Chapter 4, Problem 4.69P

To determine

The equation of motion of the pendulum.

Answer to Problem 4.69P

$\ddot{\theta }+\frac{c{L}_2^2}{I}\dot{\theta }-\frac{c{L}_2}{I}\dot{y}+\frac{(k_2{L}_2^2-k_1{L}_1^2)}{I}\theta -\frac{k_2{L}_2}{I}y=0$.

Explanation of Solution

Given:

A spring with stiffness k, and a damper with damping coefficient c, are attached to a pendulum with an input y(t)

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 rod inertia about the pivot is given as I.

$M_o$ = 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 spring element + Moments of damper with spring element.

Free body diagram of the system:

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, $x=L_1\sin \theta$

Hence the moments due to spring element = $L_1\times k_1\times L_1\sin \theta =k_1{L}_1^2\sin \theta$ anticlockwise.

Moments of the Damper together with Spring element:

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

$f=cv$

Where f = damping force

v = relative velocity

c = damping coefficient

The damper and spring element are connected parallelly, the forces of these elements can be added together to get the total force.

The distance of compression of the spring element is $y-L_2\sin \theta$ and the distance for the damping element is

Force, $f=k_2(y-L_2\sin \theta )+c\times \frac{d}{dt}(y-L_2\sin \theta )$

The distance between the pivot, O and the force applied = $L_2$.

Hence Moments = $L_2\times (k_2(y-L_2\sin \theta )+c\times \frac{d}{dt}(y-L_2\sin \theta ))$ anticlockwise.

Taking anticlockwise to be positive as the input y(t) is taken positive in the anticlockwise direction.

Substitute the above expressions to the following equation,

Total moments about O = Moments of spring element + Moments of damper with spring element.

Total moments = $k_1{L}_1^2\sin \theta +L_2\times (k_2(y-L_2\sin \theta )+c\times \frac{d}{dt}(y-L_2\sin \theta ))$

Derivation of Equation of Motion:

Substitute $I_o=I$ and $M_o=k_1{L}_1^2\sin \theta +L_2\times (k_2(y-L_2\sin \theta )+c\times \frac{d}{dt}(y-L_2\sin \theta ))$ into the equation of motion $I_o\ddot{\theta }=M_o$.

$I\ddot{\theta }=k_1{L}_1^2\sin \theta +L_2\times (k_2{L}_2\sin \theta +c\times \frac{d}{dt}(L_2\sin \theta -y))$

Assuming $\theta$ is small, we can say that $\sin \theta =\theta$.

Substituting $\theta$ for $\sin \theta$ in the equation of motion,

$I\ddot{\theta }=k_1{L}_1^2\theta +L_2\times (k_2(y-L_2\theta )+c\times \frac{d}{dt}(y-L_2\theta ))$

Simplifying the equation further:

$\begin{array}{l}I\ddot{\theta }=k_1{L}_1^2\theta +L_2\times (k_2(y-L_2\theta )+c\times \frac{d}{dt}(y-L_2\theta ))\\ I\ddot{\theta }=k_1{L}_1^2\theta +L_2\times (k_{2}y-k_2{L}_2\theta +c\times (\dot{y}-L_2\dot{\theta }))\\ I\ddot{\theta }=k_1{L}_1^2\theta +L_2\times (k_{2}y-k_2{L}_2\theta +c\dot{y}-c{L}_2\dot{\theta })\\ I\ddot{\theta }=k_1{L}_1^2\theta +k_2{L}_{2}y-k_2{L}_2^2\theta +c{L}_2\dot{y}-c{L}_2^2\dot{\theta }\\ I\ddot{\theta }-k_1{L}_1^2\theta -k_2{L}_{2}y+k_2{L}_2^2\theta -c{L}_2\dot{y}+c{L}_2^2\dot{\theta }=0\\ I\ddot{\theta }+c{L}_2^2\dot{\theta }-c{L}_2\dot{y}+\theta (k_2{L}_2^2-k_1{L}_1^2)-k_2{L}_{2}y=0\\ \ddot{\theta }+\frac{c{L}_2^2}{I}\dot{\theta }-\frac{c{L}_2}{I}\dot{y}+\frac{(k_2{L}_2^2-k_1{L}_1^2)}{I}\theta -\frac{k_2{L}_2}{I}y=0\end{array}$.

Conclusion:

The equation of motion of the pendulum is $\ddot{\theta }+\frac{c{L}_2^2}{I}\dot{\theta }-\frac{c{L}_2}{I}\dot{y}+\frac{(k_2{L}_2^2-k_1{L}_1^2)}{I}\theta -\frac{k_2{L}_2}{I}y=0$.