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Physics-31.Rotational Mechanics

5.1 DEFINITION OF CENTRE OF MASS Centre of mass: Every physical system has associated with it a certain point whose motion characterizes the motion of the whole system. When the system moves under some external forces than this point moves as if the entire mass of the system is concentrated at this point and also the external force is applied at his point of translational motion. This point is called the centre of mass of the system. Centre of mass of system of n point masses or n particles is that point about which moment of mass of the system is zero, it means that if about a particular origin the moment of mass of system of n point masses is zero, then that particular origin is the centre of mass of the system. The concept of centre of mass is a pure mathematical concept. If there are n particles having mass m1, m2 ….m n and are placed in space (x1, y1, z1), (x2, y2, z2) ……….(x n, y n, z n) then centre of mass of system is defined as (X, Y, Z) where = Y = and Z = where M = is the total mass of X the system. Locate the point with coordinates (X, Y, Z). This point is called the centre of mass of the given collection of the particles. If the position vector of the i th particle is ri, the centre of mass is defined to have the position vector Where M = m1 + m2 + ……….. + mn = where = i x cm + j y cm + k (z cm) This is equation for centre of mass of n point masses. Illustration 1: Three equal masses are situates at vertices of equilateral triangle as shown in figure. Find centre of mass of the system. Solution: Let m be the mass of three masses and XCM, YCM and ZCM be the centre of masses of along the X axis, Y axis and Z axis, then Hence coordinate of centre of mass is 5.2 CENTRE OF MASS OF CONTINUOUS BODIES If we consider the body to have continuous distribution of matter, the summation in the formula of centre of mass should be replaced by integration. If x, y, z are the coordinates of this small mass dm, we write the coordinates of the centre of mass as The integration is to be performed under proper limits so that as the integration variable goes through the limits, the elements cover the entire body. We illustrate the method with three examples. Illustration 2: Find centre of mass of a uniform straight rod of mass m and length l, Solution: Let M and L be the mass and the length of the rod respectively. Take the left end of the rod as the origin and the X–axis along the rod (figure). Consider an element of the rod between the positions A and B of the rod. Let the element be at a distance x from the centre O and its width be dx. So as x varies from 0 through L, the elements cover the entire rod. As the rod is uniform, the mass per unit length is M/L and hence the mass of the element is dm = (M/L) dx. The x–coordinate of the centre of mass of the rod is The y–coordinate is , Y = and similarly Z = 0. The centre of mass is at , i.e. at the middle

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5.1 DEFINITION OF CENTRE OF MASS
Centre of mass: Every physical system has associated with it a certain point whose motion characterizes
the motion of the whole system. When the system moves under some external forces than this point moves
as if the entire mass of the system is concentrated at this point and also the external force is applied at his
point of translational motion. This point is called the centre of mass of the system.
Centre of mass of system of n point masses or n particles is that point about which moment of mass of the
system is zero, it means that if about a particular origin the moment of mass of system of n point masses is
zero, then that particular origin is the centre of mass of the system.
The concept of centre of mass is a pure mathematical concept. If there are n
particles having mass m1, m2 ….m n and are placed in space (x1, y1, z1), (x2, y2,
z2) ……….(x n, y n, z n) then centre of mass of system is defined as (X, Y, Z)
where
= 1
M
i i
i
m x
 Y = 1
M
i i
i
m y
 and Z = 1
M
i i
i
m z
 where M =
i
i
m
 is the total
mass of X the system.
Y
Z
X
1 1 1 1
m (x ,y ,z )
0(x,y,z)













O
Locate the point with coordinates (X, Y, Z). This point is called the centre of mass of the given collection of
the particles. If the position vector of the i th
particle is ri, the centre of mass is defined to have the position
vector
CM i i
i
1
R = m r
M
 

Where M = m1 + m2 + ……….. + mn = n
i
i 1
m


where
i i i i
r x i y j z k
   
  
1 1 1 1 2 2 2 2
CM
n n n n
m (x i+ y j + z k) + m (x i+ y j+ z k) +..
1
R =
m (x i+ y j + z k)
M
  
 
 
1 1 2 2 n 1 1 2 2 n
1 1 2 2 n
i(m x m x .. mx ) j(m y m y .. my ) ..
1
k(m z m z .. mz )
       
 
  
  
 
M
= i x cm + j y cm + k (z cm)
n
i i
i=1
CM
m r
R =
M

 
This is equation for centre of mass of n point masses.
Illustration 1: Three equal masses are situates at vertices of equilateral triangle as shown in figure. Find
centre of mass of the system.
Solution: Let m be the mass of three masses and XCM, YCM and ZCM be the centre of masses of along
the X axis, Y axis and Z axis, then
1 1 2 2 3 3
CM
1 2 3
m x m x m x
X
m m m
 

 
     
CM
m 0 m a m a / 2 a
X
m m m 2
 
 
 
y
m
a
x
a
a
m
m
(0,0,0) (a,0,0)
 
a / 2, 3a / 2,0
1 1 2 2 3 3
CM
1 2 3
m y m y m y
Y
m m m
 

 
     
CM
m 0 m 0 m 3a / 2 a
Y
m m m 2 3
 
 
 
1 1 2 2 3 3
CM
1 2 3
m z m z m z
Z
m m m
 

 
Page PAGE 81 AIEEE
Chapter - 5 Rotational Motion
Hence coordinate of centre of mass is a a
, ,0
2 2 3
 
 
 
5.2 CENTRE OF MASS OF CONTINUOUS BODIES
If we consider the body to have continuous distribution of matter, the summation
in the formula of centre of mass should be replaced by integration. If x, y, z are
the coordinates of this small mass dm, we write the coordinates of the centre of
mass as
1 1 1
X = xdm,Y = y dm, Z = zdm
M M M
  
y
x
O
r

dm
The integration is to be performed under proper limits so that as the integration variable goes through the
limits, the elements cover the entire body. We illustrate the method with three examples.
Illustration 2: Find centre of mass of a uniform straight rod of mass m and length l,
Solution: Let M and L be the mass and the length of the rod respectively. Take the left end of the rod
as the origin and the X–axis along the rod (figure). Consider an element
of the rod between the positions A and B of the rod. Let the element be at a
distance x from the centre O and its width be dx. So as x varies from 0 through
L, the elements cover the entire rod.
dx
O x
(x, o, o)
x
As the rod is uniform, the mass per unit length is M/L and hence the mass of the element is dm = (M/L) dx.
The x–coordinate of the centre of mass of the rod is
L
L 2
0 0
1 1 M 1 x L
X = x dm = x dx = =
M M L L 2 2
 
