1. Classical Mechanics
Topic 7.0: Center of Mass
Motion of Center of Mass
Center of Mass and Momentum
Center of Mass and Newton’s Second Law
Based from Sears and Zemansky’s University Physics
with Modern Physics 13th ed
And Physics for Scientists and Engineers by Serway
and Jewett, 9th ed

2. • Consider a system with several particles with masses
m1 located at (x1,y1,z1), m2 located at (x2,y2,z2), and
so on. We define the center of mass as the point that
has coordinates (xcm, ycm, zcm)
• Note: The position coordinates can
be negative with respect to some
reference point.
Center of Mass
𝑥𝑐𝑚 =
𝑚1𝑥1 + 𝑚2𝑥2+. . .
𝑚1 + 𝑚2+. . .
=
σ𝑖 𝑚𝑖 𝑥𝑖
σ𝑖 𝑚𝑖
𝑦𝑐𝑚 =
𝑚1𝑦1 + 𝑚2𝑦2+. . .
𝑚1 + 𝑚2+. . .
=
σ𝑖 𝑚𝑖 𝑦𝑖
σ𝑖 𝑚𝑖
𝑧𝑐𝑚 =
𝑚1𝑧1 + 𝑚2𝑧2+. . .
𝑚1 + 𝑚2+. . .
=
σ𝑖 𝑚𝑖 𝑧𝑖
σ𝑖 𝑚𝑖
Physics for Scientists and
Engineers by Serway and
Jewett, 9th ed
Physics for Scientists and
Engineers by Serway and
Jewett, 9th ed

3. Center of Mass
Physics for Scientists and Engineers by Serway and Jewett, 9th ed

4. Center of Mass
Physics for Scientists and Engineers by Serway and Jewett, 9th ed

5. If we let the number of elements n approach infinity, the size of each element
approaches zero and 𝑥𝐶𝑀 is given precisely. In this limit, we replace the sum by
an integral and Δ𝑚𝑖 by the differential element 𝑑𝑚:
Likewise, for 𝑦𝐶𝑀 and 𝑧𝐶𝑀 we obtain
Center of Mass
We can express the vector position of the center of mass of an extended object in
the form

6. Center of Mass
Physics for Scientists and Engineers by Serway and Jewett, 9th ed

7. Center of Mass

8. 𝑥𝑐𝑚 =
𝑚1𝑥1 + 𝑚2𝑥2 + 𝑚3𝑥3
𝑚1 + 𝑚2 + 𝑚3
=
0.3 𝑘𝑔 0.2 𝑚 + 0.4 𝑘𝑔 0.1 𝑚 + 0.2 𝑘𝑔 −0.3 𝑚
0.3 𝑘𝑔 + 0.4 𝑘𝑔 + 0.2 𝑘𝑔
= 0.0444 𝑚
• Three odd-shaped blocks of chocolate have the following
masses and center-of-mass coordinates:
(1) 0.300 [kg], (0.200 [m], 0.300 [m])
(2) 0.400 [kg], (0.100 [m], – 0.400 [m])
(3) 0.200 [kg], (– 0.300 [m], 0.600 [m])
Find the coordinates of the center of mass of the system of
three chocolate blocks.
• The system's center of mass is at (0.0444 [m], 0.0556[m])
• E1. Calculate the y-coordinate of the center-of-mass. Show
your complete substitution.
𝑦𝐶𝑀 = 0.0556 𝑚
Center of Mass

9. ● The center of mass of a uniform object lies on its
geometrical center.
● The center of mass of a system composed of several
extended objects is obtained by treating each
extended object as if they are particles located at
their respective center of masses.
• An object is constructed of uniform wire bent into
shape as shown. What is the location of its center
of mass?
● Choose the center of mass of the lower
wire to be the origin.
● Replace the rods with point particles
located at their corresponding center of
masses.
Center of Mass

10. 𝑣𝑐𝑚−𝑥 =
𝑚1𝑣1x + 𝑚2𝑣2x+. . .
𝑚1 + 𝑚2+. . .
=
σ𝑖 𝑚𝑖 𝑣𝑖𝑥
σ𝑖 𝑚𝑖
𝑣𝑐𝑚−𝑦 =
𝑚1𝑣1y + 𝑚2𝑣2y+. . .
𝑚1 + 𝑚2+. . .
=
σ𝑖 𝑚𝑖 𝑣𝑖𝑦
σ𝑖 𝑚𝑖
𝑣𝑐𝑚−𝑧 =
𝑚1𝑣1z + 𝑚2𝑣2z+. . .
𝑚1 + 𝑚2+. . .
=
σ𝑖 𝑚𝑖 𝑣𝑖𝑧
σ𝑖 𝑚𝑖
• The motion of the center of mass is easier
to track than the motion of the entire
object.
Motion of the Center of Mass

