Muhammad Azeem Uddin

me******

Assigment#1

Engineering Dynamics – Dr. ***********

Rectilinear Motion If a body changes its position with respect to time in a certain frame of reference then it is termed as motion of the body. A body on its own is full of facts and properties like stress & strain but if we consider a body as a particle then its motion is termed as rectilinear motion. In such case we are not supposed to consider the properties of the body and we just observe its motion.

Problem#1

Motion of body due to Gravity

Abstract:

Most of the physical occurrences around us happen at constant rates. This gives us idea of the fact that a mathematical model can always be created about such phenomenon. A simple case has been taken under consideration about a baseball that is thrown down from certain height and we are to observe its motion and also at a point where it is about to touch ground. Numerical calculations are been done and the relation of governing species are shown.

Problem Statement:

12.1. A baseball is thrown downward from a 50-ft tower with an initial speed of 18ft/s. Determine the speed at which it hits the ground and the time of travel.

Numerical Solution:

Height= 50 ft.

Initial Velocity Vi= 18 ft. /s

Acceleration= g = 32.2 (acceleration due to gravity)

Using Newton’s equations of motion

2gh = Vf2 – Vi

2

Vf = (2gh + Vi2)0.5

Vf = {2(32.2)(50) + (18)2}0.5

Vf = 59.5 ft. /s which is the speed at which, the baseball will hit the ground

Vf = Vi + gt

t = (Vf - Vi)/g (here t is time)

t = (59.5 - 18)/32.2

t = 1.29 sec

Graphical Analysis:

*Distance covered by due to its motion, taking its highest point as zero

0

10

20

30

40

50

60

0 0.2 0.4 0.6 0.8 1 1.2 1.4

Vel

oci

ty (f

t. /

s)

Time (seconds)

Fig. 1.0: Velocity Time Graph

0

10

20

30

40

50

60

0 10 20 30 40 50 60

Vel

oci

ty (f

t /s

)

Distance Travelled* (ft)

Fig. 1.1: Velocity Distance Graph

Discussion:

Whatever goes above the ground to a certain limit comes back to ground. This certain limit can be taken as 1000 kilometers above the surface of the earth for understanding the free fall phenomena because there is very little change in this attraction under this height. Although at a height of about 50000 kilometers from surface of the earth, the value of this pull is nearly zero. The property that governs this downward motion is gravity and the force which attracts objects towards ground is called Force of gravity an acceleration thus produced is called Acceleration due to gravity.

In the above mentioned problem a baseball was thrown from a height of 50-ft. The words in the problem do not show its launch velocity instead it tells about initial velocity. So we consider this motion of ball as a Free-Fall motion. The bottom line of this downward motion is made up of two things i.e. Mass of the ball & Acceleration due to gravity. Anything that has been raised from ground has potential energy which is directly proportional to its height. If we leave that body from that position then its potential energy will start converting into kinetic energy in the form of its velocity (Fig. 1.1). Analyzing the calculations and graph made us conclude that velocity is directly proportional to time object has travelled (Fig. 1.0). In addition, we can also say that velocity of a free falling particle also depends upon its height. It increases with constant rate, so if height is more then it will be able to achieve a high magnitude of velocity before reaching the ground.

Problem#2

Motion of car moving with constant acceleration

Abstract:

Acceleration is vibrant phenomena in cars. The properties like velocity and acceleration helps us in estimating the drive time. For instance, if we want to reach any place in a particular time, we can compute the required acceleration and velocity just by knowing the distance of our destination. The problem under consideration just wants us to determine the distance a car is going to travel with a constant value of acceleration also how long it would take. Actually the distance is of between two different velocities i.e. the time and distance taken by car to achieve a velocity from another velocity.

Problem Statement:

*12-4. Travelling with an initial speed of 70km/h, a car accelerates at 6000km/h2 along a straight road. How long will it take to reach a speed of 120km/h? Also through what distance does the car travel during this time?

Numerical Solution:

Vi= 70 km/h

ac= 6000 km/h2

vF= 120 km/h

t= ? s=?

All values are of same set of units, so there is no need of converting them to SI system of units.

