ME-212: Dynamics Complex Engineering Problem, Study Guides, Projects, Research of Dynamics

Download ME-212: Dynamics Complex Engineering Problem and more Study Guides, Projects, Research Dynamics in PDF only on Docsity!

rd

June 2022

ME-212: Dynamics

Complex Engineering Problem

Instructor: Mr. Qasim Zafar

Submitted By:

Muhammad Ahmad Adil Khan

2020267

FME

Table of Contents

• Abstract
• Introduction...................................................................................................................................................................
• Problem Statement
  • Additional Instructions..........................................................................................................................................
• Literature Review
  • Inertial Frame of Reference
  • Newton's Laws of Motion
  • Equations of Motion
• Analytical Modelling
  • Assumptions
  • Derivation of the Kinetic Model
  • Derivation of the Kinematic Model
• MATLAB Code
• Plots and Results
  • Discussion of Results
• Conclusion
• References....................................................................................................................................................................

Problem Statement

A four-wheel vehicle is moving from the initial state to the goal point with some velocity

(consider motion is a curvilinear following polynomial path of order 7 ). Calculate the linear and

angular velocity of the vehicle at the goal point. Also plot velocity vs displacement.

Additional Instructions

The students having the last digit of reg no 0,1,2,3 should take polynomial order as the sum of

the last two digits. The students having the last digit of reg no 8 and 9 should take polynomial

order as the sum of the last two digits divided by 2. The polynomial order must be a whole

number. In our case, the order is 7.

Literature Review

Important theoretical information was studied, including factors affecting kinetics and kinematics

of a body, related terminologies, description of the dynamic model, and relevant formulae. These

include Newtonian laws of motion, an inertial frame of reference, and equations of motion.

Inertial Frame of Reference

An inertial reference frame is important when analyzing the

motion of objects. Whenever we apply the equations of

motion, such as the force equations or moment equations, the

acceleration must be measured about a Newtonian or inertial

reference frame. This is a coordinate system that does not

rotate and is either fixed in 3 - D space or moves in a straight

line at constant velocity / with zero acceleration.

Newton's Laws of Motion

The 3 Newtonian laws of motion form the basis of all sciences related to the analysis of forces

and their effect on bodies in the form of translation or rotation. These laws are briefly recalled

below.

1. First Law

Newton's first law states that, if no net external force acts on a body at rest, that body will

continue to remain at rest. If no net external force acts on a body moving in a straight line at

a constant velocity, that body will continue to move in a straight line at constant velocity.

2. Second Law

Newton's second law states that, when a force acts on a particle, this force is equal to mass

times acceleration. For a single particle, this takes the form of:

Where F is the applied force, m is the mass (or inertia) of the particle, and a is the acceleration

with respect to the inertial frame of reference.

Figure 2 : Inertial F.O.R

For rigid bodies, the equation is slightly modified and becomes:

𝐺

Where ΣF is the vector sum of the external forces acting on the system of particles, m is the total

mass of the system, and a G

is the acceleration of the center of mass of the system of particles.

This acceleration is in the direction of the net external force.

3. Third Law

Newton's third law states that for every action there is an equal and opposite reaction. This

means that, when two bodies are in contact, the reactions between those bodies are equal in

magnitude and opposite in direction.

Equations of Motion

Equations of motion mathematically describe the motion of a system under the influence of

forces. In particular, these equations describe the motion of particles or bodies subjected to

different forces. Since we are concerned with curvilinear motion, the relevant equations for

rectangular coordinates are:

But if we use normal-tangential coordinates , the required equations are:

For curvilinear motion, we generally prefer the n-t

system. So will be in our case.

Figure 3 : The rectangular coordinates system

Figure 4 : The n-t coordinates system

Now assuming the car travels at an angle (𝜃) to the horizontal, we can derive

expressions for

𝑛

and

𝑡

𝑡

𝐺

𝑡

𝑘

𝐴

𝑘

𝐵

𝐷

𝑛

𝐺

𝑛

𝐴

𝐵

𝐺

𝐵

𝐵

𝐴

𝐴

where 𝑓 𝐴

𝑘

𝐴

𝑏

𝑘

𝐵

Re-substituting these expressions into the original

equation, we can obtain an expression for the co-efficient

of kinectic friction (μ

k

𝑘

𝐵

𝐴

𝐴

𝐵

(NA and NB represent total normal forces per 2 back and front wheels respectively)

Derivation of the Kinematic Model

The vehicle is considered as a rigid body undergoing translational motion, following a curvilinear

path (curvilinear translation). Since the motion is translational only, the velocities and

accelerations at all the points on the body have the same value. Therefore, for the kinematic

model, the vehicle may be considered as a point mass concentrated about its center of gravity.

The vehicle is traveling along the curved path as shown below:

mg

Figure 6 : Car at an angle

Figure 7 : Direction of n-t axes and acceleration components

If normal and tangential axes are used for the analysis, then v is always in the positive t direction.

