Chapter 7
1

F21

F12

F13
2

F23
m1g
m2g

F31

F32
3
m3g
 
Internal forces : F12 , F32 etc.
External forces : The weights of the
particles
In the absence of interaction among the
particles, the problem is rather simple.
One can solve the motion of each particle
of the system separately.
In the presence of interaction, the motion
of the system gets enormously
complicated
With gravitational interaction, the motion
of a three-body system in unsolvable.
Center of mass of a system of particles
m1

r1

r2

rcm
rN
m2
mN

 mi ri
N




m1 r1  m 2 r2  .......  m N rN
rcm 

m1  m 2  .......  m N
i 1
M
The three coordinates of the center of
mass are :
x cm 
m x
i
i
M
i
; y cm 
m y
i
i
M
i
; z cm 
m z
i i
i
where, xi, yi & zi are the coordinates of
the ith particle.
M
Center of mass of a solid body
The body is divided into small masses m i ,
each of which is treated as a point particle.
So,
x cm
1

x i mi

M i
mi
In the limit that each mi goes to zero,
x cm
1
1

lim  x i mi   xdm
M m0 i
M
Similarly,
y cm
1

ydm

M
Thus,

1 
rcm   r dm
M
& z cm
1

zdm

M
1. One dimensional body (along x-axis)
dm  dl ,   mass per unit length
2. Two dimensional body (Lamina)
dm   da ,   mass per unit area
3. Three dimensional body (Solids)
dm   dv ,   mass per unit volume
Prob. 7.6
Find the centre of mass of a homogeneous
semicircular plate of radius R.
y
Soln:
R
da  r drd
dr
r

x
x  r cos  ; y  r sin 
 x da  r dr cos  d & y da  r dr sin  d
2
2

 2
x cm 
r dr  cos d  0

M0
0
R
y cm

 2

r dr  sin  d

M0
0
R
3
2
R
4R


2
2
R
3
3
Overall Translational Motion of a System of
Particles
The Centre of Mass of a system of
particles moves according to Newton’s
law, as though the entire mass of the
system were concentrated at it and the net
external force were applied to it.
Proof

Mrcm 

 mi ri
i

 Mvcm 

 mi vi
i




Ma cm   mi a i   Fi
i

  Fi ,ext 
i
i

 Fi ,int
i


Now,  Fi ,int  0 ,
i
as, internal forces occur in pairs


(for the force Fnk , there is a force Fkn )
and the sum of each pair is zero, from
Newton’s third law.

 Ma cm 


 Fi,ext  Ftot, ext
i
Principle of momentum conservation for a
system of particles

We have , M v cm 

 mi vi 
i


 Pi  P
i

where P is the total momentum of the system
That is, the total linear momentum of a
system of particles is the total mass times
the velocity of center of mass.
Differentiating w.r.t time,



dP
 Ma cm  Ftot, ext
dt
That is, the rate of change of total
momentum is the net external force acting
on the system

In the absence of external forces,
the total momentum of a system of
particles is conserved.
Double Integration in Cartesian Coordinates
y2
y2 x 2
  f (x, y) dxdy ??
f(x,y)
y1 x1
y1
x1
x2
ijth cell
f ij
j
y
x
Integral  lim lim
x 0 y 0
i
 f xy
ij
j
i
Summing along x-axis
x2
g( y)   f ( x, y) dx
x1
y
Integration along y-axis
y2
y1
x 2

f ( x , y) dxdy   g( y) dy   dy   f ( x, y) dx 
 x1

y1
y1
y1 x1
y2 x 2

y2
y2
y
R 2  y2
dy
R
y
x


x cm    xdxdy 
dy

M
M0 
R

y cm 
ydy

M0

R
R 2 y2
4R
dx

2 2 3
R y
R 2 y2
 xdx  0
R 2 y2
Double Integration in Polar Coordinates
y
da  r drd
d
r
dr
r d

x
  2
R2
D
R1
  1
fda


D
R2
2
R1
1
r
dr
f
(
r
,

)
d

 
Prob. 7.3
a) A uniform flexible chain of length L,
with mass per unit length λ, passes over
a small, frictionless peg. It is released
from a rest position with a length of
chain x hanging from one side and a
length L-x from the other side. Find the
acceleration as a function of x.
b) Find the time taken by the chain to fall
off the peg
T
y
x
Equations of motion :
yg
T
 xg
i) xg  T  xa
ii ) T  yg  ya
Adding the two equations :
2xg  Lg  La
Or,
 2x

a  g
 1
 L

dv dv dx
dv
Since a 

v ,
dt dx dt
dx
 2x

v dv  g 
 1 dx
 L

Integrating,
2
2


x  x0
v
 g 
 ( x  x 0 ) 
2
 L

2

g
2
2

( x  L / 2)  ( x 0  L / 2)
L



g 2

z  z 02 , where z  x  L / 2 & z 0  x 0  L / 2
L
dx dz
Putting v 

,
dt
dt
dy

dt
2g 2 2
z  z0
L
Integrating,
T
 dt 
0
Or, T 
L
2g
L
2g
L/ 2

y0
dz
z z
2
2
0


L
1 

cosh 
 2( x 0  L / 2) 

1.
T   as x 0  L / 2
2. L = 5m, x0 = 3L/4
1
1
T  cosh 2  0.66 s
2
x0 = 2.51m, T = 3.1s
T+dT
N
d
R
T
w

T2
T1
w  Rgd 
The net tangential force on the element is :
dT  w cos   dT  Rg cos d
Equating this force to mass of the element
times the tangential acceleration,
dT  R (g cos   a ) d
Integrating from one end of the chain to the other,
T2  T1   Ra   a
Where l is the length of the portion of the rope
that passes over the peg.
In the limit that l goes to zero, T1 = T2
Exercise :
Show that the exact acceleration of the rope is

