Doing so forms what is referred to as a right-handed coordinate system which is, by convention, the kind of
coordinate system that we use in science and mathematics. If i × j = k
− then you are dealing with a left-handed
coordinate system, something to be avoided.
1
38
Chapter 21 Vectors: The Cross Product & Torque
immediately to its left yields the negative of the next vector to the left (left-to-right “+“, but
right-to-left “−“).
Now we’re ready to look at the general case. Any vector A can be expressed in terms of unit
vectors:
A = A i + A j + A k
x
y
z
Doing the same for a vector B then allows us to write the cross product as:
A ×B = (A i + A j + A k)× (B i + B j + B k)
x
y
z
x
y
z
Using the distributive rule for multiplication we can write this as:
A ×B = A i × (B i + B j + B k) +
x
x
y
z
A j × (B i + B j + B k) +
y
x
y
z
A k × (B i + B j + B k
)
z
x
y
z
A ×B = A i × B i + A i× B j + A i× B k +
x
x
x
y
x
z
A j × B i + A j × B j + A j × B k +
y
x
y
y
y
z
A k × B i + A k × B j + A k × B
k
z
x
z
y
z
z
Using, in each term, the commutative rule and the associative rule for multiplication we can
write this as:
A ×B = A B (i×i) + A B (i ×j) + A B (i ×k) +
x
x
x
y
x
z
A B (j × i) + A B (j × j) + A B (j × k) +
y
x
y
y
y
z
A B (k × i) + A B (k × j) + A B (k × k
)
z
x
z
y
z
z
Now we evaluate the cross product that appears in each term:
A ×B = A B (0) + A B (k) + A B (−j) +
x
x
x
y
x
z
A B (−k) + A B (0) + A B (i) +
y
x
y
y
y
z
A B (j) + A B (−i) + A B (0
)
z
x
z
y
z
z
Eliminating the zero terms and grouping the terms with i together, the terms with j together,
and the terms with k together yields:
1
39
Chapter 21 Vectors: The Cross Product & Torque
A ×B = A B (i) + A B ( i
− ) +
y
z
z
y
A B (j) + A B ( j
− ) +
z
x
x
z
A B (k) + A B ( k
−
)
x
y
y
x
Factoring out the unit vectors yields:
A ×B = (A B − A B )i +
y
z
z
y
( A B − A B )j +
z
x
x
z
( A B − A B )k
x
y
y
x
which can be written on one line as:
A ×B = (A B − A B i
) + ( A B − A B j
) + ( A B − A B k
) (21-3)
y
z
z
y
z
x
x
z
x
y
y
x
This is our end result. We can arrive at this result much more quickly if we borrow a tool from
that branch of mathematics known as linear algebra (the mathematics of matrices).
We form the 3×3 matrix
i
j
k
A
A
A
x
y
z
B
B
B
x
y
z
by writing i, j, k as the first row, then the components of the first vector that appears in the
cross product as the second row, and finally the components of the second vector that appears in
the cross product as the last row. It turns out that the cross product is equal to the determinant of
that matrix. We use absolute value signs on the entire matrix to signify “the determinant of the
matrix.” So we have:
i
j
k
A ×B = A
A
A (21-4)
x
y
z
B
B
B
x
y
z
To take the determinant of a 3×3 matrix you work your way across the top row. For each
element in that row you take the product of the elements along the diagonal that extends down
and to the right, minus the product of the elements down and to the left; and you add the three
results (one result for each element in the top row) together. If there are no elements down and
to the appropriate side, you move over to the other side of the matrix (see below) to complete the
diagonal.
1
40
Chapter 21 Vectors: The Cross Product & Torque
For the first element of the first row, the i, take the product down and to the right,
i
j
k
A
A
A
x
y
z
B
B
B
x
y
z
( this yields i A B )
y
z
minus the product down and to the left
i
j
k
A
A
A
x
y
z
B
B
B
x
y
z
( the product down-and-to-the-left is i A B ).
z
y
For the first element in the first row, we thus have: i A B − i A B
y
z
z
y which can be written as:
( A B − A B i
) . Repeating the process for the second and third elements in the first row (the j
y
z
z
y
and the k) we get ( A B − A B j
) and ( A B − A B k
) respectively. Adding the three results,
z
x
x
z
x
y
y
x
to form the determinant of the matrix results in:
A ×B = (A B − A B i
) + ( A B − A B j
) + ( A B − A B k
) (21-3)
y
z
z
y
z
x
x
z
x
y
y
x
as we found before, “the hard way.”
