Document 6510250

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APPENDIX
How to Find the Center of Mass
Given a system of N point masses, the center of mass of
the system is defined as
( A•1 )
..---------'="
u
o
o
o
-;
The position vector r
c
from some reference coordinate system to
the center of mass can be written as
,.
J
J
and Eq. A.l can be split up into three equations for each of the
three components of r c ' i.e.,
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L
.1'­
(A. 2 )
"
where x., y. and z. are the components of the position vector
111
t' 1
.
o f th el.th parlce,l.e.,
~
.
L
)
--
--..
t
s
J
As an example, consider th ree point masses arranged in an
equilateral triangle with sides " a " as shown:
~l
\
o
/i--
For convenience we take the x and y axis as shown.
From
symmetry, it is obvious that Xc must lie somewhere along the
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vertical line distance a/2 from the origin 0, i.e.,
-
X-c -
I
~I l
OJ
/
'--
(
'-
/')
I
f
.
;.
~
•
...
.' t
~~,...,..
J'.~:,-
.
-,
Taking moments about the x axis, we get
~~C ­
U
If we denote the hei ght of the tri angl e
Yc
JX iL
by H, then
The position vector r c from the origin to the center
of mass can thus be written as
=
H/3.
\.
"
If we have a continuous distribution of mass such as a
sheet of metal of uniform thickness, the center of mass can be
found from Eq. A.2 by simply replacing the summation sign by an
integral, i.e.,
(A. 3)
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wh e re
~
is the density per unit area, dA is an element of
area, hence -fdA represents the mass dm of the elemental area.
The total mass of the sheet is given by
-
....
....
As an example, consider the quarter circle shown in the sketch:
)
From Eq. A.3, we write
Similarly,
Note that to get y
c
'r
we took an elemental a re a
use the same elemental area
ydx.
-f xdy.
~
We could
From inspection we see that
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the center of mass of the vertical strip
f
ydx is y/2 from
the x axis.
...­
Hence we could alternately write
d~
C
J-....\ ~) 0
"
fa r y c .
L) L~
1 2-
­
(r
)
~
Eval uating the integral, we qet
Jcr) ­
L~
J~G -:....
~-t
the same result as obtained previously.
If we know the center of mass for a number of simple
shapes such as quarter circles, triangles, rectangles, etc., we
could reduce the task of finding the CM of a complicated shape by
splitting it up into a combination of simple shapes and finding
the CM of an equivalent system of point masses.
following shape shown in the sketch:
Consider the
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c
•
This shape can be split up into an equilateral triangle of
side R, a quarter circle of radius R, and a rectangle R by 2R.
The eM of each of these simple shapes is known.
Hence the
system is equivalent to a system of three point masses, as shown:
From geometry we can easily find x.1 and y.1 for each of the three
point masses with respect to some chosen coordinate system.
For
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280 ­
example, with reference to a coordinate system taken as the
two sides of the rectangle the xi·s and Yi
masses are shown in the sketch.
IS
of the three point
The CM of this system of three
point masses can readily be found using Eq. A.2, i.e.,
o
-
~t1
lJ16 (-
~r.)
~L{b
-+
)~
-t- H
'J
1).
Sometimes we might have to find the CM of a planar area
with a hole in it.
To do that we simply find the CM of the area
by first ignoring the hole, then we take off the negative contribution
due to the hole itself.
As an example, consider a circular sheet
with a circular hole, as shown in the sketch:
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­
From symmetry it is obvious that the CM must lie on the x axis.
Using Eq. A.2, we write
L
~(_
~MiXi
-~1 AXc A and
is the total mass including the
A
hole of the big circle, Xc A is the CM of the big circle which
in this case is at the origin (i.e., X A = 0).
'Lc.m.x. = MBX B
whe re
M
C = t ? 11
c
is the negative contribution from the hOle'He,:~R is the mass
of the "hole" and Xc B = R/2 is the location of the CM of the hole
from the origin.
X-L ­-
The 01 of the system is thus
f
--
~
To find the CM of volumes we do the same thing.
We take
moments of elemental volumes dv about the appropriate axes.
For
example, consider a right circular cone as shown in the sketch:
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:r1·1l~­
For convenience we choose the x axis along the axis of the cone
and the y axis parallel to the base B.
From symmetry it is obvious
that the CM must lie along the x axis.
To find x,
"'Ie use Eq. A.3,
c
i .e. ,
f
z
But the mass of the cone is
)rt'
­
• hence
Measured from the base of the cone the CM is located at a
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283 ­
distance H/4 along the cone's axis of symmetry.
Another trick to find the CM of an area is to use
Pappus' theorem.
If we generate a volume by rotating a planar
area, the resultant volume generated is given by the product of
the area and the distance moved by the center of mass.
For
example, to find the CM of a semicircle of radius R
we rotate the area of the semicircle
axis by 2£1
~R2
about the vertical
to generate a sphere of vOlume;
\f
R3 .
Then the
CM of the semicircle from Pappus' theorem can be found as follows:
( area of semicircle) x (distanceby travelled)
CM
=
(volume
)
generated
hence
Consider another example.
To find the CM of a right angle trian9le
of base B and height H, we rotate the area of the triangle (i .e.,
}BH) by
21ll
to generate a right circular cone of volume
~
\f B2H.
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~---­
Using Pappus· theorem
x 1
hen ce
~c.-
If I've rotate about the x axis, we get Yc to be
J.
7­
hence
The
eM
--
of a right angle triangle is therefore located as shown
in the sketch:
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The center of mass of planar areas and volumes of simple
geometry are given in tabulated form in most handbooks and
mathematical tables.
To find them yourself is straightforward,
particularly if the reference axes are chosen intelligently to
simplify the integration.
The inertia tensor has already been introduced in
:'\1
Section 6.
The moments of inertia of q few simple symmetrical
bodies about the axis of rotation are derived.
It is sufficient
here to do a few more examples.
For the motion of a rigid body in a plane, the angular
-~
velocity GJ is always normal to the plane of motion.
In these
problems, it is necessary to consider only the moment of inertia
about the axis of rotation.
As an example, consider the planar
motion of a quarter circle as shown in the sketch:
The motion is in the xy plane and the rotation is about the
z axis normal to the plane of the paper.
~}
vector LJ
-"\
=
zk~
z component, i.e.,
The angular velocity
-)
The angular velocity vector L has only a