11. • The drawing shows three particles that are moving
with different velocities. Two of the particles have
mass m, and the third has a mass 2m. At the instant
shown, the center of mass (cm) of the three
particles is at the coordinate origin. What is the
velocity 𝑣𝑐𝑚 (magnitude and direction) of the center
of mass?
• Note that the objects are just moving along (or parallel)
with the x-axis.
Motion of the Center of Mass
𝑣𝑐𝑚−𝑥 =
𝑚1𝑣1x + 𝑚2𝑣2x + 𝑚3𝑣3x
𝑚1 + 𝑚2 + 𝑚3
𝑣𝑐𝑚−𝑥 =
𝑚 +4.0 Τ
𝑚 𝑠 + 𝑚 −8.0 Τ
𝑚 𝑠 + 2m +6.0 Τ
𝑚 𝑠
𝑚 + 𝑚 + 2m
𝑣𝑐𝑚−𝑥 = +2.0 Τ
𝑚 𝑠

12. • Example 8.14 of University Physics 13th ed.
• The center-of-mass does not change because there is no
net force acting on the system.
● The system is composed of James and Ramon only.
Motion of the Center of Mass

13. • The total momentum of all the particles in the
system is equal to the total mass times the velocity
of the system's center of mass.
Ԧ
𝑣𝑐𝑚 =
𝑚1 Ԧ
𝑣1 + 𝑚2 Ԧ
𝑣2+. . .
𝑚1 + 𝑚2+. . .
=
σ𝑖 𝑚𝑖 Ԧ
𝑣𝑖
σ𝑖 𝑚𝑖
• Let the sum of all the
masses be the total
M mass of the
system.
Ԧ
𝑣𝑐𝑚 =
σ𝑖 𝑚𝑖𝑣𝑖
𝑀
→→ → → →𝑀 Ԧ
𝑣𝑐𝑚 = ෍
𝑖
𝑚𝑖 Ԧ
𝑣𝑖
𝑃 = ෍
𝑖
Ԧ
𝑝𝑖
Center of Mass and Momentum

14. • A 1200-kg station wagon is moving along a straight highway at
12.0 [m/s]. Another car, with mass 1800 [kg] and speed 20.0 [m/s],
has its center of mass 40.0 [m] ahead of the center of mass of the
station wagon. (a) Find the position of the center of mass of the
systems consisting of the two automobiles. (b) Find the magnitude
of the total momentum of the system.
• Set the center of one of the object to be at the origin.
Let us choose the center of the station wagon to be at
the origin.
• E2. What is the center of mass of the system
from the station wagon?
Center of Mass and Momentum

15. • (b) Find the magnitude of the total momentum of the system.
• There are two ways to answer (b).
● Get the total momentum using the individual objects in the system.
● Calculate the velocity of the center of mass and then the total
momentum.
𝑃 = ෍
𝑖
Ԧ
𝑝𝑖 = 𝑚1200 Ԧ
𝑣1200 + 𝑚1800 Ԧ
𝑣1800
= 1200 𝑘𝑔 +12.0 Τ
𝑚 𝑠 + 1800 𝑘𝑔 +20.0 Τ
𝑚 𝑠
= 5.04 × 104
𝑘𝑔 ⋅ Τ
𝑚 𝑠
E4. and the momentum of the
system.
E3. Calculate the velocity of the center of mass 𝑣𝐶𝑀 = 16.8 Τ
𝑚 𝑠
𝑃 = 5.04 × 104 𝑘𝑔 ⋅ Τ
𝑚 𝑠
Center of Mass and Momentum

16. • If there is a net force acting on a system of
particles: ෍ Ԧ
𝐹 = 𝑀 Ԧ
𝑎𝑐𝑚
෍ Ԧ
𝐹 = ෍
𝑖
𝑚𝑖 Ԧ
𝑎𝑖
• When a collection of particles is acted on by external
forces, the center of mass moves just as though all the
mass were concentrated at that point.
Center of Mass and Newton’s Second Law

17. • Two-stage rocket. A rocket is shot into the air as shown
below. At the moment the rocket reaches its highest point, a
horizontal distance d from its starting point, a prearranged
explosion separates it into two parts of equal mass. Part I is
stopped at midair by the explosion, and falls vertically to
Earth. Where does part II land?
• Note:
● The center of
mass will just
follow the
projectile's
motion
trajectory.
● Thus, the
center of mass
will be 2d
away from the
starting point
when it hits the
ground.
• E5. How far from the starting point
will part II land?
• Results will be different if the two parts
are of different masses.
𝑥𝐼𝐼 = 3𝑑
Center of Mass and Newton’s Second Law