System of Particles: Center-of-Mass
The center-of-mass of a system of particles is the point that moves as though (1) all of the systems
mass were concentrated there and (2) all external forces were applied there.
N
M tot = ∑ mi = m1 + m2 + m3 + L + mN
N
r
r
r
r i =1 r (total massrof the system)
M tot rcm = ∑ mi ri = m1r1 + m2 r2 + m3r3 + L + mN rN
• Location of the Center-of-Mass:
Center-of-Mass
i =1
r
1
rcm =
M tot
r
r
r
r
r m1r1 + m2 r2 + m3 r3 + L + mN rN
mi ri =
∑
M tot
i =1
N
• Velocity of the Center-of-Mass:
(total momentum of the system)
N
r
r
r
r
r
r
r
M tot vcm = Ptot = ∑ pi = m1v1 + m2 v2 + m3v3 + L + mN v N
i =1
r
r
P
1
vcm = tot =
M tot M tot
r
r
r
r
r m1v1 + m2 v2 + m3v3 + L + mN v N
pi =
∑
M tot
i =1
N
• Acceleration of the Center-of-Mass:
(net force acting on the system of particles)
N
r
r
r
r
r
r
r
Fnet = M tot acm = ∑ mi ai = m1a1 + m2 a2 + m3a3 + L + mN a N
i =1
R. Field 10/10/2013
University of Florida
PHY 2053
r
Fnet
r
r
dPtot
=
= M tot acm
dt
Page 1
Center-of-Mass: Superposition
• Superposition:
rA
If object A has mass MA and its center-of-mass is located at rcmr
B
and object B has mass MB and its center-of-mass is located at rcm
then the center of mass of the system A+B is located at
r
rcmA+ B
rA
rB
M r + M B rcm
= A cm
MA + MB
Uniform density ρ
2R
R
cm
hole
Disk A
• Example:
What is the x-coordinate of the center-of-mass of the circular
disk of radius 2R with a circular hole of radius R as shown in the
figure? Assume the disk has a uniform mass density ρ.
Answer: R/3
A+ B
cm
x
A
B
M A xcm
+ M B xcm
=0=
MA + MB
Disk A+B
Let object A+B be a uniform disk of radius 2R, height h, and center at x=y=0.
Let object A be the disk with the hole in it.
Let object B be a uniform disk with radius R, height h, and center at y=0, x=-R.
R
Disk B
0
A
xcm
=
2R
M B
R
( M A + M B ) A+ B M B B
πR hρ
(
)
xcm −
xcm = − B xcm
R
=−
−
=
MA
MA
MA
(π (2 R ) 2 − πR 2 )hρ
3
R. Field 10/10/2013
University of Florida
2
PHY 2053
Page 2
Center-of-Mass: Example
z
• Fall 2011 Exam 2 Problem 22:
The figure shows a cubical box with each side consisting of a
uniform metal plate of negligible thickness. Each of the four
sides have mass, M, and the bottom has mass Mbottom. The
box is open at the top (at z = L) and has edge length L. If the
the z-coordinate of the center-of-mass is at zCM = L/3, what is
Mbottom?
y
x
Answer: 2M
% Right: 24%
0
zcm =
cm
1 1
Mz
+M z +M z +M z +M
M 1 + M 2 + M 3 + M 4 + M bottom
cm
2 2
cm
3 3
cm
4 4
cm
bottom bottom
z
=
4M ( L / 2)
2 ML
=
4 M + M bottom 4 M + M bottom
zcm (4M + M bottom ) = 2ML
M bottom =
R. Field 10/10/2013
University of Florida
2ML
2 ML
− 4M =
− 4M = 2M
( L / 3)
zcm
PHY 2053
Page 3
Collisions in 1 Dimension: Elastic
• Elastic Collision:
Lab Frame Before
An elastic collision is one in which the kinetic energy is
conserved (i.e. the initial total kinetic energy is equal to
the final total kinetic energy). If a projectile with mass
M1 and speed v1 traveling to the right along the x-axis
collides with a target particle at rest with mass M2, what
are the velocities of the two particles after they
undergo an elastic collision?
(momentum conservation)
0
pinitial = M 1v1 + M 2 v2 = p final = M 1v'1 + M 2 v'2
M 1 (v1 − v'1 ) = M 2 v'2 (1)
v1
v2 =0
M1
M2
x-axis
Lab Frame After
v'1
v'2
M1
M2
x-axis
Note: a2 - b2 = (a-b)(a+b)
(energy conservation)
0
KEinitial = 12 M 1v12 + 12 M 2 v22 = KE final = 12 M 1 (v'1 ) 2 + 12 M 2 (v'2 ) 2
1
2
M 1 (v12 − (v'1 ) 2 ) = 12 M 1 (v1 − v'1 )(v1 + v'1 ) = 12 M 2 (v'2 ) 2
Divide eq. 2 by eq. 1 and multiply by 2
Multiply eq. 3 by M1
and add it to eq. 1
v '2 =
R. Field 10/10/2013
University of Florida
v1 + v'1 = v'2
2M 1
v1
M1 + M 2
(2)
v1 = v'2 −v'1
(3)
Insert eq. 5 into eq.4
(5)
PHY 2053
v'1 =
(4)
M1 − M 2
v1
M1 + M 2
Page 4
1d Collisions: Completely Inelastic
• Completely Inelastic Collision (Lab Frame):
A projectile with mass M1 and speed v1 traveling to
the right along the x-axis collides with a target
particle at rest with mass M2. If the two particles
stick together to form a single particle of mass
M1+M2, what is its velocity? What is the velocity of
the center-of-mass of the two particles system
before the collision? What is the change in the
kinetic energy before and after the collision?
Lab Frame Before
v1
v2 =0
M1
M2
x-axis
Lab Frame After
V
M1 +M2
x-axis
0
pinitial = M 1v1 + M 2 v2 = p final = ( M 1 + M 2 )V (momentum conservation)
M1
1 M 1M 2 2
V = Vcm =
v1 ΔKE = KE f − KEi = −
v1
CM Frame Before
M1 + M 2
2 M1 + M 2
• Completely
Inelastic Collision (CM Frame):
r
r
r
r 0
r
r
p1CM = − p2CM
M tot vcm = Ptotcm = 0 = p1CM + p2CM
r cm r CM r CM r CM
Ptot = p1 + p2 = p f = ( M 1 + M 2 )V = 0
In the CM frame the initial two particles have equal and opposite momentum.
They stick together and are at rest with zero final momentum and zero
final kinetic energy. This corresponds to the maximal loss of kinetic energy
consistent with momentum conservation.
R. Field 10/10/2013
University of Florida
PHY 2053
v1
v2
M1
M2
x-axis
CM Frame After
V=0
M 1+M2
x-axis
Page 5