Transcript center of mass

Center of Mass
The point located at
the object’s
average position of
mass.
The Center of Mass and the
Center of Gravity
Are located at the same
point AS LONG AS
gravity is consistent
throughout the object.
The center of mass does not always lie
within the object itself….
When an object is in motion, its center of
mass will follow a smooth line.
Locating the Center of Gravity, CG
Method One:
It’s a balance point along the object.
Method Two:
If you suspend any object, its cg will be
located along a vertical line drawn from
the suspension point.
Center of Gravity, cg, vs Center of Mass, cm
In this course, we virtually ALWAYS use cm rather
than cg, and for almost all situations, they are
located at the same place. The cg is the
average location of the weight- the cm is the
average location of the mass. As long as the
gravitational force is the same throughout the
body, these two will be at the same place.
(Near the event horizon of a black hole, the
gravitational force could be very different in an
extended body depending on which portion of
the object is closer to the black hole! In this
case the cg would NOT be at the cm.)
The center of mass for multiple objects of
total mass M is given by
rcm
1

M
m r
i i
i
Finding the x-coordinate for the
center of mass for 2 separate
masses.
xcm
m1 x1  m2 x2

m1  m2
And the same for ycm!
Newton’s First Law: When there’s no net external
force acting on a system of masses (or a continuous
mass), if the center of mass was at rest, it will remain
at rest, if it was in motion, it continues that motion,
regardless of what the individual masses are doing!
Newton’s Second Law:
If there is a net external force acting upon a
system of masses, the center of mass will
accelerate.
When we draw free-body diagrams, we
generally draw forces acting at the cm to
determine the acceleration of the center of
mass (which may NOT be the acceleration
of every point of the body or every particle
in the system)
SF = mtacm
Newton’s Third Law- we will revisit later
when we look at recoil and collisions from
a center of mass reference frame.
The center of mass of a system moves like a
particle of mass M = S mi under the influence of
the net external force acting on the system.
Let’s look at an example….
A projectile is fired into the air over level
ground with an initial velocity of
24.5m/s at 36.9° to the horizontal. At
its highest point, it explodes into two
fragments of equal mass. One
fragment falls straight down to the
ground. Where does the other
fragment land?
Since the only external force acting on the system is gravity, the center of mass
continues on its parabolic path as if there had been no explosion. The cm lands at
R, where R is the range. The first fragment lands at 0.5R. The other fragment of
equal mass must land at 1.5R.
What if the fragment were 0.3M? At what location X will the other fragment land?
The Range (for cm) = (vo sinq)/g x 2 x vo cos q = 58.8m (the cm will fall at ½ that range)
cm 
(m1 )( x 1 )  (m 2 ) x 2 )
M
1
(0.3M )( x 58.8)  (0.7M ) X )
2
58.8 
M
X = 71.4m
The
Calculus
Connection
To find the center of mass of a continuous object, we replace the
sum
rcm
rcm
with an integral
1

m i ri

M i
1
  r dm
M
Where dm is a differential element of mass located at position r.
MEMORIZE this equation, but you probably won’t have to
evaluate the integral!
The AP objective states that the student should “use integration to
find the center of mass of a thin rod of non-uniform density”. I
have seen questions where the student was asked to write the
integration expression, but not to evaluate it, however… we’ll
look at how to do it…