Chapter 19.2, Problem 19.50P

To determine

(a)

Then the distance ‘d’ to maximise the frequency of oscillation when a small initial displacement is given.

Answer to Problem 19.50P

Distance $d=227mm$

Explanation of Solution

Given information:

Mass of collar $m_C=1{\rm K}g$

Mass of rod $m_R=3{\rm K}g$

Length of rod $L=750mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

$\begin{array}{l}\alpha =\ddot{\theta }\\ {\left({\overline{a}}_t\right)}_R=\frac{L}{2}\alpha =\frac{L}{2}\ddot{\theta }\\ {\left(a_t\right)}_C=d\alpha =d\ddot{\theta }\end{array}$

Now, from the equation of motion taking moment about A,

$\sum M_A={\sum \left(M_A\right)}_{eff}$

$\begin{array}{l}-W_R\frac{L}{2}\sin \theta -W_{C}d\sin \theta =I_R\alpha +m_R\frac{L}{2}{\left({\overline{a}}_t\right)}_R+m_{C}d{\left(a_t\right)}_C\\ I_R\ddot{\theta }+m_R\frac{L}{2}\times \frac{L}{2}\ddot{\theta }+m_{C}d\left(d\ddot{\theta }\right)+m_{R}g\frac{L}{2}\theta +m_{C}gd\theta =0\\ \left(I_R+m_R{\left(\frac{L}{2}\right)}^2+m_C{d}^2\right)\ddot{\theta }+\left(m_{R}g\frac{L}{2}+m_{C}gd\right)\theta =0\end{array}$ Since, $\sin \theta \approx \theta$ and $W=mg$

And, moment of inertia of rod is $I_R=\frac{m_R{L}^2}{12}$

$\begin{array}{l}\left(\frac{m_R{L}^2}{12}+\frac{m_R{L}^2}{4}+m_C{d}^2\right)\ddot{\theta }+\left(m_{R}g\frac{L}{2}+m_{C}gd\right)\theta =0\\ \left(\frac{m_R{L}^2+3m_R{L}^2}{12}+m_C{d}^2\right)\ddot{\theta }+\left(m_{R}g\frac{L}{2}+m_{C}gd\right)\theta =0\end{array}$

$\begin{array}{l}\left(\frac{4m_R{L}^2}{12}+m_C{d}^2\right)\ddot{\theta }+\left(m_{R}g\frac{L}{2}+m_{C}gd\right)\theta =0\\ \left(\frac{m_R{L}^2}{3}+m_C{d}^2\right)\ddot{\theta }+\left(m_{R}g\frac{L}{2}+m_{C}gd\right)\theta =0\\ \left(\frac{m_R{L}^2}{3}+m_C{d}^2\right)\ddot{\theta }+\left(m_R\frac{L}{2}+m_{C}d\right)g\theta =0\end{array}$

$\ddot{\theta }+\frac{\left(\frac{L}{2}+\frac{m_C}{m_R}d\right)g}{\left(\frac{L^2}{3}+\frac{m_C}{m_R}{d}^2\right)}\theta =0$

Compare the above equation with un-damped equation of vibration:

$M\ddot{\theta }+{\omega }_n^2\theta =0$

Natural frequency:

$\begin{array}{l}{\omega }_n^2=\frac{\left(\frac{L}{2}+\frac{m_C}{m_R}d\right)g}{\left(\frac{L^2}{3}+\frac{m_C}{m_R}{d}^2\right)}\\ {\omega }_n^2=\frac{\left(\frac{L}{2}+\frac{1}{3}d\right)g}{\left(\frac{L^2}{3}+\frac{1}{3}{d}^2\right)}\\ {\omega }_n^2=\frac{\left(\frac{3L+2d}{6}\right)g}{\left(\frac{L^2+d^2}{3}\right)}\\ {\omega }_n^2=\left(\frac{\frac{3L}{2}+d}{6}\right)g\times \left(\frac{3}{L^2+d^2}\right)\\ {\omega }_n^2=\frac{\left(\frac{3L}{2}+d\right)g}{\left(L^2+d^2\right)}\end{array}$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$\begin{array}{l}\frac{1}{g}\frac{d\left({\omega }_n^2\right)}{d\left(d\right)}=\frac{\left(L^2+d^2\right)\left(1\right)-\left(\frac{3L}{2}+d\right)\left(2d\right)}{{\left(L^2+d^2\right)}^2}=0\\ \left(L^2+d^2\right)\left(1\right)-\left(\frac{3L}{2}+d\right)\left(2d\right)=0\\ L^2+d^2-\frac{3L}{2}\times 2d-2d\times d=0\\ L^2+d^2-3Ld-2d^2=0\\ L^2-3Ld-d^2=0\\ {\left(0.75\right)}^2-3\left(0.75\right)d-d^2=0\\ 0.5625-2.25d-d^2=0or,\\ d^2+2.25d-0.5625=0\end{array}$

By solving the above equation we get,

$d=0.2270or-2.477$

Conclusion:

The distance $d=227mm.$

To determine

(b)

The period of oscillation.

