Chapter 9a - Systems of Particles
•
Center of Mass
– point masses
– solid objects
•
Newton’s Second Law for a
System of Particles
•
Linear Momentum for a System of
Particles
•
Conservation of Linear Momentum
– Rockets
– Internal Energy/External Forces
Calculating the center of mass –
point objects – 1 D
n
x cm
m1x1  m 2 x 2  ...


m1  m 2  ...
n
m x m x
i 1
n
i
m
i 1
i
i

i
i 1
M
i
Calculating the center of mass –
point objects – 2 D
n
n
rcm 
m r
i 1
i i
x cm 
M
m x
i
i 1
i
M
n
y cm 
m y
i
i 1
i
M
n
z cm 
m z
i i
i 1
M
Problem 1
• Three masses located in the x-y plane have
the following coordinates:
– 2 kg at (3,-2)
– 3 kg at (-2,4)
– 1 kg at (2,2)
• Find the location of the center of mass
Calculating the center of mass –
solid objects – 1 D
n
x cm 
x cm
m x
i
i 1
i
M
xdm


M
Calculating the center of mass –
solid objects – 2 D
rcm 
 rdm
M
x cm
xdm


y cm
ydm


z cm
zdm


M
M
M
Finding the COM
Problem 2
• What is the center of mass of the Letter “F”
shown if it has uniform density and
thickness?
2cm
2cm
20cm
5cm
10 cm
2cm
15cm
Problem 3
The blue disk has a radius 2R
The white area is a hole in the
Disk with radius R.
Where is the center of mass?
COM and translational motion
n
rcm 
m r
i 1
n
Mrcm   mi ri
i i
i 1
M
n
drcm
dr
M
  mi i
dt
dt
i 1
First time derivative
n
Mvcm   mi vi
COM Momentum
i 1
Second time derivative
Newton’s
2nd
Law
n
dv cm
dv
M
  mi i
dt
dt
i 1
n
n
i 1
i 1
Ma cm   mi a i  F1  F2  F3  ...   Fi
What this means….
• The sum of all forces acting on the system
is equal to the total mass of the system
times the acceleration of the center of mass.
• The center of mass of a system of particles
with total mass M moves like a single
particle of mass M acted upon by the same
net external force.
Conservation of Linear Momentum
• If 2 (or more) particles of masses m1, m2, … form
an isolated system (zero net external force), then
total momentum of the system is conserved
regardless of the nature of the force between them.
Problem 1
• An astronaut finds himself at rest in space
after breaking his lifeline. With only a
space tool in his hand, how can he get back
to his ship which is only 10 m away and out
of his reach.
Variable mass – Rocket propulsion
pi  p f
 M  dM v  M  v  dv    v  dv  ve  dM
small
Mv  vdM  Mv  Mdv  vdM  dvdM  vedM
Mdv  vedM
dv
dM
Thrust  M
 ve
dt
dt
Rocket thrust
m0  21000kg
mfuel  15000kg
dM
 190kg / s
dt
ve  2800m / s
dv
dM
Thrust  M
 v rel
dt
dt
Find: Thrust, initial net force,
net force as all fuel expended