{\theta }\end{array}$

The tangential component of acceleration for the rod ${\left({\overline{a}}_t\right)}_r$ is equal to the product of distance at which the weight of the rod acts and angular acceleration $\left(\frac{L}{2}\ddot{\theta }\right)$.

The tangential component of acceleration for the collar ${\left({\overline{a}}_t\right)}_c$ is equal to the product of distance at which the collar is added and angular acceleration $\left(d\ddot{\theta }\right)$.

Thus, express the moment about A due to the effective forces as:

$\begin{array}{c}{\left(\sum M_A\right)}_{effective}=-m_r\frac{L}{2}\left(\frac{L}{2}\ddot{\theta }\right)-m_{c}d\left(d\ddot{\theta }\right)-\overline{I}\ddot{\theta }\\ =-m_r{\left(\frac{L}{2}\right)}^2\ddot{\theta }-m_c{d}^2\ddot{\theta }-\overline{I}\ddot{\theta }\\ =-\left(\frac{m_r{L}^2}{4}+m_c{d}^2+\overline{I}\right)\ddot{\theta }\end{array}$

Equate the moment about A in the system for external and effective forces.

$\begin{array}{l}{\left(\sum M_A\right)}_{external}={\left(\sum M_A\right)}_{effective}\\ \left(\frac{m_{r}L}{2}+m_{c}d\right)g\theta =-\left(\frac{m_r{L}^2}{4}+m_c{d}^2+\overline{I}\right)\ddot{\theta }\\ \ddot{\theta }=-\frac{\frac{m_{r}L}{2}+m_{c}d}{\frac{m_r{L}^2}{4}+m_c{d}^2+\overline{I}}g\theta \\ \ddot{\theta }+\frac{\left(\frac{m_{r}L}{2}+m_{c}d\right)g}{\frac{m_r{L}^2}{4}+m_c{d}^2+\overline{I}}\theta =0\end{array}$ (1)

Compare the differential Equation (1) with the general differential equation of motion $\left(\ddot{x}+{\omega }_n^{2}x=0\right)$ and express the natural circular frequency of oscillation $\left({\omega }_n\right)$:

${\omega }_n^2=\frac{\left(\frac{m_{r}L}{2}+m_{c}d\right)g}{\frac{m_r{L}^2}{4}+m_c{d}^2+\overline{I}}$ (2)

Write the expression for moment of inertia of the rod:

$\overline{I}=\frac{1}{12}{m}_r{L}^2$

Substitute 1 kg for $m_c$, 3 kg for $m_r$, 750 mm for L, $\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}$ for g, and $\left(1/12\right)m_r{L}^2$ for $\overline{I}$ in Equation (2).

$\begin{array}{c}{\omega }_n^2=\frac{\left(\frac{m_{r}L}{2}+m_{c}d\right)g}{\frac{m_r{L}^2}{4}+m_c{d}^2+\overline{I}}\\ =\frac{\left(\frac{\left(3\text{kg}\right)\left(750\text{mm}\right)}{2}+\left(1\text{kg}\right)d\right)\left(\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}\right)}{\frac{\left(3\text{kg}\right){\left(750\text{mm}\right)}^2}{4}+\left(1\text{kg}\right)d^2+\left(\frac{1}{12}\left(3\text{kg}\right){\left(750\text{mm}\right)}^2\right)}\\ =\frac{\left(1.5\left(750\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)+d\right)\left(\text{9}\text{.81}\right)}{0.75{\left(750\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2+d^2+\left(\frac{1}{4}{\left(750\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2\right)}\end{array}$

$\begin{array}{c}=\frac{\left(1.5\left(0.75\text{m}\right)+d\right)\left(\text{9}\text{.81}\right)}{0.75{\left(0.75\text{m}\right)}^2+d^2+\left(\frac{1}{4}{\left(0.75\text{m}\right)}^2\right)}\\ =\frac{9.81\left(1.125+d\right)}{0.421875+d^2+0.140625}\\ =\frac{9.81\left(1.125+d\right)}{0.5625+d^2}\end{array}$ (3)

Calculate the value of d to maximize the natural frequency:

Differentiate Equation (3) with respect to d.

