Chapter 19.2, Problem 19.47P

A connecting rod is supported by a knife-edge at point A; the period of its small oscillations is observed to be 0.87 s. The rod is then inverted and supported by a knife edge at point B and the period of its small oscillations is observed to be 0.78 s. Knowing that r_a + r_b = 10 in., determine (a) the location of the mass center G, (b) the centroidal radius of gyration $\bar{k}$.

Fig. P19.47

To determine

The location $\left(r_a\text{and}{r}_b\right)$ of the mass center G.

Answer to Problem 19.47P

The location $\left(r_a\text{and}{r}_b\right)$ of the mass center G are $\underset{\_}{6.09\text{in}\text{.}}$ and $\underset{\_}{3.91\text{in}\text{.}}$.

Explanation of Solution

Given information:

The period $\left({\tau }_A\right)$ of small oscillation is 0.87 s.

The rod is then inverted and supported by a knife-edge at point B and the period $\left({\tau }_B\right)$ of small oscillation is 0.78 s.

The value of $r_a+r_b$ is 10 inch.

The acceleration due to gravity (g) is $32.2{\text{ft/s}}^{\text{2}}$.

Calculation:

Show the free body diagram of mass center G as in Figure (1).

The external forces in the system are tension in the force due to the mass of the knife. The effective couple in the system is $\overline{I}\ddot{\theta }$.

Take moment about O in the system for external forces.

${\left(\sum M_O\right)}_{external}=-mg\overline{r}\sin \theta$

Take moment about O in the system for effective forces.

${\left(\sum M_O\right)}_{effective}=\left(m\overline{r}\ddot{\theta }\right)\overline{r}+m{\overline{k}}^2\ddot{\theta }$

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

$\begin{array}{l}{\left(\sum M_O\right)}_{external}={\left(\sum M_O\right)}_{effective}\\ -mg\overline{r}\sin \theta =\left(m\overline{r}\ddot{\theta }\right)\overline{r}+m{\overline{k}}^2\ddot{\theta }\\ -mg\overline{r}\sin \theta -\left(m\overline{r}\ddot{\theta }\right)\overline{r}-+m{\overline{k}}^2\ddot{\theta }=0\\ -m\left(g\overline{r}\sin \theta +{\overline{r}}^2\ddot{\theta }+{\overline{k}}^2\ddot{\theta }\right)=0\end{array}$

$\begin{array}{l}\left({\overline{r}}^2+{\overline{k}}^2\right)\ddot{\theta }+g\overline{r}\sin \theta =0\\ \left({\overline{r}}^2+{\overline{k}}^2\right)\left(\ddot{\theta }+\frac{g\overline{r}\sin \theta }{\left({\overline{r}}^2+{\overline{k}}^2\right)}\right)=0\\ \ddot{\theta }+\frac{g\overline{r}}{\left({\overline{r}}^2+{\overline{k}}^2\right)}\sin \theta =0\end{array}$

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

$\ddot{\theta }+\frac{g\overline{r}}{\left({\overline{r}}^2+{\overline{k}}^2\right)}\theta =0$ (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 vibration:

$\begin{array}{c}{\omega }_n^2=\frac{g\overline{r}}{\left({\overline{r}}^2+{\overline{k}}^2\right)}\\ {\omega }_n=\sqrt{\frac{g\overline{r}}{\left({\overline{r}}^2+{\overline{k}}^2\right)}}\end{array}$

Write the expression for period $\left(\tau \right)$:

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

Substitute $\sqrt{\frac{g\overline{r}}{\left({\overline{r}}^2+{\overline{k}}^2\right)}}$ for ${\omega }_n$.

$\tau =\frac{2\pi }{\sqrt{\frac{g\overline{r}}{\left({\overline{r}}^2+{\overline{k}}^2\right)}}}$ (2)

Rewrite the equation (2) for rod suspended at A:

$\begin{array}{l}{\tau }_A=\frac{2\pi }{\sqrt{\frac{g{\overline{r}}_a}{\left({\overline{r}}_a^2+{\overline{k}}^2\right)}}}\\ {\tau }_A=2\pi \sqrt{\frac{\left({\overline{r}}_a^2+{\overline{k}}^2\right)}{g{\overline{r}}_a}}\\ {\left({\tau }_A\right)}^2={\left(2\pi \right)}^2\frac{\left({\overline{r}}_a^2+{\overline{k}}^2\right)}{g{\overline{r}}_a}\\ {\left({\tau }_A\right)}^{2}g{r}_a=4{\pi }^2\left(r_a^2+{\overline{k}}^2\right)\end{array}$ (3)

Rewrite the equation (2) for rod suspended at B:

$\begin{array}{l}{\tau }_B=\frac{2\pi }{\sqrt{\frac{g{\overline{r}}_b}{\left({\overline{r}}_b^2+{\overline{k}}^2\right)}}}\\ {\tau }_B=2\pi \sqrt{\frac{\left({\overline{r}}_b^2+{\overline{k}}^2\right)}{g{\overline{r}}_b}}\\ {\left({\tau }_B\right)}^2={\left(2\pi \right)}^2\frac{\left({\overline{r}}_b^2+{\overline{k}}^2\right)}{g{\overline{r}}_b}\\ {\left({\tau }_B\right)}^{2}g{r}_b=4{\pi }^2\left(r_b^2+{\overline{k}}^2\right)\end{array}$ (4)

Subtracting equation (4) from equation (3).

