Center of Mass & Collision | Physics Optional Notes for UPSC PDF Download

 Page 1


CENTER OF MASS & COLLISION 
INTRODUCTION 
The center of mass (C.M.) is a theoretical point that may or may not be within the actual 
system. In many cases in mechanics (but not always), we can consider the entire mass of 
the system to be concentrated at this point for analysis purposes. 
Illustration:- If you throw a rotating pen in the air, only one point will be going as a 
parabola. 
CENTER OF MASS 
For a system of particles center of mass is the point at which its total mass is supposed to 
be concentrated. 
 
The center of mass of an object is a point that acts as if the entire body's mass is 
concentrated there. It moves in the same manner as a single point mass with the same 
mass as the object, when it experiences the same external forces as the object. 
Center of mass (COM) of several discrete Particles: 
If coordinates of particles of mass m
1
, m
2
,….. Are 
(x
1
,y
1
,z
1
),(x
2
,y
2
,z
2
)… 
Then position vector of their center of mass is 
??? 
????
 =?? ????
??ˆ+?? ????
??ˆ+?? ????
??ˆ
 =
?? 1
(?? 1
??ˆ+?? 1
??ˆ+?? 1
??ˆ
)+?? 2
(?? 2
??ˆ+?? 2
??ˆ+?? 2
??ˆ
)+?? 3
(?? 3
??ˆ+?? 3
??ˆ+?? 3
??ˆ
)+?
?? 1
+?? 2
+?? 3
+?
 =
(?? 1
?? 1
+?? 2
?? 2
+?)??ˆ+(?? 1
?? 1
+?? 2
?? 2
…)??ˆ+(?? 1
?? 1
+?? 2
?? 2
+..)??ˆ
?? 1
+?? 2
+?? 3
+..
 
So, ?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?? 3
+?..
),?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?…….
),?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?……
) 
Example. Find COM 
Solution: 1 kg=(0,0),2 kg=(a,0) 
3 kg=(a,a),4 kg=(0,a) 
Page 2


CENTER OF MASS & COLLISION 
INTRODUCTION 
The center of mass (C.M.) is a theoretical point that may or may not be within the actual 
system. In many cases in mechanics (but not always), we can consider the entire mass of 
the system to be concentrated at this point for analysis purposes. 
Illustration:- If you throw a rotating pen in the air, only one point will be going as a 
parabola. 
CENTER OF MASS 
For a system of particles center of mass is the point at which its total mass is supposed to 
be concentrated. 
 
The center of mass of an object is a point that acts as if the entire body's mass is 
concentrated there. It moves in the same manner as a single point mass with the same 
mass as the object, when it experiences the same external forces as the object. 
Center of mass (COM) of several discrete Particles: 
If coordinates of particles of mass m
1
, m
2
,….. Are 
(x
1
,y
1
,z
1
),(x
2
,y
2
,z
2
)… 
Then position vector of their center of mass is 
??? 
????
 =?? ????
??ˆ+?? ????
??ˆ+?? ????
??ˆ
 =
?? 1
(?? 1
??ˆ+?? 1
??ˆ+?? 1
??ˆ
)+?? 2
(?? 2
??ˆ+?? 2
??ˆ+?? 2
??ˆ
)+?? 3
(?? 3
??ˆ+?? 3
??ˆ+?? 3
??ˆ
)+?
?? 1
+?? 2
+?? 3
+?
 =
(?? 1
?? 1
+?? 2
?? 2
+?)??ˆ+(?? 1
?? 1
+?? 2
?? 2
…)??ˆ+(?? 1
?? 1
+?? 2
?? 2
+..)??ˆ
?? 1
+?? 2
+?? 3
+..
 
So, ?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?? 3
+?..
),?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?…….
),?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?……
) 
Example. Find COM 
Solution: 1 kg=(0,0),2 kg=(a,0) 
3 kg=(a,a),4 kg=(0,a) 
 ? x
CM
=
(1×0)+(2×a)+(3×a)+(4×0)
1+2+3+4
 =
5a
10
=
a
2
 ? ?? ????
=
(1×0)+(2×0)+(3×?? )+(4×?? )
1+2+3+4
 =
7?? 10
 ?  COM =
?? 2
??ˆ+
7?? 10
??ˆ
 
 
Example. Three bodies of equal masses are placed at (0,0),(?? ,0) 
 
