Chapter 19.2, Problem 19.51P

(a)

To determine

The period $\left({\tau }_n\right)$ of small oscillation if the wire is suspended from point A.

Answer to Problem 19.51P

The period $\left({\tau }_n\right)$ of small oscillation if the wire is suspended from point A is $\underset{\_}{6.33\sqrt{\frac{b}{g}}}$.

Explanation of Solution

Given information:

A thin homogeneous wire is bent into the shape of an isosceles triangle of sides are b, b and 1.6b.

Calculation:

Show the position of centroid and distance as in Figure (1).

Write the equation for mass moment of inertia $\left(\overline{I}\right)$:

$\overline{I}=m{r}^2$

Here, r is distance of each particle from the axis of rotation.

Calculate the expression for mass moment of inertia $\left(\overline{I_A}\right)$ using mass by length ratio at point A:

$I_A=2\left[\frac{1}{3}\left(\frac{m}{1+1+1.6}\right)b^2\right]+\left(\frac{1.6m}{1+1+1.6}\right)\left[\frac{{\left(1.6b\right)}^2}{12}+{\left(0.6b\right)}^2\right]$

Here, $2\left[\frac{1}{3}\left(\frac{m}{1+1+1.6}\right)b^2\right]$ represents the moment of inertia of two inclined wires.

Modify the above equation,

$\begin{array}{c}{I}_A=2\left[\frac{1}{3}\left(\frac{m}{1+1+1.6}\right)b^2\right]+\left(\frac{1.6m}{1+1+1.6}\right)\left[\frac{{\left(1.6b\right)}^2}{12}+{\left(0.6b\right)}^2\right]\\ =\frac{5}{27}m{b}^2+\frac{172}{675}m{b}^2\\ =\frac{11}{25}m{b}^2\end{array}$

Calculate the centroid equation $\left(\overline{y}\right)$ about y axis:

$\begin{array}{c}\overline{y}=\frac{\left(b+b\right)\left(\frac{0.6b}{2}\right)}{b+b+1.6b}\\ =\frac{0.6b}{3.6b}\\ =\frac{b}{6}\end{array}$

Calculate the distance equation AG as below

$AG=0.6b-\overline{y}$

Substitute $\frac{b}{6}$ for $\overline{y}$.

$\begin{array}{c}AG=0.6b-\frac{b}{6}\\ =\frac{13}{30}b\end{array}$

The external forces in the system are force due mass of the thin wire and the effective restoring couple is $\overline{I}\ddot{\theta }$.

Take moment about A in the system for external forces.

${\left(\sum M_A\right)}_{external}=-mg\left(AG\right)\theta$

Substitute $\frac{13}{30}b$ for AG.

$\begin{array}{c}{\left(\sum M_A\right)}_{external}=-mg\left(\frac{13}{30}b\right)\theta \\ =-\frac{13mgb}{30}\theta \end{array}$

Take moment about A in the system for effective forces.

${\left(\sum M_A\right)}_{effective}=I_A\ddot{\theta }$

Substitute $\frac{11}{25}m{b}^2$ for $I_A$.

${\left(\sum M_A\right)}_{effective}=\frac{11}{25}m{b}^2\ddot{\theta }$

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

$\begin{array}{l}{\left(\sum M_A\right)}_{external}={\left(\sum M_A\right)}_{effective}\\ -\frac{13mgb}{30}\theta =\frac{11}{25}m{b}^2\ddot{\theta }\\ -\frac{11}{25}m{b}^2\ddot{\theta }-\frac{13mgb}{30}\theta =0\end{array}$

$\begin{array}{l}\frac{11}{25}m{b}^2\left(\ddot{\theta }+\frac{\left(\frac{13mgb}{30}\right)}{\frac{11}{25}m{b}^2}\theta \right)=0\\ \ddot{\theta }+\frac{\left(\frac{13mgb}{30}\right)}{\frac{11}{25}m{b}^2}\theta =0\end{array}$ (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 $\left({\omega }_n\right)$:

$\begin{array}{c}{\omega }_n^2=\frac{\left(\frac{13mgb}{30}\right)}{\frac{11}{25}m{b}^2}\\ {\omega }_n^2=\frac{13gb}{30}\times \frac{25}{11b^2}\\ {\omega }_n=\sqrt{0.985\frac{g}{b}}\end{array}$

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

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

Substitute $\sqrt{0.985\frac{g}{b}}$ for ${\omega }_n$.

