The document discusses the center of mass (COM) of objects. It defines COM as the point located at an object's average position of mass. For symmetrical objects like a baseball, the COM is at the exact center, but for oddly shaped objects it can be in any position depending on weight distribution. For an object to balance and not topple, its support must be directly below the COM. The location of an object's COM depends on where most of its mass is focused and does not need to be within the object itself. Objects with a lower COM are less likely to topple over.

1. Center of MASS (COM)
 Definition--The Point located at an object’s
average position of the mass
 In other words…. The center of an object’s
mass
 Symmetrical object’s, like a baseball the C of
M would be in the exact center of object
 However other oddly shaped objects will find
COM in any number of positions, depending on
weight distribution

COM

2. C.O.M.
 When objects rotate freely they must rotate
about an axis through the COM
 Basically treat the object as if all its weight is
concentrated at that one pt.

3. C.O.M. --Balancing
 For an object to balance, and not
topple… support must be
directly below C.O.M.

4. Where C.O.M. is located
 Generally found in the middle
of all the weight…
 Does not even have to be
within, the object itself
 Ex. boomerang
 Will be located toward one
side of an object where most
of its mass is focused…
 Ex. Weebles
COM
gravity

5. Weebles Wobble, but they don’t
fall down???
 Weebles have very low COM
 Whenever rolling it will roll to a stop
when its COM is as low as possible
 This occurs when it is standing upright
 Also occurs for inflatable toy clowns
 Objects with a low COM are less
likely to topple because of this
principle
 Higher COM is, the easier to topple

6. Balancing Stuff
 Again, all that has to happen to balance, is for
a support to be directly beneath COM

7. Advantage of low COM
 Athletic advantages
 wrestling—harder to takedown
 Football – “
“
“
 Both easier to drive power
through their legs
 SUV’s …. Tip over all the
time b/c COM is too high
 Farmer’s tractors
 Much more control in all
vehicles w/ low COM

8.  Deadliest Catch
 Pots on deck and freezing ice make
boat top heavy… more likely to roll
and sink
 Ballast tanks at bottom help lower
boat’s COM

9. Animals
 Low COM
High COM

10. T. Rex
& Tails

11. Humans
-
Where is our COM?
 Just below our belly
button
 Notice, support always
below COM
 Bipedalism??
 Only mammals w/ this
ability to walk on 2 legs
 Because of Evolution and
how our legs changed to
balance between steps is
why we are only mammals
to walk

12.  Because our
legs/hips evolved so
that our support
base (feet) were
close together allows
us to be bipedal
 Apes and our early
ancestors hips were
constructed
differently with a
wide set base.
 Impossible to walk
bipedal
Hip protruding
from joint….
Creates inward
angled femurs…
Which makes
feet close
together….
Providing a
stable/efficient
base for walking
upright

13. Definition
 “The center of mass (or mass center) is the
mean location of all the mass in a system”
 Marked with:

14. Examples
 The center of mass of a two-particle system
lies on the line connecting the particles
(or, more precisely, their individual centers of
mass). The center of mass is closer to the
more massive object
 The center of mass of a ring is at the center of
the ring (in the air).

16. More Examples
 The center of mass of a solid triangle lies on
all three medians and therefore at the
centroid, which is also the average of the
three vertices.

17. Applications
 Objects rotate around their center of mass
 In a uniform gravitational field, the center of
mass and center of gravity are the same.
 A projectile’s center of mass will follow a
parabolic path.
 If an object’s center of mass is outside it’s base of
support, it will topple.
 An applied force that is not through an object’s
center of mass will cause rotation.

19. CoM of Symmetrical Object
 The CoM of any symmetrical object
lies on an axis of symmetry and on
any plane of symmetry.

20. Toppling Rule of Thumb
 If the CoM of the object is
above the area of
support, the object will
remain upright.
 If the CoM is outside the
area of support the object
will topple.

21. Another look at Stability
 Stable equilibrium: when for a balanced object
a displacement raises the CoM
 Unstable equilibrium: when for a balanced
object a displacement lowers the CoM .
 Neutral equilibrium: when the height of the
CoM does not change with displacement.