 
 
 
   
 
The y–coordinate is , Y = 1
Y = y dm =0
M 
and similarly Z = 0. The centre of mass is at L
,0,0
2
 
 
 
, i.e. at the middle point of the rod.
5.3 MOTION OF CENTRE OF MASS
Motion of the Center of Mass: Let us consider the motion of a system of n particles of individual masses
m1, m2, ……., mn and total mass M. It is assumed that no mass enters or leaves the system during its motion,
so that M remains constant. Then, as we have seen, we have the relation
1 1 2 2 n n
CM
1 2 n
m r m r ...... m r
r
m m ....... m
  

   

  
1 1 2 2 n n
m r m r ...... m r
M
  

  

Or CM 1 2
1 2 n
Mr m r m r ...... m
  
   
Differentiating this expression with respect to time t, we have
CM 1 2 n
1 2 n
dr dr dr dr
M m m ..... m
dt dt dt dt

 
 
 

  
Since, dr
dt


= velocity
Therefore, CM 1 2 n
1 2 n
Mv m v m v ..... m v
   
    … (i)
Or velocity of the Center of Mass is
1 2 n
1 2 n
CM
m v m v ..... m v
v
M
  
   

Or n
i
i
i 1
CM
m v
v
M





Further, mv

= momentum of a particle P


. Therefore, Eq. (i) can be written as
Page PAGE 82
AIEEE Chapter - 5 Rotational Motion
CM 1 2 n
P P P ..... P

 
 
 

    Or n
CM i
i 1
P P

 


 
Differentiating Eq. (i) with respect to time t, we get
CM 1 2 n
1 2 n
dv dv dv dv
M m m ...... m
dt dt dt dt
   
   
Or CM 1 2 n
1 2 n
Ma m a m a ...... m a
   
    … (ii)
Or 1 2 n
1 2 n
CM
m a m a ..... m a
a
M
  
   

Or n
1
1
i 1
CM
m a
a
M





Further, in accordance with Newton’s second law of motion F ma
 
 . Hence, Eq. (ii) can be written as
CM 1 2 n
F F F ..... F
   
    Or n
CM i
i 1
F F
 

 
Thus, as pointed out earlier also, the centre of mass of a system of particles moves as though it were a
particle of mass equal to that of the whole system with all the external forces acting directly on it.
Illustration 3: Two particles A and B of mass 1 kg and 2 kg respectively are projected in the directions shown
in figure with speeds uA = 200 m/s and uB = 50 m/s. Initially they wee 90 m apart. Find the maximum height
attained by the centre of mass of the particles. Assume acceleration due to gravity to be constant. (g = 10
m/s2
).
Solution: Using mArA = mBrB
Or (1)(rA) = (2)(rB)
Or rA = 2rB … (i)
And rA + rB = 90 m … (ii)
Solving these two equations, we get
rA = 60m and rB = 30m
i.e., CM is at height 60m from the ground at time t = 0.
90m
B
A
uB
uA
Further, A B
A B
CM
A B
m a m a
a
m m
 
 


= g = 10 m/s2
(downwards)
As A B
a a g
 
  (downwards)
A B
A B
CM
A B
m u m u
u
m m
 
 


     
1 200 2 50 100
m / s
1 2 3

 

(upwards)
Let, h be the height attained by CM beyond 60 m. Using
2 2
CM CM CM
v u 2a h
 
Or
  
2
100
0 2 10 h
3
 
 
 
 
Or  2
100
h 55.55m
180
 
Therefore, maximum height attained by the centre of mass is , H = 60 + 55.55 = 115.55 m
5.4 CONSERVATION OF LINEAR MOMENTUM
The product of mass and the velocity of a particle is defined as its linear momentum
 
P


. So, P mv

 

The magnitude of linear momentum may be written as
P = mv
Or 2 2 2 2
1
P m v 2m mv 2mK
2
 
  
 
 
Page PAGE 81 AIEEE
Chapter - 5 Rotational Motion
Thus, 2
P
P 2Km or K
2m
 
Here, K is the kinetic energy of the particle. In accordance with Newton’s second law,
 
d mv
dv dv
F ma m
dt dt dt

 
 
   
Thus, dP
F
dt




In case the external force applied to a particle (or a body) be zero, we have
dP
F 0 or P constant
dt


 

  
Showing that in the absence of an external force, the linear momentum of a particle (or the body) remains
constant. This is called the law of conservation of linear momentum. The law may be extended to a system
of particles or to the centre of mass of a system of particles. For example, for a system of particles it takes
the form.
If net force (or the vector sum of all the forces) on a system of particles is zero, the vector sum of linear
momentum of all the particles remain conserved, or
If 1 2 3 n
F F F F ..... F 0
    
     
Then, 1 2 3 n
P P P ..... P

 
 
 

    = constant
The same is the case for the centre of mass of a system of particles i.e., if
CM CM
F 0,P
 

 = Constant
Thus, the law of conservation of linear momentum can be applied to a single particle, to a system of particles
or even to the centre of mass of the particles.
Illustration 4: A man of mass m1 is standing on a platform of mass m2 kept on a smooth horizontal
surface. The man starts moving on the platform with a velocity vr relative to the platform. Find the recoil
velocity of platform.
Solution: Absolute velocity of man = r
   where  = recoil velocity of platform. Taking the platform
and the man as a system, net external force on the system in
horizontal direction is zero. The linear momentum of the system remains
constant. Initially both the man and the platform were at rest.
v
vr
- v
Hence, from law of conservation of momentum
0 = m1 (vr – v) – m2v
1 r
1 2
m v
v
m m
 

IMPULSE
Consider a constant force F

which acts for a time t on a body of mass m, thus, changing its velocity from u

to v

. Because the force is constant, the body will travel with travel with constant acceleration a

where
F ma
 
 and at v u
  
 
Hence, Ft
v u
m

 
 
