Chapter 19.2, Problem 19.39P

A 6-kg uniform cylinder can roll without sliding on a horizontal surface and is attached by a pin at point C to the 4-kg horizontal bar AB. The bar is attached to two springs, each having a constant of $k=5$ kN/m, as shown. Knowing that the bar is moved 12 mm to the light of the equilibrium position and released, determine (a) the period of vibration of the system, (b) the magnitude of the maximum velocity of bar AB.
  

To determine

(a)

The period of vibration of the given system.

Answer to Problem 19.39P

Period of vibration, ${\tau }_n=0.226s$

Explanation of Solution

Given information:

Mass of cylinder $M=6{\rm K}g$

Mass of bar $m=4{\rm K}g$

Amplitude $x_m=12mm$

Spring constant $k=5{\rm K}{\rm N}/m$

Firstly we draw the free body diagram of the bar and the cylinder separately and calculate forces acting upon them. For the bar;

$F=-2kx-P$

But, $F=maor,$

$F=m\ddot{x}$

Then, $\begin{array}{l}m\ddot{x}=-2kx-P\\ \Rightarrow P=-\left(m\ddot{x}+2kx\right)\_\_\_\_\_\_\_(1)\end{array}$

Now, for the disc,

$M\ddot{x}=P-F\_\_\_\_\_\_\_(2)$

Now, moment of inertia at point C, $I\ddot{\theta }=Fr$

$\begin{array}{l}\frac{M{r}^2}{2}\ddot{\theta }=Fror,\\ \frac{Mr}{2}\ddot{\theta }=F\_\_\_\_\_\_\_(3)\end{array}$

Put the values of equation (1) and (3) in equation (2);

$\begin{array}{l}M\ddot{x}=-\left(m\ddot{x}+2kx\right)-\frac{Mr\ddot{\theta }}{2}\\ M\ddot{x}+\frac{Mr\times \frac{\ddot{x}}{r}}{2}=-\left(m\ddot{x}+2kx\right)\\ M\ddot{x}+\frac{M\ddot{x}}{2}=-\left(m\ddot{x}+2kx\right)\\ \frac{3M\ddot{x}}{2}+m\ddot{x}=-2kx\\ \left(m+\frac{3M}{2}\right)\ddot{x}+2kx=0\\ \ddot{x}+\frac{2k}{m+\frac{3M}{2}}x=0\end{array}$

Here, $\ddot{\theta }=\frac{\ddot{x}}{r}and\dot{\theta }=\frac{\dot{x}}{r}$

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

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

$\begin{array}{l}{\omega }_n^2=\frac{2k}{m+\frac{3M}{2}}\\ {\omega }_n^2=\frac{2\times 5\times 1000}{4+9}\end{array}$

$\begin{array}{l}{\omega }_n^2=769.230769\\ {\omega }_n=\sqrt{769.230769}\\ {\omega }_n=27.735rad/s\end{array}$

Thus, the time period:

$\begin{array}{l}{\tau }_n=\frac{2\pi }{{\omega }_n}\\ {\tau }_n=\frac{2\times 3.14}{27.735}\\ {\tau }_n=0.226s\end{array}$

To determine

(b)

The maximum velocity of bar AB.

Answer to Problem 19.39P

Velocity, $v=332.82mm/s$

Explanation of Solution

Given information:

Mass of cylinder $M=6{\rm K}g$

Mass of bar $m=4{\rm K}g$

Amplitude $x_m=12mm$

Spring constant $k=5{\rm K}{\rm N}/m$

Firstly, we draw the free body diagram of the bar and the cylinder separately and calculate forces acting upon them. For the bar;

$F=-2kx-P$

But, $F=maor,$

$F=m\ddot{x}$

Then, $\begin{array}{l}m\ddot{x}=-2kx-P\\ \Rightarrow P=-\left(m\ddot{x}+2kx\right)\_\_\_\_\_\_\_(1)\end{array}$

Now, for the disc,

$M\ddot{x}=P-F\_\_\_\_\_\_\_(2)$

Now, moment of inertia at point C, $I\ddot{\theta }=Fr$

$\begin{array}{l}\frac{M{r}^2}{2}\ddot{\theta }=Fror,\\ \frac{Mr}{2}\ddot{\theta }=F\_\_\_\_\_\_\_(3)\end{array}$

Put the values of equation (1) and (3) in equation (2);

$\begin{array}{l}M\ddot{x}=-\left(m\ddot{x}+2kx\right)-\frac{Mr\ddot{\theta }}{2}\\ M\ddot{x}+\frac{Mr\times \frac{\ddot{x}}{r}}{2}=-\left(m\ddot{x}+2kx\right)\\ M\ddot{x}+\frac{M\ddot{x}}{2}=-\left(m\ddot{x}+2kx\right)\\ \frac{3M\ddot{x}}{2}+m\ddot{x}=-2kx\\ \left(m+\frac{3M}{2}\right)\ddot{x}+2kx=0\\ \ddot{x}+\frac{2k}{m+\frac{3M}{2}}x=0\end{array}$

Here, $\ddot{\theta }=\frac{\ddot{x}}{r}and\dot{\theta }=\frac{\dot{x}}{r}$

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

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

$\begin{array}{l}{\omega }_n^2=\frac{2k}{m+\frac{3M}{2}}\\ {\omega }_n^2=\frac{2\times 5\times 1000}{4+9}\end{array}$

$\begin{array}{l}{\omega }_n^2=769.230769\\ {\omega }_n=\sqrt{769.230769}\\ {\omega }_n=27.735rad/s\end{array}$

Now, maximum velocity: $v={\omega }_n{x}_m$

$\begin{array}{l}v=12\times 27.735\\ v=332.82mm/s\end{array}$