Solution: ?? ????
=
0×?? +?? ×?? +
?? 2
×?? ?? +?? +?? =
?? 2
, 
 
 
?? ????
=
0×?? +0×?? +
?? v3
2
×?? ?? +?? +?? =
?? v3
6
 
Example. Calculate the position of the center of mass of a system consisting of two 
particles of masses ?? 1
 and ?? 2
 separated by a distance ?? apart, from ?? 1
. 
Solution: Treating the line joining the two particles as x axis 
x
CM
=
m
1
×0+m
2
×L
m
1
+m
2
=
m
2
 L
 m
1
+m
2
,y
CM
=0 z
CM
=0 
 
 
Page 3


CENTER OF MASS & COLLISION 
INTRODUCTION 
The center of mass (C.M.) is a theoretical point that may or may not be within the actual 
system. In many cases in mechanics (but not always), we can consider the entire mass of 
the system to be concentrated at this point for analysis purposes. 
Illustration:- If you throw a rotating pen in the air, only one point will be going as a 
parabola. 
CENTER OF MASS 
For a system of particles center of mass is the point at which its total mass is supposed to 
be concentrated. 
 
The center of mass of an object is a point that acts as if the entire body's mass is 
concentrated there. It moves in the same manner as a single point mass with the same 
mass as the object, when it experiences the same external forces as the object. 
Center of mass (COM) of several discrete Particles: 
If coordinates of particles of mass m
1
, m
2
,….. Are 
(x
1
,y
1
,z
1
),(x
2
,y
2
,z
2
)… 
Then position vector of their center of mass is 
??? 
????
 =?? ????
??ˆ+?? ????
??ˆ+?? ????
??ˆ
 =
?? 1
(?? 1
??ˆ+?? 1
??ˆ+?? 1
??ˆ
)+?? 2
(?? 2
??ˆ+?? 2
??ˆ+?? 2
??ˆ
)+?? 3
(?? 3
??ˆ+?? 3
??ˆ+?? 3
??ˆ
)+?
?? 1
+?? 2
+?? 3
+?
 =
(?? 1
?? 1
+?? 2
?? 2
+?)??ˆ+(?? 1
?? 1
+?? 2
?? 2
…)??ˆ+(?? 1
?? 1
+?? 2
?? 2
+..)??ˆ
?? 1
+?? 2
+?? 3
+..
 
So, ?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?? 3
+?..
),?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?…….
),?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?……
) 
Example. Find COM 
Solution: 1 kg=(0,0),2 kg=(a,0) 
3 kg=(a,a),4 kg=(0,a) 
 ? x
CM
=
(1×0)+(2×a)+(3×a)+(4×0)
1+2+3+4
 =
5a
10
=
a
2
 ? ?? ????
=
(1×0)+(2×0)+(3×?? )+(4×?? )
1+2+3+4
 =
7?? 10
 ?  COM =
?? 2
??ˆ+
7?? 10
??ˆ
 
 
Example. Three bodies of equal masses are placed at (0,0),(?? ,0) 
 
Solution: ?? ????
=
0×?? +?? ×?? +
?? 2
×?? ?? +?? +?? =
?? 2
, 
 
 
?? ????
=
0×?? +0×?? +
?? v3
2
×?? ?? +?? +?? =
?? v3
6
 
Example. Calculate the position of the center of mass of a system consisting of two 
particles of masses ?? 1
 and ?? 2
 separated by a distance ?? apart, from ?? 1
. 
Solution: Treating the line joining the two particles as x axis 
x
CM
=
m
1
×0+m
2
×L
m
1
+m
2
=
m
2
 L
 m
1
+m
2
,y
CM
=0 z
CM
=0 
 
 
KEY POINT: 
COM of two particles divides internally the line joining two particles in inverse ratio of 
their masses. 
Proof: 
x
CM
=
m
1
x
1
+m
2
x
2
 m
1
+m
2
(-d
1
,0) d
1
(0,0) d
2
( d
2
,0)
0=
m
1
(-d
1
)+m
2
( d
2
)
m
1
+m
2
 COM m
2
 m
1
 d
1
=m
2
 d
2
d
1
 d
2
=
m
2
 m
1
 
Question. 1 What are the co-ordinates of the center of mass of the three particles 
system shown in figure. 
 
Ans.  
16
15
 m,
20
15
 m 
Question2. Four particles of masses ?? ,2?? ,3?? ,4?? are placed at corners of a square of 
side 'a' as shown in fig. Find out co-ordinates of centre of mass. 
 