$\begin{array}{l}{\omega }_n^2=\frac{9.81\left(1.125+d\right)}{0.5625+d^2}\\ \frac{d\left({\omega }_n^2\right)}{dd}=9.81\frac{\left\{\left(0.5625+d^2\right)\left(1\right)-\left(1.125+d\right)\left(2d\right)\right\}}{{\left(0.5625+d^2\right)}^2}\\ \frac{d\left({\omega }_n^2\right)}{dd}=9.81\frac{\left\{0.5625+d^2-2.25d-2d^2\right\}}{{\left(0.5625+d^2\right)}^2}\\ \frac{d\left({\omega }_n^2\right)}{dd}=9.81\frac{\left\{0.5625-2.25d-d^2\right\}}{{\left(0.5625+d^2\right)}^2}\end{array}$

Equate $\frac{d\left({\omega }_n^2\right)}{dd}$ to zero and solve for d.

$\begin{array}{l}\frac{d\left({\omega }_n^2\right)}{dd}=0\\ 9.81\frac{\left\{0.5625-2.25d-d^2\right\}}{{\left(0.5625+d^2\right)}^2}=0\\ 0.5625-2.25d-d^2=0\\ d^2+2.25d-0.5625=0\end{array}$ (4)

Solve the above quadratic equation:

Express the roots of a quadratic equation:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute d for x, 1 for a, 2.25 for b, and -0.5625 for c to find the roots of the Equation (4).

$\begin{array}{c}d=\frac{-\left(2.25\right)\pm \sqrt{{\left(2.25\right)}^2-4\left(1\right)\left(-0.5625\right)}}{2\left(1\right)}\\ =\frac{-2.25\pm \sqrt{5.0625+2.25}}{2}\\ =\frac{-2.25\pm 2.704}{2}\\ =\frac{-2.25+2.704}{2}\text{or}\frac{-2.25-2.704}{2}\end{array}$

$\begin{array}{c}=0.227\text{m}\text{or}-\text{2}\text{.4771}\text{m}\\ =0.227\left(\frac{1,000\text{ mm}}{1\text{ m}}\right)\text{or}-\text{2}\text{.4771}\text{m}\left(\frac{1,000\text{ mm}}{1\text{ m}}\right)\\ =227\text{ mm or }-2477.1\text{ mm}\end{array}$

Therefore, the distance d to maximize the frequency of oscillation when the rod is given a small initial displacement is $\underset{\_}{227\text{mm}}$.

To determine

The corresponding period $\left({\tau }_n\right)$ of oscillation.

Answer to Problem 19.50P

The corresponding period $\left({\tau }_n\right)$ of oscillation is $\underset{\_}{1.352\sec }$.

Explanation of Solution

Given information:

The mass $\left(m_C\right)$ of the collar is 1 kg.

The mass $\left(m_R\right)$ of the rod AB is 3 kg.

The length (L) of the rod AB is 750 mm.

The acceleration due to gravity (g) is $\text{9}\text{.81}{\text{m/s}}^{\text{2}}$.

Calculation:

Calculate the value of maximum natural circular frequency ${\omega }_n$:

Substitute 0.227 m for d in Equation (3)

$\begin{array}{l}{\omega }_n^2=\frac{9.81\left(1.125+d\right)}{0.5625+d^2}\\ {\omega }_n^2=\frac{9.81\left(1.125+\left(0.227\text{m}\right)\right)}{0.5625+{\left(0.227\text{m}\right)}^2}\\ {\omega }_n^2=\frac{13.26312}{0.614029}\\ {\omega }_n=\sqrt{21.6}\\ {\omega }_n=4.6476\text{rad}/\text{s}\end{array}$

Calculate the time period of oscillation $\left({\tau }_n\right)$ using the relation:

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

Substitute $4.6476\text{rad}/\text{s}$ for ${\omega }_n$.

$\begin{array}{c}{\tau }_n=\frac{2\pi }{4.6476\text{rad}/\text{s}}\\ =1.352\text{s}\end{array}$

Therefore, the corresponding period $\left({\tau }_n\right)$ of oscillation is $\underset{\_}{1.352\sec }$.