Or Ft mv mu
  
 
The product of constant force F

and the time t for which it acts is called the impulse
 
J

of the force and this
is equal to the change in linear momentum which it produces.
Impulse  i
J F P P P
    
    
t f
Instantaneous Impulse: There are many occasions when a force acts for such a short time that the effect
is instantaneous, e.g., a bat striking a ball. In such cases, although the magnitude of the force and the time
for which it acts may each be unknown but the value of their product (i.e., impulse) can be known by
measuring the initial and final momenta. Thus, we can write
Page PAGE 82
AIEEE Chapter - 5 Rotational Motion
i
J F P P P
    
    
 dt f
Regarding the impulse it is important to note that impulse applied to an object in a given time interval can
also be calculated from the area under force–time (F – t) graph in the same time interval.
Impulse momentum relation: The change in momentum of a particle during a time interval equals the
impulse of the net force that acts on the particle during that interval.
2 1
J p p
  
  (impulse – momentum relation)
Illustration 5: A ball of mass m, travelling with velocity ˆ ˆ
2i 3j
 receives an impulse – 3m î . What is the
velocity of the ball immediately afterwards?
Solution: Using
f i
J m(v v )
  
 
–3m î = m
 
f
ˆ ˆ
v 2i 3j

 
 
 
Or
 
f
ˆ ˆ ˆ
v 3i 2i 3j

    Or
f
ˆ ˆ
v i 3j

  
5.5 VARIABLE MASS
In our discussion of the conservation of linear momentum, we have so far dealt with system whose mass
remains constant. We now consider those systems whose mass is variable, i.e., those in which mass enters
or leaves the system. A typical case is that of the rocket from which hot gases keep on escaping, thereby
continuously decreasing its mass.
In such problems you have nothing to do but apply a thrust force
 
t
F

to the main mass in addition to the all
other forces acting on it. This thrust force is given by,
t rel
dm
F v
dt
   
  
 
Here, rel
v

is the velocity of the mass gained or mass ejected relative to
the main mass. In case of rocket this is sometimes called the exhaust
velocity of the gases. dm
dt
is the rate at which mass is increasing or
decreasing.
m - dm
dm
r
v

v

v dv

 
system
The expression for the thrust force can be derived from the conservation of linear momentum in the absence
of any external forces on a system as follows :
Suppose at some moment t = t mass of a body is m and its velocity is v

. After some time at t = t + dt its mass
becomes (m = dm) and velocity becomes v dv
 
 . The mass dm is ejected with relative velocity r
v

, Absolute
velocity of mass ‘dm’ is therefore
 
r
v v dv
  
  . If no external forces are acting on the system, the linear
momentum of the system will remain conserved,
Or i f
P P

 

 Or     
r
mv m dm v dv dm v v dv
     
     
Or   
mv mv mdv dmv dm dv
    
      
r
dm v v dm dm dv
  
  
r
mdv v dm
 
  
Or
r
dv dm
m v
dt dt


   
 
 
   
 
Here,
 
1
dv
m thrust force F
dt


 

 
 
And dm
dt

= rate at which mass is ejecting
Problems related to variable mass can be solved in following three steps:
1. Make a list of all the forces acting on the main mass and apply them on it.
Page PAGE 81 AIEEE
Chapter - 5 Rotational Motion
2. Apply an additional thrust force t
F

on the mass, the magnitude of which is
r
dm
v
dt
  

 
 
and direction
is given by the direction of r
v

in case the mass is increasing and otherwise the direction of r
v

 if it is
decreasing.
3. Find net force on the mass and apply
net
dv
F m
dt



(m = mass at that particular instant)
Rocket Propulsion: Let m0 be the mass of the rocket at time t = 0. m its mass at any time t and v its velocity
a that the velocity of the rocket is u.
v = u
m = m0
At t = 0
m = m
v = v
At t = t
Exhaust velocity = vr
Further, let dm
dt

 
 
 
be the mass of the gas ejected per unit time and vr the exhaust velocity of the gases.
Usually dm
dt

 
 
 
and vr are kept constant throughout the journey of the rocket. Now, let us write few equations
which can be used in the problems of rocket propulsion. At time t = t,
1. Thrust force on the rocket
t r
dm
F v
dt

 
  
 
(upwards)
2. Weight of the rocket
W = mg (downwards)
3. New force on the rocket
Fnet = Ft – W (upwards)
Or
net r
dm
F v mg
dt

 
 
 
 
4. Net acceleration of the rocket F
a
m

Or r
v
dv dm
g
dt m dt

 
 
 
 
Or
r
dm
dv v gdt
m

 
 
 
 
Or
0
v m t
r
u m 0
dm
dv v g dt
m

 
  
Or 0
r
m
v u v In gt
m
 
  
 
 
Thus, 0
r
m
v u gt v In
m
 
    
 
… (i)
Illustration 6: (A) A rocket set for vertical firing weighs 50 kg and contains 450 kg of fuel. It can have
a maximum exhaust velocity of 2 km/s. What should be its minimum rate of fuel consumption
(i) to just lift it off the launching pad.
(ii) to give it an acceleration of 20 m/s2
.
(B) What will be the speed of the rocket when the rate of consumption of fuel is 10 kg/s after
whole of the fuel is consumed ? Take g = 9.8 m/s2
.
Solution: (A) To just lift it off the launching pad Weight = thrust force

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Physics-31.Rotational Mechanics