22. Stability

23. Center of Mass of:
 System of Particles
 Extended Object

24. Center of Mass of a System of
Particles in motion
vCM =
Smivi
SM
 mi is the mass of each particle
 M is the sum of the masses of all particles
 Momentum
vCM =
Spcm
SM

26. MOTION of center of mass(CoM)
Refer pg 37 of course book the grey out area.
1. Consider just the single motion of the center of
mass(CoM)
2. Motion of the vcm does not change in magnitude or
direction
3. ptotal is always conserved(pbefore = pafter = ptotal
4. CoM momentum and velocity is unaffected by
interactions and collisions)
=
pcm )

27. Level 3 COURSE book solutions page 38
Page 38 number 2 and 4
2a p =mv = 260 x 3.2 = 832kgm/s
b vcm = 0m/s CoM velocity and momentum does
not change.
c 2nd dogem must stop as well
3a 100m/s right
b. 100m/s left
*** NO. 4 VERY IMPORTANT TO grasp CoM
4a vcm = (m1v +m2v /(m1 +m2) = 15000x 5 /
(10000+15000) = 3m/s left
b 3.0m.s

28. Level 3 COURSE book solutions page 38
pg 38 number 1 at the bottom and pg 39 no 2
1d period T = ¼ =0.25
x component of vcm = 2cm/0.25 = 8cm/s
e pt = pcm = 0.08 x 0.3 = 0.024 kg m/s
4×3 = 12
2a
8x2=16
122 + 162 = 20
b 20kgm/s
c vcm = Spcm = 20 = 1.7m/s
SM
12
d tan ϴ = 16/12
ϴ = 53 to x axis

29. Level 3 COURSE book solutions page 39
pg 39 no 3
3a
100002 + 120002 =
800x15=12 000
1000×10 = 10 000
b 15600kgm/s
c vcm = Spcm = 15600
SM
1800
d 8.7m/s
e tan ϴ = 12000/10000
= 8.7m/s
ϴ = 50 to x axis

30. Example 1: Center of Mass in
one Dimension
 Find the CM of a system of four particles that
have a mass of 2 kg each. Two are located 3cm
and 5 cm from the origin on the + x-axis and two
are 2 and 4 cm from the origin on the – x-axis
 Answer: 0.5cm

31. Level 3 Text book solutions for activity 7D page 101
number 1
a
CoM
(90 x X)
90X
200X
X
=
=
=
=
{110 x (15-X)}
1650 -110X
1650
1650/200 = 8.3m
bi CoM has not moved
ii 8.25/ 15 x (15-5.5)
=5.2m from A
iii t =8.5s d =3.1
MX
90X
90X
200X
X
OR
= m(d-X)
= 110(9.5-X)
= 1045 – 110X
= 1045
= 5.2m from A
v = d/t 3.1/8.5 = 0.36m/s
iv pa = mv = 90 x 0.36 = 32.4kgm/s
pb = 32.4kgm/s = 90 x vb
vb
= 32.4/90 = 0.29m/s

32. Level 3 Text book solutions for activity 7D page 101
number 4
ai
o m/s
ii
5m
bi use rotational torque
Ta = Fa x d = (12.6x9.8) x 3 = 370.44Nm
Tc = 370.44Nm
Fc = Ta ÷ d
= 370.44 ÷ 2 = 185.22N
m = 185.22 ÷ 9.8 = 19kg
bii d = 5m
v = 2m/s t = d/v 5/2 = 2.5s
iii zero no external forces therefore vcm = 0m/s
*** iv
center of mass moved 3m in 2.5s
therefore vcm = 3/2.5 = 1.2m/s

33. Level 3 Text book solutions for activity 7D page 101
number 5
a
vcm = 3.4m/s
b mm = mass of Moana
mj = mass of Joe
before: ∑ pcm = (mm + mj ) x 3.4 = 3.4 mm + 3.4 mj
After: pm = mv = 3.1 x mm
pj = 4.0 x mj
∑ pcm = (3.1 x mm + 4.0 x mj )
∑ pcm = 3.4 mm + 3.4 mj
(3.1 x mm + 4.0 x mj ) = 3.4 mm + 3.4 mj
4.0 mj - 3.4 mj = 3.4 mm - 3.1 mm
0.6 mj
=
0.3mm
mm = 0.6/0.3 mj = 2mj
Mass of Moana = 2 mass of Joe