Ans.  (
?? 2
,
7
10
a) 
Question3. A rigid body consists of a 3 kg mass connected to a 2 kg mass by a massless 
rod. The 3 kg mass is located at ?? 
1
=(2??ˆ+5??ˆ)m and the 2 kg mass at ?? 
2
=(4??ˆ+2??ˆ)m . 
Find the coordinates of the centre of mass. 
Ans.  (
14
5
iˆ+
19
5
jˆ)m 
Center of Mass of Continuous Distribution of Mass: 
If the system has continuous distribution of mass, considering the mass clement dm at 
position r  as a point mass and replacing summation by integration. R
?? 
CM
=
1
M
?r dm . 
Page 4


CENTER OF MASS & COLLISION 
INTRODUCTION 
The center of mass (C.M.) is a theoretical point that may or may not be within the actual 
system. In many cases in mechanics (but not always), we can consider the entire mass of 
the system to be concentrated at this point for analysis purposes. 
Illustration:- If you throw a rotating pen in the air, only one point will be going as a 
parabola. 
CENTER OF MASS 
For a system of particles center of mass is the point at which its total mass is supposed to 
be concentrated. 
 
The center of mass of an object is a point that acts as if the entire body's mass is 
concentrated there. It moves in the same manner as a single point mass with the same 
mass as the object, when it experiences the same external forces as the object. 
Center of mass (COM) of several discrete Particles: 
If coordinates of particles of mass m
1
, m
2
,….. Are 
(x
1
,y
1
,z
1
),(x
2
,y
2
,z
2
)… 
Then position vector of their center of mass is 
??? 
????
 =?? ????
??ˆ+?? ????
??ˆ+?? ????
??ˆ
 =
?? 1
(?? 1
??ˆ+?? 1
??ˆ+?? 1
??ˆ
)+?? 2
(?? 2
??ˆ+?? 2
??ˆ+?? 2
??ˆ
)+?? 3
(?? 3
??ˆ+?? 3
??ˆ+?? 3
??ˆ
)+?
?? 1
+?? 2
+?? 3
+?
 =
(?? 1
?? 1
+?? 2
?? 2
+?)??ˆ+(?? 1
?? 1
+?? 2
?? 2
…)??ˆ+(?? 1
?? 1
+?? 2
?? 2
+..)??ˆ
?? 1
+?? 2
+?? 3
+..
 
So, ?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?? 3
+?..
),?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?…….
),?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?……
) 
Example. Find COM 
Solution: 1 kg=(0,0),2 kg=(a,0) 
3 kg=(a,a),4 kg=(0,a) 
 ? x
CM
=
(1×0)+(2×a)+(3×a)+(4×0)
1+2+3+4
 =
5a
10
=
a
2
 ? ?? ????
=
(1×0)+(2×0)+(3×?? )+(4×?? )
1+2+3+4
 =
7?? 10
 ?  COM =
?? 2
??ˆ+
7?? 10
??ˆ
 
 
Example. Three bodies of equal masses are placed at (0,0),(?? ,0) 
 
Solution: ?? ????
=
0×?? +?? ×?? +
?? 2
×?? ?? +?? +?? =
?? 2
, 
 
 
?? ????
=
0×?? +0×?? +
?? v3
2
×?? ?? +?? +?? =
?? v3
6
 
Example. Calculate the position of the center of mass of a system consisting of two 
particles of masses ?? 1
 and ?? 2
 separated by a distance ?? apart, from ?? 1
. 
Solution: Treating the line joining the two particles as x axis 
x
CM
=
m
1
×0+m
2
×L
m
1
+m
2
=
m
2
 L
 m
1
+m
2
,y
CM
=0 z
CM
=0 
 
 
KEY POINT: 
COM of two particles divides internally the line joining two particles in inverse ratio of 
their masses. 
Proof: 
x
CM
=
m
1
x
1
+m
2
x
2
 m
1
+m
2
(-d
1
,0) d
1
(0,0) d
2
( d
2
,0)
0=
m
1
(-d
1
)+m
2
( d
2
)
m
1
+m
2
 COM m
2
 m
1
 d
1
=m
2
 d
2
d
1
 d
2
=
m
2
 m
1
 
Question. 1 What are the co-ordinates of the center of mass of the three particles 
system shown in figure. 
 
Ans.  
16
15
 m,
20
15
 m 
Question2. Four particles of masses ?? ,2?? ,3?? ,4?? are placed at corners of a square of 
side 'a' as shown in fig. Find out co-ordinates of centre of mass. 
 