$\begin{array}{l}{\left({\tau }_A\right)}^{2}g{r}_a-{\left({\tau }_B\right)}^{2}g{r}_b=4{\pi }^2\left(r_a^2+{\overline{k}}^2\right)-4{\pi }^2\left(r_b^2+{\overline{k}}^2\right)\\ {\left({\tau }_A\right)}^{2}g{\overline{r}}_a-{\left({\tau }_B\right)}^{2}g{r}_b=4{\pi }^2\left(r_a^2-r_b^2\right)\\ {\left({\tau }_A\right)}^{2}g{r}_a-{\left({\tau }_B\right)}^{2}g{r}_b=4{\pi }^2\left(r_a+r_b\right)\left(r_a-r_b\right)\end{array}$

Substitute 0.87 s for ${\tau }_A$, 0.78 for ${\tau }_B$, $32.2{\text{ft/s}}^{\text{2}}$ for g, and 10 in. for $r_a+r_b$.

$\begin{array}{l}{\left(0.87\right)}^2\left(32.2\right)r_a-{\left(0.78\right)}^2\left(32.2\right)r_b=4{\pi }^2\left(10\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\left(r_a-r_b\right)\\ 24.372r_a-19.590r_b=32.899\left(r_a-r_b\right)\\ 24.372r_a-19.590r_b=32.899r_a-32.899r_b\end{array}$

$\begin{array}{l}\left(-19.590r_b+32.899r_b\right)=\left(32.899r_a-24.372r_a\right)\\ 13.309r_b=8.527r_a\\ r_b=\frac{8.527r_a}{13.309}\\ r_b=0.641r_a\end{array}$ (5)

Write the relationship $\left(r_a+r_b\right)$:

$r_a+r_b=10\text{in}\text{.}$ (6).

Calculate the value $\left(r_a\right)$:

Substitute $0.641r_a^{}$ for $r_b^{}$ in equation (6).

$\begin{array}{l}{r}_a+\left(0.641r_a\right)=10\text{in}\text{.}\\ r_a=\frac{10\text{in}\text{.}}{1.641}\\ r_a=6.09385\text{in}\text{.}\\ r_a\simeq 6.09\text{in}\text{.}\end{array}$

Calculate the value $\left(r_b\right)$:

Substitute $6.09\text{in}\text{.}$ for $r_a$ in equation (5).

$\begin{array}{c}{r}_b=0.641\left(6.09385\text{in}\text{.}\right)\\ =3.91\text{in}\text{.}\end{array}$

Therefore, the location $\left(r_a\text{and}{r}_b\right)$ of the mass center G are $\underset{\_}{6.09\text{in}\text{.}}$ and $\underset{\_}{3.91\text{in}\text{.}}$.

To determine

The centroidal radius of gyration $\left(\overline{k}\right)$.

Answer to Problem 19.47P

The centroidal radius of gyration $\left(\overline{k}\right)$ is $\underset{\_}{2.83\text{in}\text{.}}$

Explanation of Solution

Given information:

The period $\left({\tau }_A\right)$ of small oscillation is 0.87 s.

The rod is then inverted and supported by a knife-edge at point B and the period $\left({\tau }_B\right)$ of small oscillation is 0.78 s.

The value of $r_a+r_b$ is 10 in.

The acceleration due to gravity (g) is $32.2{\text{ft/s}}^{\text{2}}$.

Calculation:

Calculate the centroidal radius of gyration $\overline{k}$:

Substitute $32.2{\text{ft/s}}^{\text{2}}$ for g,0.87 s for ${\tau }_A$ and $6.09\text{in}\text{.}$ for $r_B$ in equation (3).

$\begin{array}{c}{\left(0.87\right)}^2\left(32.2\times \left(6.09385\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\right)=4{\pi }^2\left({\left(6.09385\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2+{\overline{k}}^2\right)\\ 12.377=4{\pi }^2\left(0.2579+{\overline{k}}^2\right)\\ 12.377=\left(4{\pi }^2\times 0.2579\right)+4{\pi }^2{\overline{k}}^2\end{array}$

$\begin{array}{c}4{\pi }^2{\overline{k}}^2=12.377-\left(4{\pi }^2\times 0.2579\right)\\ \overline{k}=\sqrt{\frac{2.1955}{4{\pi }^2}}\\ =0.2358\text{ft}\times \frac{\text{12}\text{in}\text{.}}{1\text{ft}}\\ =2.83\text{in}\text{.}\end{array}$

Therefore, the centroidal radius of gyration $\left(\overline{k}\right)$ is $\underset{\_}{2.83\text{in}\text{.}}$