This document reveals the physics behind the Center of Mass and the rocket propulsion. Furthermore theories pertaining to the rocket propulsion (Newton's 3rd Law) has been discussed here.

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CENTER OF MASS
&
ROCKET
PROPULSION
COMPILED BY: GROUP VII
YEAR: 1 ST
YEAR 1 ST
SEMESTER

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CONTENT
1. Center of Mass (CM)
1.1 History
1.2 CM-Definition
1.3 Deriving CM mathematically
1.4 Applications in CM
2. Rocket propulsion

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1. Center of Mass(CM)
1.1.History
The concept of CM was first introduced by Greek physicist, mathematician and an engineer
Archimedes of Syracuse. He considered about an amount of uniform field, there by arrived at
mathematical properties of what we called CM. Archimedes also showed that the torque exerted on a
lever by weights resting at various points along the lever is same as what it would be if all the weights
were moved to one single point-their CM. Also in working with floating objects he showed that the
orientation of the floating object is dependent on its CM.
After Archimedes many scientists such as Pappus of Alexandria, Guido Ubaldi, Francesco Maurolico,
Fedrico Commandino, Simon Stevin, Luka Valerio etc.
Figure 1-Archimedes of Syracuse
1.2. CM-Definition
“Point at which the Center of Mass in space that has the property that the weighted position vectors
relative to this point sums zero.”
Every response to a net external force or a torque is acting on this point, hence in physics, CM acts
a major role. It should be noted that when applying different types of physics concept it simplified
to the CM.
When considering about the CM of a rigid body, it is relative to the body, and in the case of an
object with a uniform density the CM resides on the Centroid. In the case of a hollowed object like
a horseshoe the CM may be locate at outside.
CM is a useful reference point for calculations in mechanics that involves masses distributed in
space. This concept helps to build up concepts such as linear and angular momentum and also rigid
body dynamics.
Let’s consider the mathematical phenomenon behind the CM.

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1.3. Deriving of CM mathematically
I. CM of system of masses aligned in a straight line.
Above shown is a system of mass points along a straight line where Mr denotes n number of
mass points along the line. If the CM of the system of masses is G and OG=𝑥̅,
𝑥̅ can be defined as,
𝑥̅ =
𝑚1 𝑥1+𝑚2 𝑥2+𝑚3 𝑥3+⋯+𝑚 𝑟 𝑥 𝑟+⋯𝑚 𝑛 𝑥 𝑛
𝑚1+𝑚2+𝑚3+⋯+𝑚 𝑟+⋯+𝑚 𝑛
II. CM of system of masses placed in a 2D plane.
*If the system of masses consist of n no. of mass points then,
O
MrM4M3M2M1 Mn

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*In the case of 3D system,
III. CM of two identical masses.
The CM is located at the middle point of line joining the two identical masses.
IV. CM of a uniform rod
Position vector of CM is 𝑟̅,
𝑟̅ = 𝑥𝑖̅ + 𝑦𝑗̅ + 𝑧𝑘̅
𝑥 =
Σ𝑚𝑖 𝑥𝑖
𝑀
𝑦 =
Σ𝑚𝑖 𝑦𝑖
𝑀
𝑧 =
Σ𝑚𝑖 𝑧𝑖
𝑀
𝑟̅ =
Σ𝑚𝑖
𝑀
(𝑥𝑖̅ + 𝑦𝑗̅ + 𝑧𝑘̅)
𝑟̅ =
1
𝑀
Σ𝑚𝑖 𝑟̅𝑖
Where 𝑟𝑖̅ is the position vector of 𝑖 𝑡ℎ
mass of the
system.

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V. CM of a solid triangle plate
Y
X
x
dx G
O
OG=𝑥̅
Linear density=𝜌
Length of the rod=𝑙
Considering O, taking moments,
𝑥̅ = ∫
𝑥𝑑𝑚
𝑑𝑚
𝑑𝑚 = 𝜌𝑑𝑥
𝑥̅ = ∫
𝑥𝜌𝑑𝑥
𝑀
𝑙
0
𝑥̅ =
𝜌[𝑥2
]0
𝑙
2𝑀
𝑥̅ =
𝜌𝑙2
2𝑀
=
𝜌𝑙 ∗ 𝑙
2𝑀
=
𝑀𝑙
2𝑀
=
𝑙
2
The CM of a uniform rod is located
a distance of 𝒍
𝟐⁄ from O.
Y
Y
Y
X
Y
O
The height of the triangle is h and this
is an equilateral triangle.
Considering a small mass strip dx,
which is x distance away from the OY
axis,
OG=𝑥̅ and area density is 𝜌
𝑑𝑥
O
X h-x
300
Y
300G