The acceleration has two components. The tangential component, 𝒂 𝒕

, accounts for the change

in the magnitude of the velocity; a slowing down is in the negative t direction, and a speeding

up is in the positive t direction. The normal component 𝒂

𝒏

accounts for the change in the

direction of the velocity. This component is always in the positive n direction. The equations

primarily used in this analysis:

Where 𝜌 = radius of curvature of the path at a point and is given by:

[ 1 +

2

]

3

2

2

2

𝑆𝑖𝑛𝑐𝑒

=

6

𝑎𝑛𝑑

2

2

=

5

,

𝜌 =

[ 1 +(

7 𝑥

6

267

2

]

3

2

14 𝑥

5

89

Angle of inclination, 𝜃, at any point on the curve can be calculated by the relation:

tan(𝜃) =

6

Since this is a polynomial path, the displacement (shortest distance) will be the curve length.

This curve length is given by:

2

𝑥

0

6

2

𝑥

0

Since acceleration is constant, we can use one of the equations of motion:

2

2

𝑡

MATLAB Code

This section holds the code used in MATLAB with the comments that explain the working step-

by-step:

clear;

clc;

syms x; %symbolic toolbox to define variable for calculus operations

v_o = 10.68; %initial velocity multiple of last 3 digits of reg. no.

a_tan = 2.67; %tangential acceleration assumed to be constant at 2.67 m/s^

a(x) = (1/267)*x.^7; %fraction co-efficient used to expand graph

b(x) = diff(a(x)); %obtaining dy/dx for use in s and rho formulae

c(x) = diff(b(x)); %finding second derivative to use in rho formula

ds = sqrt(1+(b(x))^2); %displacement formula (curve length)

s = ones(1,534);

v = ones(1,534); %defining arrays to store values for later use in plots

omega = ones(1,534);

for k = 1:1:534 %storing values of s,v,and w using for-loop and arrays

v(k) = sqrt(int(2a_tands,x,0,(k/100))+v_o^2); %equation of motion

s(k) = int(ds,x,0,k/100);

p = vpa(subs(c(x),x,(k/100))); %extracting numeric values using subs & vpa function

q = vpa(subs(b(x),x,(k/100)));

rho=((1+p^2)^(3/2))/q;

omega(k) = v(k)/rho;

end

x = 0.01:0.01:5.34; %setting up plots and x and y axes

subplot(2,2,1); %subplot function to simultaneously make several plots

plot(x,s);

title("Displacement (s) v/s x");

xlabel('x-distance (m)','FontWeight','bold');

ylabel('Displacement, s (m)','FontWeight','bold');

subplot(2,2,2);

plot(x,v);

title("Linear Velocity (v) v/s x");

xlabel('x-distance (m)','FontWeight','bold');

ylabel('Linear Velocity, v (m/s)','FontWeight','bold');

subplot(2,2,3);

plot(x,omega);

title("Angular Velocity (w) v/s x");

xlabel('x-distance (m)','FontWeight','bold');

ylabel('Angular Velocity, w (rad/s)','FontWeight','bold');

subplot(2,2,4);

plot(s,v);

title("Velocity (v) v/s Displacement (s)");

xlabel('Displacement, s (m)','FontWeight','bold');

ylabel('Linear Velocity, v (m/s)','FontWeight','bold');

Plots and Results

The above code was run on MATLAB and the following plots were generated. Since the subplot

command was used, all plots were generated simultaneously in a single window. But for clearer

visualization each trend, separate plots are attached:

Discussion of Results

Assuming goal point to be x = 150 m , the following values were found:

v = 11.049 m/s

The graphs of displacement v/s x and of linear velocity v/s x follow very similar trends. This is

because the linear velocity is the slope of the displacement curve at each point, and by

observation, it is seen that this slope continually increases. Hence, the theoretical deductions are

confirmed by the 2 plots.

The angular velocity increases up to a certain point, reaches a maximum value, and then starts

to decrease and then almost flattens out. This is because the radius of curvature is minimum at

that point and does not depend on linear velocity as it is constantly increasing.

Linear velocity v/s displacement graph shows that both quantities increase together, initially

velocity increases at a greater rate but then displacement does. This is because the graph of

displacement v/s x is flatter, meaning displacement is less responsive to changes in x-values, but

then as its slope starts to increase, changes in displacement w.r.t x are greater than those in

linear velocity, so the graph starts to incline towards the displacement axis.

Conclusion

This report hints that if the curvilinear path is described by a polynomial of order n , then the

results of angular velocity and linear velocity v/s distance or displacement follow similar

trends , as highlighted in the Plots and Results section. The plots confirmed the theoretical

expectations gained from observing the kinematic model, which indicates that the analysis was

executed successfully. We conclude that MATLAB is indeed a powerful tool that can solve

complex problems and equations in a very short time. All calculus equations and plots in this

report were solved using MATLAB in a total processing time of less than 2 minutes.

References

[1] R.C. Hibbeler; Engineering mechanics, Dynamics, 13th Edition, Pearson Prentice Hall, 2015.

[2] J.L Meriam, L.G Kraige, “Engineering Mechanics: Dynamics”, 7th ed., Wiley.

[3] F. P. Beer, E. R. Johnston, “Vector Mechanics for Engineers: Dynamics”, 8th ed., McGraw-Hill.

[4] P. Garnier, et al. A Hybrid Motion Controller for a Real Car-Like Robot Evolving in a Multi-

Vehicle Environment. Grenoble Cedex 1, France.

[5] Introduction to MATLAB: For Engineering Students, David Hoque, Northwestern, 2005.