System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 4, Problem 4.68P
To determine

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

Expert Solution & Answer
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Answer to Problem 4.68P

θ¨+22L(kL12mLg)=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: Ioω˙=Mo

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

ω = Angular velocity of the mass about an axis through a point O.

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

Let the angular displacement be θ.

The angular velocity, ω=dθdt=θ˙

ω˙=dωdt=d2θdt2=θ¨

Hence, the equation of motion of this object can be rewritten by substituting, ω˙=θ¨:

Ioθ¨=Mo

To find the equation of motion, the required unknowns are Io and Mo.

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

I=r2dm

Where r = distance from the reference axis to mass element

Mass moment of Inertia of a rotating pendulum = r2dm=mr2

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

Io=mL2

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

Moments = Perpendicular Force × 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:

System Dynamics, Chapter 4, Problem 4.68P

Moments of mass, m:

The force mg can be resolved to two components, mgsin(45θ) and mgcos(45θ).

The force causing the moments will be mgsin(45θ) and its distance from the point O is L.

Moments = mgLsin(45θ) 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=L1sin(45θ)

Hence the moments due to spring element = L1×k×L1sin(45θ)=kL12sin(45θ) 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×ddt(L1sin(45θ))

The distance between the pivot, O and the force applied = L1.

Hence Moments of the damper = L1×c×ddt(L1sin(45θ)) 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 = mgLsin(45θ)kL12sin(45θ)L1×c×ddt(L1sin(45θ))

Derivation of Equation of Motion:

Substitute Io=mL2 and Mo=mgLsin(45θ)kL12sin(45θ)L1×c×ddt(L1sin(45θ)) into the equation of motion Ioθ¨=Mo.

mL2θ¨=mgLsin(45θ)kL12sin(45θ)L12cddtsin(45θ)

Assuming θ is small, we can say sin(45θ)sin45 and the value sin45=22 can be substituted for sin(45θ).

mL2θ¨=mgL×22kL1222L12cddt22

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

Simplifying the equation further:

mL2θ¨=mgL×22kL1222mL2θ¨mgL22+kL1222=0θ¨mgL22L2+kL1222mL2=0θ¨mgL22mL2+kL1222mL2=0θ¨g22L+kL1222mL2=0θ¨+22L(g+kL12mL)=0θ¨+22L(kL12mLg)=0

The equation of motion of the system is θ¨+22L(kL12mLg)=0. In this if kL12mL=g or kL12mL>g we can say that the system is neutrally stable.

When kL12mL=g, the roots obtained from the equation of motion will both be zero which indicates a neutral stability.

When kL12mL>g, the roots obtained from the equation of motion are imaginary and hence the system is neutrally stable.

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Chapter 4 Solutions

System Dynamics

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