  • 1. 5.1 DEFINITION OF CENTRE OF MASS Centre of mass: Every physical system has associated with it a certain point whose motion characterizes the motion of the whole system. When the system moves under some external forces than this point moves as if the entire mass of the system is concentrated at this point and also the external force is applied at his point of translational motion. This point is called the centre of mass of the system. Centre of mass of system of n point masses or n particles is that point about which moment of mass of the system is zero, it means that if about a particular origin the moment of mass of system of n point masses is zero, then that particular origin is the centre of mass of the system. The concept of centre of mass is a pure mathematical concept. If there are n particles having mass m1, m2 ….m n and are placed in space (x1, y1, z1), (x2, y2, z2) ……….(x n, y n, z n) then centre of mass of system is defined as (X, Y, Z) where = 1 M i i i m x  Y = 1 M i i i m y  and Z = 1 M i i i m z  where M = i i m  is the total mass of X the system. Y Z X 1 1 1 1 m (x ,y ,z ) 0(x,y,z)              O Locate the point with coordinates (X, Y, Z). This point is called the centre of mass of the given collection of the particles. If the position vector of the i th particle is ri, the centre of mass is defined to have the position vector CM i i i 1 R = m r M    Where M = m1 + m2 + ……….. + mn = n i i 1 m   where i i i i r x i y j z k        1 1 1 1 2 2 2 2 CM n n n n m (x i+ y j + z k) + m (x i+ y j+ z k) +.. 1 R = m (x i+ y j + z k) M        1 1 2 2 n 1 1 2 2 n 1 1 2 2 n i(m x m x .. mx ) j(m y m y .. my ) .. 1 k(m z m z .. mz )                   M = i x cm + j y cm + k (z cm) n i i i=1 CM m r R = M    This is equation for centre of mass of n point masses. Illustration 1: Three equal masses are situates at vertices of equilateral triangle as shown in figure. Find centre of mass of the system. Solution: Let m be the mass of three masses and XCM, YCM and ZCM be the centre of masses of along the X axis, Y axis and Z axis, then 1 1 2 2 3 3 CM 1 2 3 m x m x m x X m m m            CM m 0 m a m a / 2 a X m m m 2       y m a x a a m m (0,0,0) (a,0,0)   a / 2, 3a / 2,0 1 1 2 2 3 3 CM 1 2 3 m y m y m y Y m m m            CM m 0 m 0 m 3a / 2 a Y m m m 2 3       1 1 2 2 3 3 CM 1 2 3 m z m z m z Z m m m     
  • 2. Page PAGE 81 AIEEE Chapter - 5 Rotational Motion Hence coordinate of centre of mass is a a , ,0 2 2 3       5.2 CENTRE OF MASS OF CONTINUOUS BODIES If we consider the body to have continuous distribution of matter, the summation in the formula of centre of mass should be replaced by integration. If x, y, z are the coordinates of this small mass dm, we write the coordinates of the centre of mass as 1 1 1 X = xdm,Y = y dm, Z = zdm M M M    y x O r  dm The integration is to be performed under proper limits so that as the integration variable goes through the limits, the elements cover the entire body. We illustrate the method with three examples. Illustration 2: Find centre of mass of a uniform straight rod of mass m and length l, Solution: Let M and L be the mass and the length of the rod respectively. Take the left end of the rod as the origin and the X–axis along the rod (figure). Consider an element of the rod between the positions A and B of the rod. Let the element be at a distance x from the centre O and its width be dx. So as x varies from 0 through L, the elements cover the entire rod. dx O x (x, o, o) x As the rod is uniform, the mass per unit length is M/L and hence the mass of the element is dm = (M/L) dx. The x–coordinate of the centre of mass of the rod is L L 2 0 0 1 1 M 1 x L X = x dm = x dx = = M M L L 2 2               The y–coordinate is , Y = 1 Y = y dm =0 M  and similarly Z = 0. The centre of mass is at L ,0,0 2       , i.e. at the middle point of the rod. 5.3 MOTION OF CENTRE OF MASS Motion of the Center of Mass: Let us consider the motion of a system of n particles of individual masses m1, m2, ……., mn and total mass M. It is assumed that no mass enters or leaves the system during its motion, so that M remains constant. Then, as we have seen, we have the relation 1 1 2 2 n n CM 1 2 n m r m r ...... m r r m m ....... m             1 1 2 2 n n m r m r ...... m r M         Or CM 1 2 1 2 n Mr m r m r ...... m        Differentiating this expression with respect to time t, we have CM 1 2 n 1 2 n dr dr dr dr M m m ..... m dt dt dt dt            Since, dr dt   = velocity Therefore, CM 1 2 n 1 2 n Mv m v m v ..... m v         … (i) Or velocity of the Center of Mass is 1 2 n 1 2 n CM m v m v ..... m v v M         Or n i i i 1 CM m v v M      Further, mv  = momentum of a particle P   . Therefore, Eq. (i) can be written as
  • 3. Page PAGE 82 AIEEE Chapter - 5 Rotational Motion CM 1 2 n P P P ..... P             Or n CM i i 1 P P        Differentiating Eq. (i) with respect to time t, we get CM 1 2 n 1 2 n dv dv dv dv M m m ...... m dt dt dt dt         Or CM 1 2 n 1 2 n Ma m a m a ...... m a         … (ii) Or 1 2 n 1 2 n CM m a m a ..... m a a M         Or n 1 1 i 1 CM m a a M      Further, in accordance with Newton’s second law of motion F ma    . Hence, Eq. (ii) can be written as CM 1 2 n F F F ..... F         Or n CM i i 1 F F      Thus, as pointed out earlier also, the centre of mass of a system of particles moves as though it were a particle of mass equal to that of the whole system with all the external forces acting directly on it. Illustration 3: Two particles A and B of mass 1 kg and 2 kg respectively are projected in the directions shown in figure with speeds uA = 200 m/s and uB = 50 m/s. Initially they wee 90 m apart. Find the maximum height attained by the centre of mass of the particles. Assume acceleration due to gravity to be constant. (g = 10 m/s2 ). Solution: Using mArA = mBrB Or (1)(rA) = (2)(rB) Or rA = 2rB … (i) And rA + rB = 90 m … (ii) Solving these two equations, we get rA = 60m and rB = 30m i.e., CM is at height 60m from the ground at time t = 0. 90m B A uB uA Further, A B A B CM A B m a m a a m m       = g = 10 m/s2 (downwards) As A B a a g     (downwards) A B A B CM A B m u m u u m m             1 200 2 50 100 m / s 1 2 3     (upwards) Let, h be the height attained by CM beyond 60 m. Using 2 2 CM CM CM v u 2a h   Or    2 100 0 2 10 h 3         Or  2 100 h 55.55m 180   Therefore, maximum height attained by the centre of mass is , H = 60 + 55.55 = 115.55 m 5.4 CONSERVATION OF LINEAR MOMENTUM The product of mass and the velocity of a particle is defined as its linear momentum   P   . So, P mv     The magnitude of linear momentum may be written as P = mv Or 2 2 2 2 1 P m v 2m mv 2mK 2         
