Chapter 4, Problem 4.68P

To determine

Whether the equation of motion of the pendulum and the system is stable, neutrally stable or unstable.

Answer to Problem 4.68P

$\ddot{\theta }+\frac{\sqrt{2}}{2L}(\frac{k{L}_1^2}{mL}-g)=0$ and the system is neutrally stable.

Explanation of Solution

Given:

A spring with stiffness k, and a damper with damping coefficient c, are attached to a pendulum of mass, m.

Concept used:

For an objects’ planar motion which rotates only about an axis perpendicular to the plane, the equation of motion can be written down using Newton’s Second Law.

Equation of Motion: $I_o\dot{\omega }=M_o$

Where $I_o$ = Mass moment of Inertia of the body about point O.

$\omega$ = Angular velocity of the mass about an axis through a point O.

$M_o$ = Sum of the moments applied to the body about the point O.

Let the angular displacement be $\theta$.

The angular velocity, $\omega =\frac{d\theta }{dt}=\dot{\theta }$

$\Rightarrow \dot{\omega }=\frac{d\omega }{dt}=\frac{d^2\theta }{d{t}^2}=\ddot{\theta }$

Hence, the equation of motion of this object can be rewritten by substituting, $\dot{\omega }=\ddot{\theta }$:

$I_o\ddot{\theta }=M_o$

To find the equation of motion, the required unknowns are $I_o$ and $M_o$.

The mass moment of Inertia, I about a specified reference axis is given as:

$I={\displaystyle \int r^{2}dm}$

Where r = distance from the reference axis to mass element

Mass moment of Inertia of a rotating pendulum = ${\displaystyle \int r^{2}dm=m{r}^2}$

In this question, the distance from the reference axis to the mass element, r = L. Substituting this to the above equation gives:

$I_o=m{L}^2$

$M_o$ = Sum of the moments applied to the body about the point O.

Moments = Perpendicular Force $\times$ Distance from the Pivot.

In this question, the pivot is point O.

Total moments about O = Moments of mass, m + Moments of spring element + Moments of the damper

Free body diagram of the system:

Moments of mass, m:

The force mg can be resolved to two components, $mg\sin (45-\theta )$ and $mg\cos (45-\theta )$.

The force causing the moments will be $mg\sin (45-\theta )$ and its distance from the point O is L.

Moments = $mgL\sin (45-\theta )$ clockwise

Moments of the spring element:

Using the Hooke’s Law, a linear force-deflection model can be written, $f=kx$.

Where f = restoring force

x = compression or extension distance

k = Spring constant or stiffness

Here the extension distance, $x=L_1\sin (45-\theta )$

Hence the moments due to spring element = $L_1\times k\times L_1\sin (45-\theta )=k{L}_1^2\sin (45-\theta )$ anticlockwise.

Moments of the Damper:

The linear model for the force applied by the damper is:

$f=cv$

Where f = damping force

v = relative velocity

c = damping coefficient

Here the force, $f=c\times \frac{d}{dt}(L_1\sin (45-\theta ))$

The distance between the pivot, O and the force applied = $L_1$.

Hence Moments of the damper = $L_1\times c\times \frac{d}{dt}(L_1\sin (45-\theta ))$ anticlockwise.

Taking clockwise to be positive and substitute the above expressions to the following equation,

Total moments about O = Moments of mass, m + Moments of spring element + Moments of the damper

Total moments = $mgL\sin (45-\theta )-k{L}_1^2\sin (45-\theta )-L_1\times c\times \frac{d}{dt}(L_1\sin (45-\theta ))$

Derivation of Equation of Motion:

Substitute $I_o=m{L}^2$ and $M_o=mgL\sin (45-\theta )-k{L}_1^2\sin (45-\theta )-L_1\times c\times \frac{d}{dt}(L_1\sin (45-\theta ))$ into the equation of motion $I_o\ddot{\theta }=M_o$.

$m{L}^2\ddot{\theta }=mgL\sin (45-\theta )-k{L}_1^2\sin (45-\theta )-L_1^{2}c\frac{d}{dt}\sin (45-\theta )$

Assuming $\theta$ is small, we can say $\sin (45-\theta )\approx \sin 45$ and the value $\sin 45=\frac{\sqrt{2}}{2}$ can be substituted for $\sin (45-\theta )$.

$m{L}^2\ddot{\theta }=mgL\times \frac{\sqrt{2}}{2}-k{L}_1^2\frac{\sqrt{2}}{2}-L_1^{2}c\frac{d}{dt}\frac{\sqrt{2}}{2}$

The differentiation of a constant is 0, hence the moment due to the force applied by the damper becomes 0.

Simplifying the equation further:

$\begin{array}{l}m{L}^2\ddot{\theta }=mgL\times \frac{\sqrt{2}}{2}-k{L}_1^2\frac{\sqrt{2}}{2}\\ m{L}^2\ddot{\theta }-\frac{mgL\sqrt{2}}{2}+\frac{k{L}_1^2\sqrt{2}}{2}=0\\ \ddot{\theta }-\frac{mgL\sqrt{2}}{2L^2}+\frac{k{L}_1^2\sqrt{2}}{2m{L}^2}=0\\ \ddot{\theta }-\frac{\bcancel{m}g\bcancel{L}\sqrt{2}}{2\bcancel{m}{L}^{\bcancel{2}}}+\frac{k{L}_1^2\sqrt{2}}{2m{L}^2}=0\\ \ddot{\theta }-\frac{g\sqrt{2}}{2L}+\frac{k{L}_1^2\sqrt{2}}{2m{L}^2}=0\\ \ddot{\theta }+\frac{\sqrt{2}}{2L}(-g+\frac{k{L}_1^2}{mL})=0\\ \ddot{\theta }+\frac{\sqrt{2}}{2L}(\frac{k{L}_1^2}{mL}-g)=0\end{array}$

The equation of motion of the system is $\ddot{\theta }+\frac{\sqrt{2}}{2L}(\frac{k{L}_1^2}{mL}-g)=0$. In this if $\frac{k{L}_1^2}{mL}=g$ or $\frac{k{L}_1^2}{mL}>g$ we can say that the system is neutrally stable.

When $\frac{k{L}_1^2}{mL}=g$, the roots obtained from the equation of motion will both be zero which indicates a neutral stability.

When $\frac{k{L}_1^2}{mL}>g$, the roots obtained from the equation of motion are imaginary and hence the system is neutrally stable.