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Therefore,
𝑦 = (ℎ − 𝑥) tan 300
𝑥̅ = ∫
𝑥𝑑𝑚
𝑀
𝑑𝑚 = 𝜌 ∗ 2(ℎ − 𝑥) tan 300
∗ 𝑑𝑥
𝑥̅ = ∫
2𝜌(ℎ − 𝑥) tan 300
𝑀
𝑑𝑥
ℎ
0
𝑥̅ = ∫
2𝜌 tan 300
ℎ
𝑀
𝑑𝑥
ℎ
0
− ∫
2𝜌 tan 300
𝑥
𝑀
𝑑𝑥
ℎ
0
𝑥̅ =
2ℎ𝜌 tan 300[𝑥]0
ℎ
𝑀
−
2𝜌 tan 300[𝑥2]0
ℎ
2𝑀
𝑥̅ =
2𝜌ℎ2
tan 300
2𝑀
𝑥̅ =
√3ℎ
2
The CM of an equilateral triangle is √𝟑𝒉/𝟐 from the base of the triangle.
VI. CM of a circle
O
Y
X
r
𝑥̅
𝑑𝑥
O x
r y
𝜃

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Area density of the circle is 𝜌, distance from O to the CM is 𝑥̅.
Let’s consider a small mass 𝑑𝑚; 𝑤𝑖𝑑𝑡ℎ 𝑖𝑠 𝑑𝑥 which is 𝑥 distance away from O.
Therefore,
𝑥̅ = ∫
𝑥𝑑𝑚
𝑑𝑚
𝑑𝑚 = 𝜌 ∗ 2𝑟 cos 𝜃 ∗ 𝑑𝜃
𝑥 = 𝑟 sin 𝜃
𝑥̅ = ∫
𝑟 sin 𝜃 . 2𝜌𝑟 cos 𝜃
𝑀
2𝜋
0
𝑑𝜃
𝑥̅ = ∫
𝑟2
𝜌 sin 2𝜃
𝑀
2𝜋
0
𝑑𝜃
𝑥̅ =
𝑟2
𝜌[− cos 2𝜃]0
2𝜋
2𝑀
𝑥̅ =
𝑟2
𝜌(−1 + 1)
2𝑀
𝑥̅ = 0
i.e. the CM of a circle is located at the geometric center of the circle.
1.4. Applications in CM
Applications of CM are everywhere around us. Engineers especially use this concept to create balance
over their physical creations like buildings, cars etc. Even for an excellent example the manufacturing
of sports cars draw the main consideration over the position of CM; there the CM should be as lower
as possible, so the car will not turn other way round once the velocity increases. Not only in those
areas, even in sport activities like gymnastics, high jumping, the CM is the critical performance factor.
Therefore it is critical to identify the areas where CM plays the major role.
I. Aeronautics
CM is an important factor when building an aircraft. In order to stabilize the aircraft while
flying the CM must be change within specified limits. If the CM is more forward than it
would be, it is less stable, even if the CM is so much backward the same instability happens.
Not only in aircrafts even in helicopters the CM acts a major role. If it is not so stable it
occurs nose down motion in helicopters causing a crash.
Figure 2-CM of an aircraft

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II. Astronomy
III. Kinesiology-The bio mechanism of studying of human body locomotion
In here the CM is considered as major because, the balanced motion of body is always
due to the shift of the CM in specified limit.
Figure 3-Fosbury flop
SUMMARY
 The location where all the mass of system could be concentrated to be is the Center of
Mass.
 CM make responses to the external forces and torques.
 Distribution of mass is balanced around CM.
 Weighted relative position coordinates defines the position of CM.
 The CM of two identical masses lies on the mid-point of the line joining the two masses.
 The CM of a uniform rod is located on the middle of the uniform rod.
 The CM of an equilateral triangle is √3ℎ/2 from the base of the triangle.
 The CM of a uniform circle (geometrical circle) is on its geometrical center.
 The CM is an important concept as it influences everywhere in our life, industries etc.
Figure 4-Balancing of a baseball bat from
its CM
Figure 5-Balancing hawk bird where
CM is on its beak tip

10. 10 | P a g e
2. Rocket propulsion
According to Sir Isaac Newton, he had stated in his third law of motion that “for every
action there is an equal and opposite reaction”. The propulsion of rockets are totally based
on this concept.
If briefly explained the propellants; combustible fuels are combined in the combustion
chamber, and there all the propellants combine together and chemically reacted. Thereby,
those reactions produce a hot gas that will be ejected via nozzles. This gas ejected in such
a way that has greater acceleration. Due to this acceleration it creates an external force on
the ejected gas resulting another reaction up thrust on the rocket’s engine. This will cause
the propulsion of the rocket.
Let’s discuss the deep physics in the rocket propulsion.
The main force acting on the rocket during the propulsion is the thrust. This is the force
which facilitates motion within the air. The thrust generates due to the pressure which is
exerted on the combustion chamber.
Combustion chamber has an opening to the outside known as a nozzle; the pathway to eject
gas. Due to the asymmetric pressure inside the chamber, the pressure difference is zero in
any given point, i.e. the pressure is constant inside the chamber. But at the
nozzle the conditions are way different as the pressure is comparatively low.
The forced due to the gas pressure on the bottom of the chamber is not
compensated for from the outside. So the resultant force F due to the internal
and external pressure difference, the thrust is opposite to the direction of the
gas ejected. This results a push on the rocket upwards.
this F/thrust equals to,
𝐹 = 𝑞𝑣𝑒 + (𝑃𝑒 − 𝑃𝑎) ∗ 𝐴 𝑒 where 𝑞 is the rate of ejection of fuel, 𝑣𝑒is velocity
of ejecting fuel and 𝑃𝑎, 𝑃𝑒 are ambient atmospheric pressure and ejection
pressure of fuel respectively.
During the whole mechanism the Conservation of Linear Momentum
Concept plays a major role.
Let’s consider a rocket of mass 𝑀 moving in 𝑉 velocity with respect to the earth at time 𝑡.
Figure 6-How a thrust
generates inside the
combustion chamber