  • 4. Page PAGE 81 AIEEE Chapter - 5 Rotational Motion Thus, 2 P P 2Km or K 2m   Here, K is the kinetic energy of the particle. In accordance with Newton’s second law,   d mv dv dv F ma m dt dt dt          Thus, dP F dt     In case the external force applied to a particle (or a body) be zero, we have dP F 0 or P constant dt         Showing that in the absence of an external force, the linear momentum of a particle (or the body) remains constant. This is called the law of conservation of linear momentum. The law may be extended to a system of particles or to the centre of mass of a system of particles. For example, for a system of particles it takes the form. If net force (or the vector sum of all the forces) on a system of particles is zero, the vector sum of linear momentum of all the particles remain conserved, or If 1 2 3 n F F F F ..... F 0            Then, 1 2 3 n P P P ..... P             = constant The same is the case for the centre of mass of a system of particles i.e., if CM CM F 0,P     = Constant Thus, the law of conservation of linear momentum can be applied to a single particle, to a system of particles or even to the centre of mass of the particles. Illustration 4: A man of mass m1 is standing on a platform of mass m2 kept on a smooth horizontal surface. The man starts moving on the platform with a velocity vr relative to the platform. Find the recoil velocity of platform. Solution: Absolute velocity of man = r    where  = recoil velocity of platform. Taking the platform and the man as a system, net external force on the system in horizontal direction is zero. The linear momentum of the system remains constant. Initially both the man and the platform were at rest. v vr - v Hence, from law of conservation of momentum 0 = m1 (vr – v) – m2v 1 r 1 2 m v v m m    IMPULSE Consider a constant force F  which acts for a time t on a body of mass m, thus, changing its velocity from u  to v  . Because the force is constant, the body will travel with travel with constant acceleration a  where F ma    and at v u      Hence, Ft v u m      Or Ft mv mu      The product of constant force F  and the time t for which it acts is called the impulse   J  of the force and this is equal to the change in linear momentum which it produces. Impulse  i J F P P P           t f Instantaneous Impulse: There are many occasions when a force acts for such a short time that the effect is instantaneous, e.g., a bat striking a ball. In such cases, although the magnitude of the force and the time for which it acts may each be unknown but the value of their product (i.e., impulse) can be known by measuring the initial and final momenta. Thus, we can write
  • 5. Page PAGE 82 AIEEE Chapter - 5 Rotational Motion i J F P P P            dt f Regarding the impulse it is important to note that impulse applied to an object in a given time interval can also be calculated from the area under force–time (F – t) graph in the same time interval. Impulse momentum relation: The change in momentum of a particle during a time interval equals the impulse of the net force that acts on the particle during that interval. 2 1 J p p      (impulse – momentum relation) Illustration 5: A ball of mass m, travelling with velocity ˆ ˆ 2i 3j  receives an impulse – 3m î . What is the velocity of the ball immediately afterwards? Solution: Using f i J m(v v )      –3m î = m   f ˆ ˆ v 2i 3j        Or   f ˆ ˆ ˆ v 3i 2i 3j      Or f ˆ ˆ v i 3j     5.5 VARIABLE MASS In our discussion of the conservation of linear momentum, we have so far dealt with system whose mass remains constant. We now consider those systems whose mass is variable, i.e., those in which mass enters or leaves the system. A typical case is that of the rocket from which hot gases keep on escaping, thereby continuously decreasing its mass. In such problems you have nothing to do but apply a thrust force   t F  to the main mass in addition to the all other forces acting on it. This thrust force is given by, t rel dm F v dt          Here, rel v  is the velocity of the mass gained or mass ejected relative to the main mass. In case of rocket this is sometimes called the exhaust velocity of the gases. dm dt is the rate at which mass is increasing or decreasing. m - dm dm r v  v  v dv    system The expression for the thrust force can be derived from the conservation of linear momentum in the absence of any external forces on a system as follows : Suppose at some moment t = t mass of a body is m and its velocity is v  . After some time at t = t + dt its mass becomes (m = dm) and velocity becomes v dv    . The mass dm is ejected with relative velocity r v  , Absolute velocity of mass ‘dm’ is therefore   r v v dv      . If no external forces are acting on the system, the linear momentum of the system will remain conserved, Or i f P P      Or      r mv m dm v dv dm v v dv             Or    mv mv mdv dmv dm dv             r dm v v dm dm dv       r mdv v dm      Or r dv dm m v dt dt                 Here,   1 dv m thrust force F dt          And dm dt  = rate at which mass is ejecting Problems related to variable mass can be solved in following three steps: 1. Make a list of all the forces acting on the main mass and apply them on it.
  • 6. Page PAGE 81 AIEEE Chapter - 5 Rotational Motion 2. Apply an additional thrust force t F  on the mass, the magnitude of which is r dm v dt         and direction is given by the direction of r v  in case the mass is increasing and otherwise the direction of r v   if it is decreasing. 3. Find net force on the mass and apply net dv F m dt    (m = mass at that particular instant) Rocket Propulsion: Let m0 be the mass of the rocket at time t = 0. m its mass at any time t and v its velocity a that the velocity of the rocket is u. v = u m = m0 At t = 0 m = m v = v At t = t Exhaust velocity = vr Further, let dm dt        be the mass of the gas ejected per unit time and vr the exhaust velocity of the gases. Usually dm dt        and vr are kept constant throughout the journey of the rocket. Now, let us write few equations which can be used in the problems of rocket propulsion. At time t = t, 1. Thrust force on the rocket t r dm F v dt         (upwards) 2. Weight of the rocket W = mg (downwards) 3. New force on the rocket Fnet = Ft – W (upwards) Or net r dm F v mg dt          4. Net acceleration of the rocket F a m  Or r v dv dm g dt m dt          Or r dm dv v gdt m          Or 0 v m t r u m 0 dm dv v g dt m       Or 0 r m v u v In gt m          Thus, 0 r m v u gt v In m          … (i) Illustration 6: (A) A rocket set for vertical firing weighs 50 kg and contains 450 kg of fuel. It can have a maximum exhaust velocity of 2 km/s. What should be its minimum rate of fuel consumption (i) to just lift it off the launching pad. (ii) to give it an acceleration of 20 m/s2 . (B) What will be the speed of the rocket when the rate of consumption of fuel is 10 kg/s after whole of the fuel is consumed ? Take g = 9.8 m/s2 . Solution: (A) To just lift it off the launching pad Weight = thrust force