Answer to Problem 19.50P

Period of vibration, ${\tau }_n=1.3512s$

Explanation of Solution

Given information:

Mass of collar $m_C=1{\rm K}g$

Mass of rod $m_R=3{\rm K}g$

Length of rod $L=750mm$

The forces corresponding to the rod and collar are shown in the free body diagram below:

$\begin{array}{l}\alpha =\ddot{\theta }\\ {\left({\overline{a}}_t\right)}_R=\frac{L}{2}\alpha =\frac{L}{2}\ddot{\theta }\\ {\left(a_t\right)}_C=d\alpha =d\ddot{\theta }\end{array}$

Now, from the equation of motion taking moment about A,

$\sum M_A={\sum \left(M_A\right)}_{eff}$

$\begin{array}{l}-W_R\frac{L}{2}\sin \theta -W_{C}d\sin \theta =I_R\alpha +m_R\frac{L}{2}{\left({\overline{a}}_t\right)}_R+m_{C}d{\left(a_t\right)}_C\\ I_R\ddot{\theta }+m_R\frac{L}{2}\times \frac{L}{2}\ddot{\theta }+m_{C}d\left(d\ddot{\theta }\right)+m_{R}g\frac{L}{2}\theta +m_{C}gd\theta =0\\ \left(I_R+m_R{\left(\frac{L}{2}\right)}^2+m_C{d}^2\right)\ddot{\theta }+\left(m_{R}g\frac{L}{2}+m_{C}gd\right)\theta =0\end{array}$ Since, $\sin \theta \approx \theta$ and $W=mg$

And, moment of inertia of rod is $I_R=\frac{m_R{L}^2}{12}$

$\begin{array}{l}\left(\frac{m_R{L}^2}{12}+\frac{m_R{L}^2}{4}+m_C{d}^2\right)\ddot{\theta }+\left(m_{R}g\frac{L}{2}+m_{C}gd\right)\theta =0\\ \left(\frac{m_R{L}^2+3m_R{L}^2}{12}+m_C{d}^2\right)\ddot{\theta }+\left(m_{R}g\frac{L}{2}+m_{C}gd\right)\theta =0\end{array}$

$\begin{array}{l}\left(\frac{4m_R{L}^2}{12}+m_C{d}^2\right)\ddot{\theta }+\left(m_{R}g\frac{L}{2}+m_{C}gd\right)\theta =0\\ \left(\frac{m_R{L}^2}{3}+m_C{d}^2\right)\ddot{\theta }+\left(m_{R}g\frac{L}{2}+m_{C}gd\right)\theta =0\\ \left(\frac{m_R{L}^2}{3}+m_C{d}^2\right)\ddot{\theta }+\left(m_R\frac{L}{2}+m_{C}d\right)g\theta =0\end{array}$

$\ddot{\theta }+\frac{\left(\frac{L}{2}+\frac{m_C}{m_R}d\right)g}{\left(\frac{L^2}{3}+\frac{m_C}{m_R}{d}^2\right)}\theta =0$

Compare the above equation with un-damped equation of vibration;

$M\ddot{\theta }+{\omega }_n^2\theta =0$

Natural frequency:

$\begin{array}{l}{\omega }_n^2=\frac{\left(\frac{L}{2}+\frac{m_C}{m_R}d\right)g}{\left(\frac{L^2}{3}+\frac{m_C}{m_R}{d}^2\right)}\\ {\omega }_n^2=\frac{\left(\frac{L}{2}+\frac{1}{3}d\right)g}{\left(\frac{L^2}{3}+\frac{1}{3}{d}^2\right)}\\ {\omega }_n^2=\frac{\left(\frac{3L+2d}{6}\right)g}{\left(\frac{L^2+d^2}{3}\right)}\\ {\omega }_n^2=\left(\frac{\frac{3L}{2}+d}{6}\right)g\times \left(\frac{3}{L^2+d^2}\right)\\ {\omega }_n^2=\frac{\left(\frac{3L}{2}+d\right)g}{\left(L^2+d^2\right)}\end{array}$

To maximise the frequency, we need to take the derivative with respect to d and set it equal to zero.

$\begin{array}{l}\frac{1}{g}\frac{d\left({\omega }_n^2\right)}{d\left(d\right)}=\frac{\left(L^2+d^2\right)\left(1\right)-\left(\frac{3L}{2}+d\right)\left(2d\right)}{{\left(L^2+d^2\right)}^2}=0\\ \left(L^2+d^2\right)\left(1\right)-\left(\frac{3L}{2}+d\right)\left(2d\right)=0\\ L^2+d^2-\frac{3L}{2}\times 2d-2d\times d=0\\ L^2+d^2-3Ld-2d^2=0\\ L^2-3Ld-d^2=0\\ {\left(0.75\right)}^2-3\left(0.75\right)d-d^2=0\\ 0.5625-2.25d-d^2=0or,\\ d^2+2.25d-0.5625=0\end{array}$

By solving the above equation we get,

$d=0.2270or-2.477$

Thus, $d=0.2270mord=227mm$

Now, natural frequency:

$\begin{array}{l}{\omega }_n^2=\frac{\left(\frac{3L}{2}+d\right)g}{\left(L^2+d^2\right)}\\ {\omega }_n^2=\frac{\left(\frac{3\left(0.75\right)}{2}+0.2270\right)9.81}{\left({\left(0.75\right)}^2+{\left(0.2270\right)}^2\right)}\\ {\omega }_n^2=\frac{\left(1.125+0.2270\right)9.81}{\left(0.5625+0.051529\right)}\\ {\omega }_n^2=21.6001524\\ {\omega }^2=4.6476rad/s\end{array}$

Then, Time period ${\tau }_n=\frac{2\pi }{{\omega }_n}$

$\begin{array}{l}{\tau }_n=\frac{2\times 3.14}{4.6476}\\ {\tau }_n=1.3512s\end{array}$