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When time=𝑡 + ∆𝑡, the rocket ejects fuel of mass ∆𝑚 in 𝑣𝑒velocity with respect to the
rocket, thereby rocket accelerates, so in 𝑡 + ∆𝑡 time the velocity of the rocket is 𝑣 + ∆𝑣
with respect to the earth.
Therefore the velocity of the fuel ejected with respect to the earth.
𝑣(𝑅,𝐸)̅̅̅̅̅̅̅ = 𝑣 + ∆𝑣⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑣(𝐹,𝑅)̅̅̅̅̅̅̅̅ = 𝑣𝑒⃖⃗⃗⃗ 𝑣(𝐹,𝐸)̅̅̅̅̅̅̅ =?
By considering the relative velocity concept, velocity of ejected fuel with respect to earth is,
𝑣(𝐹,𝐸)̅̅̅̅̅̅̅ = (𝑣 + ∆𝑣) − 𝑣𝑒
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Then, applying Conservation of Linear Momentum Principle to the rocket with respect to earth, (as
at the moment of ejecting fuel, there is no any external force acting on the rocket)
𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑓𝑖𝑛𝑎𝑙 𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝑀𝑣 = ∆𝑚([𝑣 + ∆𝑣] − 𝑣𝑒) + 𝑀(𝑣 + ∆𝑣)
𝑀𝑣 = ∆𝑚. 𝑣 + ∆𝑚. ∆𝑣 − ∆𝑚. 𝑣𝑒 + 𝑀𝑣 + 𝑀. ∆𝑣
Here, ∆𝑚, ∆𝑣 are small quantities, therefore,
∆𝑚. ∆𝑣 ≈ 0 & ∆𝑚. 𝑣 ≈ 0
0 = 𝑀. ∆𝑣 − ∆𝑚. 𝑣𝑒
𝑀. ∆𝑣 = ∆𝑚. 𝑣𝑒
Dividing both sides by ∆𝑡; time period where fuel ejection takes place
𝑀.
∆𝑣
∆𝑡
= 𝑣𝑒.
∆𝑚
∆𝑡
∆𝑣
∆𝑡
; The rate of increment of rocket’s velocity
∆𝑚
∆𝑡
; The rate of ejection of fuel
Therefore, considering limits on both sides,
lim
∆𝑡→0
𝑀.
∆𝑣
∆𝑡
= lim
∆𝑡→0
𝑣𝑒.
∆𝑚
∆𝑡

12. 12 | P a g e
𝑀. 𝑑𝑣 = 𝑣𝑒. 𝑑𝑚
1
𝑣𝑒
. 𝑑𝑣 =
1
𝑀
. 𝑑𝑚
1
𝑣𝑒
∫ 𝑑𝑣
𝑣 𝑓
𝑣 𝑖
= ∫
1
𝑀
𝑚𝑖
𝑚 𝑓
𝑑𝑚
1
𝑣𝑒
[𝑣] 𝑣 𝑖
𝑣 𝑓
= [ln 𝑀] 𝑚 𝑓
𝑚𝑖
1
𝑣𝑒
[𝑣𝑓 − 𝑣𝑖] = [ln 𝑚𝑖 − ln 𝑚 𝑓]
𝑣𝑓 − 𝑣𝑖 = 𝑉
Therefore,
𝑉 = 𝑣𝑒. ln[
𝑚𝑖
𝑚 𝑓
]
So it is clear that when the rocket propels, the mass of the rocket gradually decreased as fuel
combustion happens throughout the rocket’s propulsion. Hence the CM of the rocket tends to
fluctuate which may result in some torques to taking place. In order to stop this unwanted external
forces and torques rocket manufacturers introduce spare nozzles by the side of the rocket which there
by helps to balance the rocket while in the propulsion mechanism.
Figure 7-Nozzle of a rocket Figure 8-Propulsion of
rocket

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References:
http://physics.tutorvista.com/rotational-dynamics/center-of-mass.html 25/10/2015 7.52 PM
https://en.wikipedia.org/wiki/Center_of_mass 25/10/2015 8.19 PM
https://www.grc.nasa.gov/www/k-12/airplane/rocket.html 26/10/2015 10.39 PM
http://www.braeunig.us/space/propuls.htm 26/10/2015 11.07 PM
Group members:
S.M.F.Abeywickrama 13/AS/090 EP 1767
P.D.Sandaruwan 13/AS/111 EP 1783
H.P.G.Y.Lochana 13/AS/124 EP 1792
R.D.P.M.Prabhodika 13/AS/127 EP 1795
W.M.H.G.T.C.K.Weerakoon 13/AS/134 EP 1801