We learned yesterday that the center of mass of a system of particles can be found
using
1 n 

rcm 
 mi ri
M i 1

rcm 
How might the center of mass of a rigid, continuous object be found?
y
1 n

 mi ri
M i 1
n
1



rcm 
lim   mi ri 
M m0 i 1

1 

rcm   r dm
M
mi

ri
cm

rcm
x
z
Example 1
Show that the center of mass of a uniform thin rod of mass M and length L
lies midway between its ends.
y
L
dm
Think of the thin rod as 1-dimensional
x
x
dx
1 

xcm   xdm
M
Linear Density   - Mass per unit length
1 L

xcm  0 xdx
M
 L

xcm  0 xdx
M

xcm 
 1
2
L
x 

M  2 0
M

L
dm

dx
dm  dx
Example 1
Show that the center of mass of a uniform thin rod of mass M and length L
lies midway between its ends.
y
L
dm
Think of the thin rod as 1-dimensional
x
x
dx
 1 2 

xcm   x 
M  2 0
L
Linear Density   - Mass per unit length
 1 2 1 2 

xcm   L  0 
M 2
2

L2

xcm 
2M
M L2

xcm  L
2M
L

xcm 
2
M

L
dm

dx
dm  dx
Consider a system of particles.
1 n 

xcm   mi xi
M i 1




Mxcm  m1x1  m2 x2      mn xn
d 
d



Mxcm  m1x1  m2 x2      mn xn 
dt
dt
d 
d 
d 
d 
M xcm  m1 x1  m2 x2      mn xn
dt
dt
dt
dt




Mvcm  m1v1  m2v2      mnvn




Mvcm  m1v1  m2v2      mnvn
d 
d



Mvcm  m1v1  m2v2      mnvn 
dt
dt
d 
d 
d 
d 
M vcm  m1 v1  m2 v2      mn vn
dt
dt
dt
dt




Macm  m1a1  m2a2      mnan
But Newton’s Second Law says


F  ma
 


Macm  F1  F2      Fn


Macm   F
For this system of particles there could be two types of forces.
1. Forces between the particles.
2. External forces from outside the system.
However, by Newton’s Third Law, any force between two particles would always have an
equal and opposite pair and would therefore sum to 0!


 F external  Macm
Newton’s Second Law
for a system of particles
Where M  total mass of the system
The total mass must remain constant, known as a closed system
Example 1
A 50. kg person stands on the edge of a 3.0 m long cart of mass 40. kg
which is on a frictionless floor as in the diagram below. If the person walks
forward 1.2 m with respect to the floor, how far does the cart move?
50. kg
40. kg
3.0 m
System = Cart, Person, Earth
This is a closed, isolated system.


 F external  Macm

 acm  0
Since the center of mass was initially at rest, it
stays at rest!
Example 1
A 50. kg person stands on the edge of a 3.0 m long cart of mass 40. kg
which is on a frictionless floor as in the diagram below. If the person walks
forward 1.2 m with respect to the floor, how far does the cart move?
50. kg
40. kg
3.0 m


xi cm  x f cm




mc xci  m p x pi mc xcf  m p x pf

M
M



mc xci  mc xcf  m p x pf
Example 1
A 50. kg person stands on the edge of a 3.0 m long cart of mass 40. kg
which is on a frictionless floor as in the diagram below. If the person walks
forward 1.2 m with respect to the floor,
howThat
far does the cart move?
Whew!
was close!
50. kg
40. kg
3.0 m

xcf 


mc xci  m p x pf
mc
40. kg 1.5 m  50. kg 1.2 m

xcf 
40. kg

xcf  0 m
 

x f  xi  x  1.5 m