$\begin{array}{c}{\tau }_n=\frac{2\pi }{\sqrt{0.985\frac{g}{b}}}\\ =6.33\sqrt{\frac{g}{b}}\end{array}$

Therefore, the period $\left({\tau }_n\right)$ of small oscillation if the wire is suspended from point A is $\underset{\_}{6.33\sqrt{\frac{b}{g}}}$.

(b)

To determine

The period $\left({\tau }_n\right)$ of small oscillation if the wire is suspended from point B.

Answer to Problem 19.51P

The period $\left({\tau }_n\right)$ of small oscillation if the wire is suspended from point B is $\underset{\_}{6.67\sqrt{\frac{b}{g}}}$.

Explanation of Solution

Given information:

A thin homogeneous wire is bent into the shape of an isosceles triangle of sides are b, b and 1.6b.

Calculation:

Calculate the expression for mass moment of inertia $\left(\overline{I_A}\right)$ using mass by length ratio at point B:

$I_B=I_A-m{\left(AG\right)}^2+m\left({\left(0.8b\right)}^2+{\overline{y}}^2\right)$

Substitute $\frac{11}{25}m{b}^2$ for $I_A$, $\frac{13}{30}b$ for $AG$, and $\frac{b}{6}$ for $\overline{y}$.

$\begin{array}{c}{I}_B=\frac{11}{25}m{b}^2-m{\left(\frac{13}{30}b\right)}^2+m\left({\left(0.8b\right)}^2+{\left(\frac{b}{6}\right)}^2\right)\\ =\frac{11}{25}m{b}^2-\frac{169}{900}m{b}^2+\frac{601}{900}m{b}^2\\ =\frac{23}{25}m{b}^2\end{array}$

Calculate the equation for distance GB by using the Pythagoras theorem:

$GB=\sqrt{{\overline{y}}^2+{\left(0.8b\right)}^2}$

Substitute $\frac{b}{6}$ for $\overline{y}$.

$\begin{array}{c}GB=\sqrt{{\left(\frac{b}{6}\right)}^2+{\left(0.8b\right)}^2}\\ =\sqrt{{\frac{b}{36}}^2+0.64b^2}\\ =\sqrt{\frac{b^2+23.04b^2}{36}}\\ =\frac{b}{6}\sqrt{24.04}\end{array}$

The external forces in the system are force due mass of the thin wire and the effective restoring couple is $\overline{I}\ddot{\theta }$.

Take moment about B in the system for external forces.

${\left(\sum M_B\right)}_{external}=-mg\left(GB\right)\theta$

Substitute $\frac{b}{6}\sqrt{24.04}$ for GB.

$\begin{array}{c}{\left(\sum M_B\right)}_{external}=-mg\left(\frac{b}{6}\sqrt{24.04}\right)\theta \\ =-0.817mgb\theta \end{array}$

Take moment about B in the system for effective forces.

${\left(\sum M_B\right)}_{effective}=I_B\ddot{\theta }$

Substitute $\frac{23}{25}m{b}^2$ for $I_B$.

${\left(\sum M_B\right)}_{effective}=\frac{23}{25}m{b}^2\ddot{\theta }$

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

$\begin{array}{l}{\left(\sum M_B\right)}_{external}={\left(\sum M_B\right)}_{effective}\\ -0.817mgb\theta =\frac{23}{25}m{b}^2\ddot{\theta }\\ -\frac{23}{25}m{b}^2\ddot{\theta }-0.817mgb\theta =0\end{array}$

$\begin{array}{l}\frac{23}{25}m{b}^2\left(\ddot{\theta }+\frac{0.817mgb}{\frac{23}{25}m{b}^2}\theta \right)=0\\ \ddot{\theta }+\frac{0.817mgb}{\frac{23}{25}m{b}^2}\theta =0\end{array}$ (2)

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

$\begin{array}{c}{\omega }_n^2=\frac{0.817mgb}{\frac{23}{25}m{b}^2}\\ {\omega }_n^2=0.817g\times \frac{25}{23b}\\ {\omega }_n=\sqrt{0.888\frac{g}{b}}\end{array}$

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

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

Substitute $\sqrt{0.888\frac{g}{b}}$ for ${\omega }_n$.

$\begin{array}{c}{\tau }_n=\frac{2\pi }{\sqrt{0.888\frac{g}{b}}}\\ =6.67\sqrt{\frac{g}{b}}\end{array}$

Therefore, the period $\left({\tau }_n\right)$ of small oscillation if the wire is suspended from point B is $\underset{\_}{6.67\sqrt{\frac{b}{g}}}$.