  • 7. Page PAGE 82 AIEEE Chapter - 5 Rotational Motion Or r dm mg v dt         Or r dm mg dt v         Substituting the values, we get,    3 450 50 9.8 dm 2.45kg/ s dt 2 10            (ii) Net acceleration a = 20 m/s2 ma = Ft – mg Or t F a g m   Or r v dm a g m dt          This gives   r m g a dm dt v          Substituting the values, we get,    3 450 50 9.8 20 dm 7.45kg/ s dt 2 10             (B) The rate of fuel consumption is 1.0 kg/s. So, the time for the consumption of entire fuel is, 450 t 45 10   second Using Eq. (i), i.e., 0 r m v u gt v In m          Here, u = 0, vr = 2  103 m/s, m0 = 500 kg and m = 50 kg Substituting the values, we get, 3 500 v 0 (9.8)(45) (2 10 )In 50           Or v = – 441 + 4605.17 Or v = 4164.17 m/s Or v = 4.164 km/s 5.6 COLLISIONS When two particles approach each other, their motion changes or their momentum changes due to their mutual interactions. This phenomenon is called collision. During collision (i) an impulse (a large force for a relatively short time) acts on each colliding particle (ii) the total momentum of the particles remain conserved. The collision is infact a redistribution of total momentum of the particles. Physical interaction is not necessary for collision. Generally, the collisions are of two types: (1) Elastic collision (2) Inelastic collision 1. Elastic collision: A collision is said to be elastic if kinetic energy is also conserved along with the linear momentum. There is no loss or transformation of kinetic energy in this collision Consider two particles of masses m1 and m2 moving with velocities 1 u   and 2 u   respectively. They collide and their velocities after the collision become 1 v   and 2 v   respectively. By conservation of the linear momentum, Total linear momentum before collision = Total linear momentum after collision , 1 1 2 2 1 1 2 2 m u + m u = m v + m v         or     1 1 1 2 2 2 m u - v = m v - u         … (1) By conservation of kinetic energy, Total kinetic energy before collision = Total kinetic energy after collision 2 2 1 1 2 2 1 1 m u + m u 2 2 = 2 2 1 1 2 2 1 1 m v + m v 2 2 or     2 2 2 2 1 1 1 2 2 2 m u - v = m v - u … (2) ' 1 u ' 2 u 1 u 2 u v 2 v 1 v ' 1 v ' 2 v          During collision Long before collision Long after collision and only for one dimension collision or head on collision On solving the above equation (1) and (2),   1 2 1 2 u - u = - v - v         … (3)
  • 8. Page PAGE 81 AIEEE Chapter - 5 Rotational Motion i. e. Relative velocity before collision = Relative velocity after collision. Thus in elastic collision, the relative velocity of approach of particles before collision is equal to the relative velocity with which the particles recedes after collision. i.e. the magnitude of relative velocity remains the same, but the direction is reversed. On solving the equation (1) and (3) we get 1 2 2 1 1 2 1 2 1 2 m - m 2m v = u + u m +m m + m             and 1 2 1 2 1 2 1 2 1 2 2m m - m v = u + u m +m m + m             Illustration 7: A ball of 0.4kg mass and a speed of 3 m/s has a head-on, completely elastic collision with a 0.6-kg mass initially at rest. Find the speeds of both bodies after the collision: Solution: By Conservation of momentum: (0.4  3) + 0 = 0.4v + 0.6V or v + 1.5V = 3 We know that: velocity of separation = –velocity of approach or –v + V = 3 We solve by adding the two equations to yield 2.5V = 6 V = 2.4 m/s v = –0.6 m/s 2. Inelastic collision: A collision in which the linear momentum is conserved, but a part of kinetic energy change into the other forms (such as heat, vibration, excitation energy etc) is called the inelastic collision. In other words, the kinetic energy is not conserved.(Actually there is no violation of the law of conservation of energy, but a part of the kinetic energy, changes into a useless form.) In this, the particles do not regain their shape and size completely after collision. Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. However, in the absence of external forces, law of conservation of linear momentum still holds good. Let two particles of masses m1 and m2 moving with initial velocities 1 u   and 2 u   Collide and travel with velocities 1 v   and 2 v   respectively after the collision. By conservation of linear momentum     1 1 2 2 1 1 2 2 1 1 1 2 2 2 m u + m u = m v + m v or m u - v = m v - u                      and by conservation of kinetic energy, 2 2 1 1 2 2 1 1 m u m u 2 2  = 2 2 1 1 2 2 1 1 m v m v 2 2  + E Where E is the part of energy which changes into the useless form due to inelastic collision; For example, if two particles coalesce after collision and the combined system travel with a velocity v  after the collision in same direction, then     1 1 2 2 1 1 2 2 1 2 1 2 m u m u m u m u m m v or v m m       And loss in kinetic energy,   2 2 2 1 1 2 2 1 2 1 1 1 E = m u m u m m v 2 2 2    or     2 2 2 1 1 2 2 1 1 2 2 1 2 m u m u 1 1 E = m u m u 2 2 m m           2 1 2 1 2 1 2 m m 1 (u u ) 2 (m m )    5.7 CONCEPT OF COEFFICIENT OF RESTITUTION When two bodies collide head–on, the ratio of their relative velocities after collision and their relative velocities before collision is called the coefficient of restitution e. Thus 2 1 1 2 v - v e = u - u     … (i) Or seperation speed e approach speed  (along line of impact) … (ii) Where v1 = The speed of first body after collision. v2 = The speed of second body after collision. u1 = the speed of first body before collision.
