Transcript Center of Mass

Center of Mass
Physics
Montwood High School
R. Casao
 The center of mass is the point at
which all of the mass of an object or
system may be considered to be
concentrated, for the purposes of
linear or translational motion only.
 Newton’s second law for the motion of
the center of mass: Fnet = m∙acm
 The momentum of the center of mass
does not change if there are no
external forces on the system.
 The center of mass of a flat object
can be found by suspension.
 The center of mass may be located
outside a solid object.
 The principle of conservation of momentum can
be restated in terms of the center of mass.
 For several particles of mass m1, m2, etc., let the
coordinates of the center of mass of m1 be (x1,
y1), those of m2 be (x2, y2), etc.
 We define the center of mass of the system as
the point having coordinates (xcm, ycm) given by:
x cm
m1  x1  m2  x 2  m3  x 3

m1  m2  m3  ...
m x

 ...

m
i
i
i
i
i
y cm
m1  y1  m2  y 2  m3  y 3

m1  m2  m3  ...
m y

 ...

m
i
i
i
i
i
 When a homogeneous
body has a geometric
center, such as a billiard
ball or a sugar cube, the
center of mass is at the
geometric center.
 When a body has an axis
of symmetry, such as a
wheel or a pulley, the
center of mass lies on that
axis.
 There is no law that says
the center of mass has to
be within the body. See
the donut!
 When the center of mass of a particle moves’
the x and y components of velocity of the
center of mass, vcm-x and vcm-y, is the time
derivative of xcm and ycm. Keep in mind
velocity = dx/dt.
m1  v1x  m2  v 2x  m3  v 3x  ...
v cm x 
m1  m2  m3  ...
m1  v1y  m2  v 2y  m3  v 3y  ...
v cm y 
m1  m2  m3  ...
 If vx and vy are used to determine the
resultant vector v;




m1  v1  m2  v 2  m3  v 3  ...
v cm 
m1  m2  m3  ...
 Expressing m1 + m2 + m3 + … as M
and multiplying both sides by M
gives us:





M  vcm  m1  v1  m2  v2  m3  v3  ...  P
 The total momentum is equal to the total
mass times the velocity of the center of
mass.
 For a system of particles on which
the net external force is 0 N, the total

momentum is constant and the

P
velocity of the center of mass vcm  M
is constant.
 If the net external force on a system
of particles in not 0 N, then total
momentum is not conserved and the
velocity of the center of mass will
change.
 Taking the time derivative of the
velocity,we get acceleration, acm =
dvcm/dt.




M  a cm  m1  a1  m2  a 2  m3  a 3  ...


ΣFext  M  a cm
 When a body or a collection of particles is acted
on by external forces, the center of mass moves
just as tho all the mass were concentrated at
that point and it were acted on by a net force
equal to the sum of the external forces on the
system.
 Suppose a cannon shell traveling in a parabolic
trajectory (neglecting air friction) explodes in flight,
splitting into two fragments of equal mass.
 The fragments follow new parabolic paths, but the
center of mass continues on the original parabolic path
as if all the mass were still concentrated at that point.
 One more useful way of describing the motion of
a system of particles:


dv cm
a cm 
dt




dv cm d(M  v cm ) dP
M  a cm  M 


dt
dt
dt


dP
ΣFext 
dt
Practice Problem 1
 Three odd-shaped blocks have the following masses and center-ofmass coordinates: (1) 0.3 kg, (0.2 m, 0.3 m); (2) 0.4 kg, (0.1 m, -0.4
m); (3) 0.2 kg, (-0.3 m, 0.6 m). Find the coordinates of the center of
mass of the system of three blocks.
x cm
x cm
x cm
x cm
m1  x1  m2  x 2  m3  x 3

m1  m2  m3
0.3kg  0.2m  0.4kg  0.1m  0.2kg  -0.3m

0.3kg  0.4kg  0.2kg
0.06kg  m  0.06kg  m  0.06kg  m

0.9kg
0.04kg  m

 0.0444m
0.9kg
y cm
y cm
y cm
y cm
m1  y1  m2  y 2  m3  y 3

m1  m2  m3
0.3kg  0.3m  0.4kg  0.4m  0.2kg  0.6m

0.3kg  0.4kg  0.2kg
0.09kg  m  0.16kg  m  0.12kg  m

0.9kg
0.05kg  m

 0.0556m
0.9kg
 Center of mass coordinates: x = 0.044 m, y = 0.0556 m.
Practice Problem 2
 A 1200 kg station wagon is moving along a straight highway at 12
m/s. Another car of mass 1800 kg and speed 20 m/s has its center
of mass 40 m ahead of the center of mass of the station wagon.
 A. Find the center of mass of the system consisting of the two
automobiles.
 Since the cars are on a horizontal surface, you can ignore the y component.
The center of mass of the two auto system will lie entirely on the x axis.
1200kg  0m  1800kg  40m
x cm 
1200kg  1800kg
72000kg  m
x cm 
 24m
3000kg
 Center of mass lies 24 m in front of the center of mass of the station wagon.
 B. Find the magnitude of the total momentum of the system
from the given data.
p = 1200kg∙12 m/s + 1800 kg·20 m/s = 14400 kg·m/s + 36000
kg∙m/s
p = 50400 kg∙m/s
 C. Find the speed of the center of mass of the system.
v cm
v cm
v cm
1200kg  12 m  1800kg  20 m
s
s

1200kg  1800kg
14400kg  m  36000kg  m
s
s

3000kg
50400kg  m
s  16.8 m

s
3000kg

Find the total momentum of the system, using the speed of the
center of mass. Compare this value to the answer in part B.
D.
P= (1200 kg + 1800 kg)∙16.8 m/s = 50400 kg·m/s
 The two calculations for the momentum of the system, either
individually or using the center of mass of the system, return the
same value.