Ans.  (
?? 2
,
7
10
a) 
Question3. A rigid body consists of a 3 kg mass connected to a 2 kg mass by a massless 
rod. The 3 kg mass is located at ?? 
1
=(2??ˆ+5??ˆ)m and the 2 kg mass at ?? 
2
=(4??ˆ+2??ˆ)m . 
Find the coordinates of the centre of mass. 
Ans.  (
14
5
iˆ+
19
5
jˆ)m 
Center of Mass of Continuous Distribution of Mass: 
If the system has continuous distribution of mass, considering the mass clement dm at 
position r  as a point mass and replacing summation by integration. R
?? 
CM
=
1
M
?r dm . 
So that x
cm
=
1
M
?xdm ,y
cm
=
1
M
?ydm and ?? cm
=
1
M
?zdm 
 
Symmetrical Bodies 
Let's start by considering a system of two identical particles to understand the concept 
easily. Then, we can expand this idea to apply to a straight uniform rod, uniform 
symmetric plates, and uniform symmetric solid objects. 
Center of mass of a system of two identical particles 
Center of mass of a system of two identical particles lies at the midpoint between them 
on the lie joining them. 
Center of mass of a system of a straight uniform rod 
 Consider two identical particles  
A And B 
 At equal distances 
 
From the center C of the rod. Mass center of system these two particles is at C. The whole 
rod can be assumed to be made of large number of such systems each having its center of 
mass at the mid point C of the rod. Therefore center of mass of the whole rod must be at 
its mid-point. 
Center of mass of a system of a uniform symmetric curved rod 
 Consider two identical particles A and B located at equal distances from the line of 
symmetry. Center of mass of system of these two particles is at  
C 
. 
 
Line of symmetry: The whole rod can be assumed to be made of large number of such 
systems each having its center of mass at the mid-point C of the line joining them. 
Therefore center of mass of the whole rod must be on the axis of symmetry. 
Page 5


CENTER OF MASS & COLLISION 
INTRODUCTION 
The center of mass (C.M.) is a theoretical point that may or may not be within the actual 
system. In many cases in mechanics (but not always), we can consider the entire mass of 
the system to be concentrated at this point for analysis purposes. 
Illustration:- If you throw a rotating pen in the air, only one point will be going as a 
parabola. 
CENTER OF MASS 
For a system of particles center of mass is the point at which its total mass is supposed to 
be concentrated. 
 
The center of mass of an object is a point that acts as if the entire body's mass is 
concentrated there. It moves in the same manner as a single point mass with the same 
mass as the object, when it experiences the same external forces as the object. 
Center of mass (COM) of several discrete Particles: 
If coordinates of particles of mass m
1
, m
2
,….. Are 
(x
1
,y
1
,z
1
),(x
2
,y
2
,z
2
)… 
Then position vector of their center of mass is 
??? 
????
 =?? ????
??ˆ+?? ????
??ˆ+?? ????
??ˆ
 =
?? 1
(?? 1
??ˆ+?? 1
??ˆ+?? 1
??ˆ
)+?? 2
(?? 2
??ˆ+?? 2
??ˆ+?? 2
??ˆ
)+?? 3
(?? 3
??ˆ+?? 3
??ˆ+?? 3
??ˆ
)+?
?? 1
+?? 2
+?? 3
+?
 =
(?? 1
?? 1
+?? 2
?? 2
+?)??ˆ+(?? 1
?? 1
+?? 2
?? 2
…)??ˆ+(?? 1
?? 1
+?? 2
?? 2
+..)??ˆ
?? 1
+?? 2
+?? 3
+..
 
So, ?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?? 3
+?..
),?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?…….
),?? ????
=(
?? 1
?? 1
+?? 2
?? 2
+?….
?? 1
+?? 2
+?……
) 
Example. Find COM 
Solution: 1 kg=(0,0),2 kg=(a,0) 
3 kg=(a,a),4 kg=(0,a) 
 ? x
CM
=
(1×0)+(2×a)+(3×a)+(4×0)
1+2+3+4
 =
5a
10
=
a
2
 ? ?? ????
=
(1×0)+(2×0)+(3×?? )+(4×?? )
1+2+3+4
 =
7?? 10
 ?  COM =
?? 2
??ˆ+
7?? 10
??ˆ
 
 
Example. Three bodies of equal masses are placed at (0,0),(?? ,0) 
 