 2x
a  g
1 
L
 L
y  y / 2
y


L
2L
2
y cm
y
CM
y dy
v cm 
L dt
2
2 

1   dy 
d y
a cm     y 2
L   dt 
dt 
1 2 2
1 3 2 2
1
2
 g t  g( L  2 gt  ( 2 g t  gL )
L
L


 N  Lg  La cm  g t  Lg
3
2
2 2
 N  g t  3xg
3
2
2 2
SYSTEMS OF VARIABLE MASS
1. Rocket
For the entire system S’ :


Initial momentum  Pi  Mv
Final momentum =




Pf  ( M  M)( v  v)  Mu
  

 

 P  Pf  Pi  Mv  ( v  u )M  Mv

 Fext


P
dv   dM
 lim
 M  (v  u)
t 0 t
dt
dt
 
 dM
dv
Or, M
 Fext  v rel
dt
dt

 
where , v rel  u  v
2. Conveyor Belt

v

u
M
M
Time t


v  v
M  M
Time t  t





P  M  M v  v   Mv  Mu

 
 Mv  Mv  u 
 
 dM
dv
 M
 Fext  v rel
dt
dt
Rocket Motion
dv
dM
M
  Mg  v rel
dt
dt
I
Dividing by M and integrating over
time,
v2
t2
M2
dM
v dv   t gdt  v rel M M
1
1
1
Or,
v 2  v1   g( t 2  t1 )  v rel ln
N
D
I
A
 
M2
M1
Ex. 22
During a lunar mission, it is necessary to make
a mid-course correction of 22.6 m/s in the speed
of the spacecraft, which is moving at 388 m/s.
The exhaust speed of the rocket engine is 1230
m/s. What fraction of the initial mass of the
rocket must be discarded as exhaust ?
Ans. We have here,
v 2  v1  22.6 m / s, Fext  0, v rel 1230 m / s
 ln
M
 2
M
 1




22.6

  0.0183
1230
M2
 1
 0.0182
M1
Prob. 12 A flexible in extensible string of
length L is threaded into a smooth tube. The
tube contains a right-angled bend and is
positioned in the vertical plane so that one arm
is vertical and the other is horizontal. Initially a
length y0 of the string hangs in the vertical arm.
The string is released and slides through the
tube. a) Show that in terms of the variable mass
problem vrel = 0 so
that
the
equation
of
motion
2
d
of the string is y2  gy. b) Show that
dt

y0
y( t ) 
e
2
is the solution
L
g/L t
 e
g/L t

L-y
N
T
y
T
mg
(M-m)g
Subsystem II
Subsystem I
Observation : Masses are added to each
subsystem with a relative velocity of zero
Subsystem I :
dv
m
 Fext  mg  T
dt
dv
Subsystem II : ( M  m)
T
dt
Adding the above two equations :
dv
M
 mg
dt
Or,
dv m
gy

g
dt
M
L
2
d y gy
Or,

2
dt
L
y 0  g / Lt


b. y  
e
e
 2 

g / Lt

2
d y gy
 2 
dt
L
Moreover,
dy
y(0)  y 0 &
(0)  0
dt
c.
y(t)
y0
t
Prob. 11 Two long barges are floating in the
same direction in still water, one with speed of
9.65 km/hr and the other with a speed of 21.2
km/hr. While they are passing each other, coal
is shoveled from the slower to the faster barge
at a rate of 925 kg/min; how much force must
be provided by the engine of each barge, if
they were to maintain their speeds.
Equations of motion of the two barges are :
dv
dM
1
1
Slower barge : M1
 F1,ext  v1,rel
dt
dt
dv 2
dM 2
Faster barge : M 2
 F2,ext  v 2,rel
dt
dt
Here, v1,rel, the relative velocity of shoveled
coal w.r.t. the slower barge, is zero, and v2,rel is
v 2,rel  v1  v 2   11.55 km / hr   3.2 m / s
dM 2
dM1
Moreover,

 925 kg / min 15.4 kg / s
dt
dt
Since the speed of neither is to change, ( dv1
dv 2

 0)
dt
dt
F1,ext  0 & F2 ,ext   v 2 ,rel
dM 2
 49.5 N
dt
Falling Raindrop
r
v
dM
2
 4kvr ;
dt
v rel   v
Equation of Motion :
dv
dM
M
 Mg  v
dt
dt
d
Mv   Mg
Or,
dt
Or,
 
d 3
3
r v  gr
dt
dM
d  4 3 
2 dr
Since
  r   4r
,
dt
dt  3
dt

dM
equating it to the given
dt
dr k
 v
dt 
Claim : The solution, with the condition that
r = v = 0 at t = 0, is :
v
2g
r
7k
dv
a

dt

2g 1 dr
7k 2 r dt
2g 1 kv g

7k 2 r  7
Full Solution
g d 7
g d 7 dt
Writing gr  3 (r )  3 (r )
7r dr
7r dt
dr
3
dr
and substituti ng for
,
dt
d 3
g d 7
r v 
(r )
3
dt
7 kr v dt
 
d 3
g d 7
Or, r v
r v 
(r )
dt
7k dt
3
 
 
1 d 3
Or,
r v
2 dt
2
g d 7

(r )
7 k dt
Integrating both the sides,
1 6 2 g 7
r v 
r C
2
7k
Using the condition that at t = 0, r = v = 0,
we get C = 0.
v
2g
r
7k
dv
a

dt

2g 1 dr
7k 2 r dt
2g 1 kv g

7k 2 r 
7