  • 9. Page PAGE 82 AIEEE Chapter - 5 Rotational Motion u2 = the speed of second body before collision. The ratio e is called the coefficient of restitution and is constant for two particular objects. In general, 0 e 1   e = 0, for completely inelastic collision, as both the objects stick together. So, their separation speed is zero or e = 0. e = 1, for an elastic collision, as we can show that from equation (i) v2 – v1 = u1 – u2 … (iii) Or separation speed = approach speed Or e = 1 Let us now find the velocities of two particles after collision if they collide directly and the coefficient of restitution between them is given as e. v2 v1 Before Collision m2 m1 v2 ' v1 ' After Collision Applying conservation of linear momentum, ' ' 1 1 2 2 1 1 2 2 m v m v m v m v    … (iv) Further, separation speed = e (approach speed) or u1 – u2 = e (v2 – v1) … (v) Solving equations (iv) and (v) we get , ' 1 2 1 2 1 1 2 1 2 1 2 m em m em v v v m m m m                   … (vi) and ' 2 1 1 1 2 2 1 1 2 1 2 m em m em v v v m m m m                   … (vii) Illustration 8: A moving particle of mass m, makes a head–on collision with a particle of mass 2m, which is initially at rest. Show that the colliding particle loses (8/9) of its energy after collision. Solution: Let u be the initial velocity of particle of mass m and ν its velocity after the collision. Let V be the velocity of particle of mass 2m after the collision. From the principle of conservation of linear momentum, we have mu = mv + (2m) V Or u – v = 2 V … (i) The conservation of kinetic energy gives 1 2 mu2 = 1 2 mv2 + 1 2 (2m)V2 Or u2 – v2 = 2V2 … (ii) Or (u – v) (u + v) = 2V2 Using Eq. (1) in Eq. (2) we have 2V (u + v) = 2V2 or u + v = V Or 2(u + v) = 2V … (iii) Comparing (1) and (3) we get, u – v = 2 (u + v) Or v = u 3  … (iv) Now, initial kinetic energy of the colliding mass is, Ki = 1 2 mu2 Final kinetic energy, Kf = 1 2 mv2 Loss in kinetic energy is, ΔK = Ki – Kf = 1 2 mu2 – 1 2 mv2 ∴ Fractional loss = i K K = 2 2 2 1 1 mu - mv 2 2 1 mu 2 = 2 2 2 u - v u = 1 – 2 v u       = 1 – 2 1 3        = 8 9 ( v = – u/3)
  • 10. Page PAGE 81 AIEEE Chapter - 5 Rotational Motion 5.8 OBLIQUE COLLISION When two bodies collide such that there velocities are not in the line of action of contact force then the collision is known as oblique collision. We can divide also oblique collision in elastic, and inelastic collision. It is important to notice that momentum will conserve in each direction curing oblique collision also. When two bodies collide obliquely, their relative velocity resolved along their common normal after impact is in a constant ratio to their relative velocity before impact (resolved along common normal) and is in the opposite direction. 1 2 1 2 v cos v cos e u cos u cos           1 2 1 2 v cos v cos e u cos u cos         common normal u2 u1     v2 v1 Illustration 9: A ball of mass m moving at a speed v makes a head on collision with an identical ball at rest. The energy of the balls after the collision is 3/4th of the original. Find the coefficient of restitution. v2 v1 Before Collision m2 m1 v2 ' v1 ' After Collision Solution: As we have seen in the above discussion, that under the given conditions : ' ' 1 2 1 e 1 e v v and v v 2 2                 Given that f i 3 K K 4  Or 2 ' 2 1 1 3 1 mv mv 2 4 2        Substituting the value, we get, 2 2 1 e 1 e 3 2 2 4                 Or (1 + e)2 + (1 – e)2 = 3 Or 2 + 2e2 = 3 Or 2 1 e 2  Or 1 e 2  5.9 RIGID BODY A rigid body is a body with a definite and unchanged shape and size i.e. a body is said to be rigid if the distance between any two particles of the body remains invariant. MOTION OF A RIGID BODY Translational: If a body moves such that its orientation does not change with respect to time then body is said to move in translational motion. ROTATION: If a body is rotating about the fix axis of rotation then its motion is known as pure rotational motion. MIX – MOTION: If a body moves such that its motion neither be pure rotational nor be pure translational then its motion is known as mix motion. 5.10 ANGULAR DISPLACEMENT
  • 11. Page PAGE 82 AIEEE Chapter - 5 Rotational Motion Consider a rigid body undergoing rotation about an axis, perpendicular to the plane of the paper and passing through O. Suppose that A and B are any two particles of the rigid body at the position 1 while A′ and B′ are their subsequent locations when the body is at the position 2. Since the body undergoes rotation, OA = OA′ and OB = OB′ Further AB = A′B′, since the body is rigid. ⇒ ΔOAB ≅ ΔOA′B′(congruent) O A B B 2 1 A i.e. ∠ AOB = ∠A′OB′ Adding ∠ AOB′ to both sides of the above equation, we get ∠ BOB′ = ∠AOA′ = θ (say) This implies that in a given interval of time the angular displacements of all particles of the rigid body undergoing rotation are identical. Therefore, a single variable, viz. angular displacement (θ) can be used to describe the rotational motion of the rigid body. Angular displacement is not a vector quantity 5.11 ANGULAR VELOCITY() The rate of change of angular displacement with respect to time is known as angular velocity. Average angular velocity: The rate of change of angular displacement with respect to time is known as angular velocity. To define average angular velocity it is necessary to specify the interval in which we are talking about average. ω = 2 1 2 1 t t     Instantaneous angular velocity: Instantaneous velocity means angular velocity at a particular instant. It is mathematically defined as , 2 1 2 1 1 t t 2 1 d lim at t t t t dt              5.12 ANGULAR ACCELERATIONS ( ) Angular accelerations (α): The rate of change of angular velocity with respect to time is known as angular acceleration. Average angular acceleration: It is necessary to specify the time interval to in which we are talking about average. In a given time interval t1 to t2 the average angular acceleration is defined as , 2 1 2 1 t t          Instantaneous angular acceleration: Instantaneous angular acceleration means angular acceleration at a particular dot instant at t = t1 mathematically it is defined as, 1 2 2 2 1 t t 2 1 t t d limit t t dt                 Direction of angular acceleration: If magnitude of ω increasing then direction of α will be same as direction of ω and vice–versa. 5.13 EQUATION OF ANGULAR MOTION ω(t) = ωo + αt θ(t) = θo + ωot + 1 2 αt2 ω   2 t = 2 0 2    Here ωo = magnitude of the initial angular velocity
  • 12. Page PAGE 81 AIEEE Chapter - 5 Rotational Motion ω(t) = magnitude of the angular velocity after time t. θo = Initial angular position. θ(t) = Angular position after time t. Illustration 10: A disc starts rotating with constant angular acceleration of π radian/s2 about a fixed axis perpendicular to its plane and passing through its centre. Find (a) The angular velocity of the disc after 4 sec. (b) The angular displacement of the disc after 4 sec and (c) Number of turns accomplished by the disc in 4 sec. Solution: Here = rad/sec2 ω0 = 0 t = 4 sec (a) ω(4 sec) = 0 + (π rad/sec2 ) × 4 sec = 4π rad/sec. (b) θ(4 sec) = 0 + 1 2 (π rad/sec2 ) × (16 sec2 ) = 8 π radian. (c) Let the number of turns be n ⇒ n 2 π rad = 8 π rad ⇒ n = 4 5.14 RELATION BETWEEN LINEAR AND ANGULAR VARIABLES Consider a particle A of a rigid body undergoing rotation about a fixed axis-n̂ , the particle A describing an arc ABA′ of a circle with its centre O on the axis of rotation. Taking the origin at O, the position vector of A, r  ≡ OA  . OA = OA′ = constant (radius of the circle) ∠A′OA = θ (t) (say) The arc length, ABA′, S = rθ O A A B  2 1 n The tangential velocity, vA = dS d r dt dt   = rω The direction of the angular velocity vector   be taken along the axis of rotation:   = ωn̂ , n̂ being the unit vector along the axis of rotation. Then, v  , instantaneous velocity of A with respect to the axis of rotation, can be written as , v r       The acceleration of the point A with respect to the axis of rotation is ⇒ a  = dv d dr r dt dt dt            ⇒ a r v            If   is constant, then   = 0 and a v       =   ×  r     = − ω 2 r  5.15 TORQUE Torque of a force about the axis of rotation: The turning effect of a force about the axis of rotation is called moment of force or torque due to the force and is measured as the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the axis of rotation. O P F r Fig. (a) It is denoted by the Greek letter τ (tou). Thus, torque = force × perpendicular distance from the axis of rotation Its unit is N m in SI. Its dimensional formula is [M L2 T–2 ], same as that of work. It may be pointed out that no doubt, N m is equivalent to joule (the unit of work); but joule is not used as the unit of torque. Consider a force F  acting on a particle P. Choose an origin O and let r  be the position vector of the particle experiencing the force. We define the torque of the force F about O as , P r F       … (1) This is a vector quantity having its direction perpendicular to r  and F  according to the rule of cross product. Now consider a rigid body rotating about a given axis of rotation AB (fig b). Let F be a force acting on the particle O of the body. F may not be in the plane ABP. Take the origin O somewhere on the axis of rotation. A B O1 O P r F Fig. (b)
  • 13. Page PAGE 82 AIEEE Chapter - 5 Rotational Motion The torque of F about O is P r F       . Its component along OA is called the torque of F  about OA. To calculate it, we should find the vector r F    and then find out the angle θ it makes with OA. The torque about OA is then r F cos     . The torque of a force about a line. This can be shown as given below. Let O1 be any point on the line AB (Figure 10.4). The torque of F about O1 is 1 1 1 O P F (O O OP) F O O F OP F                  . As 1 1 O P F O O,      this term will have no component along AB. Thus, the component of 1 O P F    is equal to that of 1 O P F    . Illustration 11: A particle of mass m is dropped at point A, find the torque about O. Solution:    r F        r F sin n̂ = (r sin) Fn̂   = b mg The direction of torque is directed inward the paper or in other words, rotation about O is clockwise.  r mg O m A b 5.16 KINETIC ENERGY OF A RIGID BODY Let a rigid body is purely rotating about an axis AB with angular velocity ω consider a general particle Δm2 which is at a distance of r2 from axis of rotation. V2 = r2.ω So energy associated with this Δm2 is ΔK · E2 ΔK · E2 = 2 2 2 2 2 2 2 1 1 m v m r 2 2     K·Etotal = 2 2 2 2 2 2 2 2 2 1 1 k·E m ·r m r 2 2           K·Etotal = 2 1 I 2  A  2 r 1 m  Where I is called as moment of inertia of body about an given axis of rotation. In this case I is about AB. Moment of inertia is also called as rotational mass of object. Illustration 12: Four point masses each of value m are placed at points A, B, C and D at distances a 3a ,a, 2 2 and 2a from the free and of a mass-less rod. (i) Find the M.I. of the system about an axis perpendicular to the point B. (ii) Will the mass on the left side contribute a negative term? Solution: Suppose that the four masses each of mass m are placed at the points A, B, C and D of a mass-less rod (fig.) (i) M.I. of the rod about the axis through the point B, I = m (BA)2 + m (BB)2 + m (BC)2 + m (BD)2 = m (a/2)2 + m (0)2 + m (a/2)2 + m (a)2 = 2 2 2 2 1 1 3 ma 0 ma ma ma 2 4 4     (ii) The distance occurs with power 2 in the formula for moment of inertia. Therefore, if the distance of the mass on left of the axis is taken as negative, its moment of inertia will still contribute a positive term. Hence, the mass m on the left side of the axis will not contribute a negative term to the moment of inertia of the system. a/2 3a/2 2a m m m m AXIS A B C D 5.17 MOMENT OF INERTIA
  • 14. Page PAGE 81 AIEEE Chapter - 5 Rotational Motion Like the centre of mass, the moment of inertia is a property of an object that is related to its mass distribution. The moment of inertia (denoted by I) is an important quantity in the study of system of particles that are rotating. The role of the moment of inertia in the study of rotational motion is analogous to that of mass in the study of linear motion. Moment of inertia gives a measurement of the resistance of a body to a change in its rotational motion. If a body is at rest, the larger the moment of inertia of a body, the more difficult it is to put that body into rotational motion. Similarly, the larger the moment of inertia of a body, the more difficult it is to stop its rotational motion. Moment of Inertia of a Single Particle: For a very simple case the moment of inertia of a single particle about an axis is given by, I = mr2 Here, m is the mass of the particle and r its distance from the axis under consideration. m r Moment of Inertia of a System of Particles The moment of inertia of a system of particles about an axis is given by : 2 i i i I m r   m1 r1 m2 r2 m3 r3 Where ri is the perpendicular distance from the axis to the ith particle, which has a mass mi. Moment of Inertia of Rigid Bodies: For a continuous mass distribution such as found in a rigid body, we replace the summation of Eq. (ii) by an integral. If the system is divided into infinitesimal elements of mass dm and if r is the distance from a mass element to the axis of rotation, the moment of inertia is, 2 I r dm   Where the integral is taken over the system. r dm Radius of Gyration: Radius of gyration may be defined as the distance from the axis at which, if the whole mass of the body were to be concentrated, the moment of inertia would be the same about the given axis as with its actual distribution of mass. Suppose a rigid body consists of n particles of each of the mass m. Let r1, r2, ...... rn be the perpendicular distances of these particles from the axis of rotation. Then 2 2 2 1 2 n I mr mr ..... mr     2 2 2 1 2 n m r r ..... r         2 2 2 1 2 n r r ..... r m n n            2 2 2 1 2 n r r ..... r M n           (where M = m × n) … (i) If whole mass of the body is regarded to be concentrated at a perpendicular distance K (radius of gyration), then I = M K2 … (ii) From eqs. (1) and (2), 1 2 2 2 2 1 2 n r r ..... r K n           … (iii) Therefore, radius of gyration of a body about an axis is equal to the root mean square distance of the constituent particles from the given axis. THEOREM OF PERPENDICULAR AXIS Illustration 13: The moment of inertia of a thin square plate ABCD (as shown in figure)