Using Newton’s Equations of motion,

Vf = Vi + at

t= 0.0083 hrs

2as = Vf2-Vi

2

s = 0.792 km

Graphical Analysis:

0

0.2

0.4

0.6

0.8

1

1.2

0 0.002 0.004 0.006 0.008 0.01 0.012

Dis

tan

ce C

ove

red

(km

)

Time passed (h)

Fig. 2.0: Distance Time Graph

Discussion:

Due to the fact that we cannot achieve any absolute value of velocity all of a sudden then there is always some acceleration. The reason for this acceleration is the required driving force i.e. the force required to make a static body move. Although there are many systems which are commonly referred as uniformly moving devices like fans. But even such devices also possess some acceleration at the time they are suddenly brought to work and after sometime there acceleration turns zero.

The above mentioned problem just discusses about the motion of a car in a straight line. Contextually, the value of constant acceleration of the car is given to be 6000km/h. Its initial velocity is also given i.e. 70km/h. As the car accelerates, its velocity increases, finally reaches 120km/h. During this process the car has travelled through a distance S and taken some time t. From the results it was found that distance covered is directly proportional to time passed (Fig. 2.0) also the velocity was found to linearly increasing with passing time (Fig. 2.1). We can also manipulate this case to find out how long car would take to reach to a velocity of 120km/h if we start from rest.

t= 0.02 hrs

The answer clearly shows that car took 24x more time in reaching 120km/h if it starts from 0km/h. So it is easy to reach to certain speed if existing speed is near.

0

20

40

60

80

100

120

140

0 0.002 0.004 0.006 0.008 0.01 0.012

Ve

loci

ty (

km/h

)

Time passed (h)

Fig. 2.1: Velocity Time Graph

Problem#3

Train’s relative motion with kilometer marks

Abstract:

A group of students were subjected to monitor the velocity profile of a train. They were taking readings of velocity and time at the kilometer marks. A bunch of information was gathered and it was required to find out the pattern for the increment of velocity. Combining all the information, it was possible to obtain a velocity graph and total time taken while passing by three kilometer marks which make a total distance of 2 kilometers.

Problem Statement:

12-30. As a rain accelerates uniformly it passes successive kilometer marks while travelling at velocities of 2 m/s and 10 m/s. Determine the train’s velocity when it passes the next kilometer mark and the time it takes to travel the 2-km distance.

Numerical Solution:

Motion A to B

Vi = 2m/s

Vf = 10m/s

S = 1km or 1000 m

Using Newton’s equation of Motion,

2as = Vf2 –Vi

2

a

a = 0.048 m/s2

Motion B to C

Vi = 10m/s

a = 0.048 m/s2

S = 1km or 1000 m

Using Newton’s equation of Motion,

2as = Vf2 –Vi

2

Vf = √

Vf = 14 m/s

Motion A to C

Vi = 2m/s

Vf = 14m/s

S = 2km or 2000 m

Vf = Vi + at

t =

t = 250 sec

Graphical Analysis:

0

2

4

6

8

10

12

14

16

0 50 100 150 200 250 300

Ve

loci

ty (

m/s

)

Time (Seconds)

Fig. 3.0: Velocity Increment Profile

Discussion:

On railway tracks, usually there are kilometer marks for the reference of the train driver and passengers. In given problem, we were subjected to glance over the motion of the train that is passing through such kilometer marks. Three kilometer marks are considered namely A, B & C. Velocity at point A & B is given.

The Newton’s equations of motion are too much helpful in studying a particle’s motion. It is simple to evaluate the acceleration if a body is moving with constant acceleration. Therefore in our problem the velocities at point A & B evaluated the constant acceleration. The main relation i.e. the relation of Velocity and time shows that, velocity is directly proportional to time in this case. Overall, another important thing can be seen that the plotted points between velocity and time formed a straight line which clearly shows the constant acceleration. Finally we can say that there are several methods for solving a problem. So in this case, if we considered the whole path from A to C then it would be impossible to calculate required stuff. In many cases, we can split a long path into several small portions so that analysis can be carried out easily.

Image Courtesy: http://www.clipart-fr.com/images.php?id=5528