Solution: ?? ????
=
0×?? +?? ×?? +
?? 2
×?? ?? +?? +?? =
?? 2
, 
 
 
?? ????
=
0×?? +0×?? +
?? v3
2
×?? ?? +?? +?? =
?? v3
6
 
Example. Calculate the position of the center of mass of a system consisting of two 
particles of masses ?? 1
 and ?? 2
 separated by a distance ?? apart, from ?? 1
. 
Solution: Treating the line joining the two particles as x axis 
x
CM
=
m
1
×0+m
2
×L
m
1
+m
2
=
m
2
 L
 m
1
+m
2
,y
CM
=0 z
CM
=0 
 
 
KEY POINT: 
COM of two particles divides internally the line joining two particles in inverse ratio of 
their masses. 
Proof: 
x
CM
=
m
1
x
1
+m
2
x
2
 m
1
+m
2
(-d
1
,0) d
1
(0,0) d
2
( d
2
,0)
0=
m
1
(-d
1
)+m
2
( d
2
)
m
1
+m
2
 COM m
2
 m
1
 d
1
=m
2
 d
2
d
1
 d
2
=
m
2
 m
1
 
Question. 1 What are the co-ordinates of the center of mass of the three particles 
system shown in figure. 
 
Ans.  
16
15
 m,
20
15
 m 
Question2. Four particles of masses ?? ,2?? ,3?? ,4?? are placed at corners of a square of 
side 'a' as shown in fig. Find out co-ordinates of centre of mass. 
 
Ans.  (
?? 2
,
7
10
a) 
Question3. A rigid body consists of a 3 kg mass connected to a 2 kg mass by a massless 
rod. The 3 kg mass is located at ?? 
1
=(2??ˆ+5??ˆ)m and the 2 kg mass at ?? 
2
=(4??ˆ+2??ˆ)m . 
Find the coordinates of the centre of mass. 
Ans.  (
14
5
iˆ+
19
5
jˆ)m 
Center of Mass of Continuous Distribution of Mass: 
If the system has continuous distribution of mass, considering the mass clement dm at 
position r  as a point mass and replacing summation by integration. R
?? 
CM
=
1
M
?r dm . 
So that x
cm
=
1
M
?xdm ,y
cm
=
1
M
?ydm and ?? cm
=
1
M
?zdm 
 
Symmetrical Bodies 
Let's start by considering a system of two identical particles to understand the concept 
easily. Then, we can expand this idea to apply to a straight uniform rod, uniform 
symmetric plates, and uniform symmetric solid objects. 
Center of mass of a system of two identical particles 
Center of mass of a system of two identical particles lies at the midpoint between them 
on the lie joining them. 
Center of mass of a system of a straight uniform rod 
 Consider two identical particles  
A And B 
 At equal distances 
 
From the center C of the rod. Mass center of system these two particles is at C. The whole 
rod can be assumed to be made of large number of such systems each having its center of 
mass at the mid point C of the rod. Therefore center of mass of the whole rod must be at 
its mid-point. 
Center of mass of a system of a uniform symmetric curved rod 
 Consider two identical particles A and B located at equal distances from the line of 
symmetry. Center of mass of system of these two particles is at  
C 
. 
 
Line of symmetry: The whole rod can be assumed to be made of large number of such 
systems each having its center of mass at the mid-point C of the line joining them. 
Therefore center of mass of the whole rod must be on the axis of symmetry. 
Center of mass of a uniform plate (lamina) 
Consider a symmetric uniform plate. It can be assumed to be composed of several thin 
uniform parallel rods like rod ???? shown in the figure. All of these rods have center of 
mass on the line of symmetry, therefore the whole lamina has its center of mass on the 
line of symmetry. 
 
Centre of mass of a uniform symmetric solid object 
 
Let's imagine a uniform symmetric solid object that's made by rotating a symmetric shape 
180 degrees about its line of symmetry. Now, let's consider this object as if it's made up of 
many thin, uniform disks stacked on top of each other. These disks have their center of 
mass aligned along the line of symmetry. As a result, the entire solid object also has its 
center of mass located on this line of symmetry. 
 
 
 
Center of mass of uniform bodies 
Following the similar reasoning, it can be shown that center of mass of uniform bodies ’ 
lies on their geometric centers. 
Example. Locate COM 
 
 
Coordinate of COM=(
L
2
,
L
3
) 
COM OF CONTINUOUS OBJECT: