Chapter 19.2, Problem 19.41P

To determine

(a)

The period of vibration.

Answer to Problem 19.41P

Period of vibration, $\tau =0.491s$

Explanation of Solution

Given information:

Weight of rod $W=15lb$

Weight of disc $W_{disc}=12lb$

Spring constant $k=30lb/in.=360lb/ft$

Length $L=36in.=3ft$

Radius $r=10in.=0.83333ft$

The free body diagram of the given bar is as follows:

Now taking moment about point B,

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

$mg\frac{L}{2}\cos \theta -kxr=I_{AB}\alpha +m\left(\alpha \times \frac{L}{2}\right)\frac{L}{2}+I_{disc}\alpha$

Here, $x=r\theta +{\delta }_{st}$

And also from the statics of the diagram; $mg\frac{L}{2}=k{\delta }_{st}r$

Assuming small angles $\cos \theta \approx 1$ then the above equation becomes,

$\begin{array}{l}mg\frac{L}{2}-kr\left(r\theta +{\delta }_{st}\right)=\left(I_{AB}+m{\left(\frac{L}{2}\right)}^2+I_{disc}\right)\alpha \\ kr{\delta }_{st}-k{r}^2\theta -kr{\delta }_{st}=\left(I_{AB}+m{\left(\frac{L}{2}\right)}^2+I_{disc}\right)\alpha \\ \left(I_{AB}+\left(\frac{m{L}^2}{4}\right)+I_{disc}\right)\ddot{\theta }+k{r}^2\theta =0\end{array}$

First we calculate the moment of inertia for AB,

$\begin{array}{l}{I}_{AB}=\frac{m{L}^2}{12}\\ I_{AB}=\frac{{\left(3\right)}^2}{12}\left(\frac{15}{32.2}\right)\\ I_{AB}=0.349lb.s^2.ft\end{array}$

Now, for disc the moment of inertia is,

$\begin{array}{l}{I}_{disc}=\frac{m{r}^2}{2}\\ I_{disc}=\frac{{\left(0.83333\right)}^2}{2}\left(\frac{12}{32.2}\right)\\ I_{disc}=0.1294lb.s^2.ft\end{array}$

By putting all the values in the above equation we get,

$\begin{array}{l}\left(I_{AB}+\left(\frac{m{L}^2}{4}\right)+I_{disc}\right)\ddot{\theta }+k{r}^2\theta =0\\ \left[0.349+\frac{1}{4}\left(0.46584\right){\left(3\right)}^2+0.129\right]\ddot{\theta }+360{\left(0.83333\right)}^2\theta =0\\ 1.5269\ddot{\theta }+250\theta =0or,\\ \ddot{\theta }+163.73\theta =0\end{array}$

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

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

Then, Natural frequency: $\begin{array}{l}{\omega }_n^2=163.73{\left(rad/s\right)}^{2}or,\\ {\omega }_n=12.796rad/s\end{array}$

And, Period $\tau =\frac{2\pi }{{\omega }_n}$

$\begin{array}{l}\tau =\frac{2\times 3.14}{12.796}\\ \tau =0.491s\end{array}$

To determine

(b)

Maximum velocity at point A.

Answer to Problem 19.41P

Maximum velocity, $v_m=9.60in./s$

Explanation of Solution

Given information:

Weight of rod $W=15lb$

Weight of disc $W_{disc}=12lb$

Spring constant $k=30lb/in.=360lb/ft$

Length $L=36in.=3ft$

Radius $r=10in.=0.83333ft$

The free body diagram of the given bar is as follows:

Now taking moment about point B,

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

$mg\frac{L}{2}\cos \theta -kxr=I_{AB}\alpha +m\left(\alpha \times \frac{L}{2}\right)\frac{L}{2}+I_{disc}\alpha$

Here, $x=r\theta +{\delta }_{st}$

And also, from the statics of the diagram; $mg\frac{L}{2}=k{\delta }_{st}r$

Assuming small angles $\cos \theta \approx 1$ then the above equation becomes,

$\begin{array}{l}mg\frac{L}{2}-kr\left(r\theta +{\delta }_{st}\right)=\left(I_{AB}+m{\left(\frac{L}{2}\right)}^2+I_{disc}\right)\alpha \\ kr{\delta }_{st}-k{r}^2\theta -kr{\delta }_{st}=\left(I_{AB}+m{\left(\frac{L}{2}\right)}^2+I_{disc}\right)\alpha \\ \left(I_{AB}+\left(\frac{m{L}^2}{4}\right)+I_{disc}\right)\ddot{\theta }+k{r}^2\theta =0\end{array}$

First we calculate the moment of inertia for AB,

$\begin{array}{l}{I}_{AB}=\frac{m{L}^2}{12}\\ I_{AB}=\frac{{\left(3\right)}^2}{12}\left(\frac{15}{32.2}\right)\\ I_{AB}=0.349lb.s^2.ft\end{array}$

Now, for disc the moment of inertia is,

$\begin{array}{l}{I}_{disc}=\frac{m{r}^2}{2}\\ I_{disc}=\frac{{\left(0.83333\right)}^2}{2}\left(\frac{12}{32.2}\right)\\ I_{disc}=0.1294lb.s^2.ft\end{array}$

By putting all the values in the above equation we get,

$\begin{array}{l}\left(I_{AB}+\left(\frac{m{L}^2}{4}\right)+I_{disc}\right)\ddot{\theta }+k{r}^2\theta =0\\ \left[0.349+\frac{1}{4}\left(0.46584\right){\left(3\right)}^2+0.129\right]\ddot{\theta }+360{\left(0.83333\right)}^2\theta =0\\ 1.5269\ddot{\theta }+250\theta =0or,\\ \ddot{\theta }+163.73\theta =0\end{array}$

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

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

Then, Natural frequency: $\begin{array}{l}{\omega }_n^2=163.73{\left(rad/s\right)}^{2}or,\\ {\omega }_n=12.796rad/s\end{array}$

Maximum velocity: $v_m={\omega }_n{x}_m$

$\begin{array}{l}{v}_m=\left(12.796\right)\left(0.75\right)\\ v_m=9.60in./s\end{array}$