EGB211 Computer Lab Assignment: Analytical and Numerical Solutions to Non-Linear Equation of Motion

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Assessment No: 3
Assessment Type: Individual Computer Lab Assignment
Due Date: Monday, 21st May 2018
Student ID:
Name:

EGB211 – Computer Lab Assignment
Dynamics is the subdivision of applied arithmetic (precisely classical mechanics) that deals
with the learning of forces and torques and their impact on motion. The learning of dynamics
falls further down to two groups: linear and rotational. Linear dynamics apply to bodies
moving in a line and encompasses such capacities as force, mass/inertia, displacement (in
units of distance), velocity , acceleration and momentum . Rotational dynamics apply to
bodies that are in rotary or moving in a curved track and comprises such capacities
as torque, moment of inertia/gyratory inertia, angular movement (in radians or less
repeatedly, degrees), angular velocity (radians per unit time), angular deceleration (radians
per unit of time squared) and angular motion (moment of inertia times unit of angular speed).
Over and over again, bodies display both linear and rotational motion (Goodman & Warner,
2013).
A dynamic problem can be divided in two phases: attaining the equation of motion and
answering it. In order to attain the equation of motion we use corporate dynamics concept,
such as Newton’s second law. The equation of motion can then be solved via analytical or
numerical methods. So far, you have frequently considered the analytical explanations to
problems, on the other hand even for simple dynamics problems these can promptly turn into
complex and unwieldy (think drag acting on a projectile). In the real world, the occurrence of
a true analytical solution is rare in many cases, and we rather rely on simplifications and
assumptions (e.g. neglecting drag). Numerical solutions on the other hand can be readily
used to solve simple and complex dynamics problems, without simplification (Forrester,
2013). In numerical solutions the main challenges become computational resources and the
elegance of the model.
In this computer lab assignment, we will explore the analytical and numerical solution to the
non-linear equation of motion of the inverted pendulum system illustrated in Figure. 1. For
small oscillations, we can apply the small angle theorem, linearizing the equation of motion,
allowing us to obtain an analytical solution. We will explore the impact of non-linearity and
the limits of linear assumption by comparison against the numerical solution for the non-
linear equation of motion. Finally, we explore the use of numerical methods as a tool to
investigate the real-world behaviour and characteristics of non-linear dynamics.
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Figure 1. Diagram of the system (not to-scale)
Property Symbol Quantity Unit
Pendulum
Mass m 0.5 kg
Initial angle
with the vertical
axis (CCW)
θ 0 10 degrees
Initial angular
velocity ̇θ0 0.01 rad / s
Length l 0.4 m
Cross-sectional
area of mass A 0.05 m2
Spiral spring
constant k 19.62 Nm
rad
Drag
Air density ρ 1.183 kg /m3
Drag coefficient CD 0.47
Table 1. Quantities
For a dynamic analysis of the inverted pendulum system illustrated in Figure. 1 we will
consider the following.
 We will consider this as a one-dimensional problem in θ radians.
 The inverted pendulum rotates about O (x=0, y=0) and there is no restriction to the
rotation (i.e., number of full 360 ° turns).
 The spiral spring produces a moment kθ [N ⋅m] at O.
 The bar connecting the spherical mass and spiral spring is assumed to be rigid and
have no mass.
 Unless situated in a complete vacuum, due to surrounding air drag force with a of
FD =−1
2 CD ρA v2 ⋅sign(v ) is exerted on the spherical mass at the end of bar.
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Equation of Motion
1. Define the equation of motion for this inverted pendulum spring system in Figure 1.
Show all steps including a free-body-diagram (include drag).
Answer within this box. Change the size of the box if needed.
For a free body
m- mass of the body
torque = moment of inertia – angular acceleration
τnet = IӪ
Torque due to gravity
τnet = mgl sin θ
θ – angle in radians from the equilibrium position
Hence
IӪ = mgl sin θ
Moment of inertia
I = ml2
Therefore,
ml2Ӫ = mgl sin θ
after solving we obtain,
Ӫ = g
l sin θ
The equation above assumes no friction or any other resistance to movement.
The inverted pendulum in space is similar to the uninverted pendulum.
FD =−1
2 CD ρA v2 ⋅sign(v )
Equation for the inverted pendulum spring system
IӪ = mgl sin θ – Kθ - FD
K- elastic coefficient of the spiral spring
I – mass moment of inertia
m – weight of the translating mass
g – gravity
l – length
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