1
41
Chapter 22 Center of Mass, Moment of Inertia
22 Center of Mass, Moment of Inertia
A mistake that crops up in the calculation of moments of inertia, involves the Parallel
Axis Theorem. The mistake is to interchange the moment of inertia of the axis through
the center of mass, with the one parallel to that, when applying the Parallel Axis
Theorem. Recognizing that the subscript “CM” in the parallel axis theorem stands for
“center of mass” will help one avoid this mistake. Also, a check on the answer, to make
sure that the value of the moment of inertia with respect to the axis through the center of
mass is smaller than the other moment of inertia, will catch the mistake.
Center of Mass
Consider two particles, having one and the same mass m, each of which is at a different position
on the x axis of a Cartesian coordinate system.
y
#1
#2
m
m
x
Common sense tells you that the average position of the material making up the two particles is
midway between the two particles. Common sense is right. We give the name “center of mass”
to the average position of the material making up a distribution, and the center of mass of a pair
of same-mass particles is indeed midway between the two particles.
How about if one of the particles is more massive than the other? One would expect the center
of mass to be closer to the more massive particle, and again, one would be right. To determine
the position of the center of mass of the distribution of matter in such a case, we compute a
weighted sum of the positions of the particles in the distribution, where the weighting factor for a
given particle is that fraction, of the total mass, that the particle’s own mass is. Thus, for two
particles on the x axis, one of mass m , at x , and the other of mass m , at x ,
1
1
2
2
y
( x ,0
)
( x ,0 )
1
2
m
m
x
1
2
1
42
Chapter 22 Center of Mass, Moment of Inertia
the position x of the center of mass is given by
m
m
1
2
x =
x +
x (22-1)
1
2
m + m
m + m
1
2
1
2
Note that each weighting factor is a proper fraction and that the sum of the weighting factors is
always 1. Also note that if, for instance, m is greater than m , then the position x of particle 1
1
2
1
will count more in the sum, thus ensuring that the center of mass is found to be closer to the
more massive particle (as we know it must be). Further note that if m = m , each weighting
1
2
1
factor is
, as is evident when we substitute m for both m and m in equation 22-1:
2
1
2
m
m
x =
x +
x
1
2
m + m
m + m
1
1
x =
x +
x
1
2
2
2
x + x
1
2
x =
2
The center of mass is found to be midway between the two particles, right where common sense
tells us it has to be.
The Center of Mass of a Thin Rod
Quite often, when the finding of the position of the center of mass of a distribution of particles is
called for, the distribution of particles is the set of particles making up a rigid body. The easiest
rigid body for which to calculate the center of mass is the thin rod because it extends in only
one dimension. (Here, we discuss an ideal thin rod. A physical thin rod must have some non-
zero diameter. The ideal thin rod, however, is a good approximation to the physical thin rod as
long as the diameter of the rod is small compared to its length.)
In the simplest case, the calculation of the position of the center of mass is trivial. The simplest
case involves a uniform thin rod. A uniform thin rod is one for which the linear mass density µ,
the mass-per-length of the rod, has one and the same value at all points on the rod. The center of
mass of a uniform rod is at the center of the rod. So, for instance, the center of mass of a
uniform rod that extends along the x axis from x = 0 to x = L is at (L/2, 0).
The linear mass density µ, typically called linear density when the context is clear, is a measure
of how closely packed the elementary particles making up the rod are. Where the linear density
is high, the particles are close together.
1
43
Chapter 22 Center of Mass, Moment of Inertia
To picture what is meant by a non-uniform rod, a rod whose linear density is a function of
position, imagine a thin rod made of an alloy consisting of lead and aluminum. Further imagine
that the percentage of lead in the rod varies smoothly from 0% at one end of the rod to 100% at
the other. The linear density of such a rod would be a function of the position along the length of
the rod. A one-millimeter segment of the rod at one position would have a different mass than
that of a one-millimeter segment of the rod at a different position.
People with some exposure to calculus have an easier time understanding what linear density is
than calculus-deprived individuals do because linear density is just the ratio of the amount of
mass in a rod segment to the length of the segment, in the limit as the length of the segment goes
to zero. Consider a rod that extends from 0 to L along the x axis. Now suppose that ms(x) is the
mass of that segment of the rod extending from 0 to x where x ≥ 0 but x < L. Then, the linear
dm
density of the rod at any point x along the rod, is just
s evaluated at the value of x in
dx
question.
Now that you have a good idea of what we mean by linear mass density, we are going to
illustrate how one determines the position of the center of mass of a non-uniform thin rod by
means of an example.
Example 22-1
Find the position of the center of mass of a thin rod that extends from 0 to .890 m
along the x axis of a Cartesian coordinate system and has a linear density given
kg
by
2
µ(x) = 0 650
.
x .
3
m
In order to be able to determine the position of the center of mass of a rod with a given length
and a given linear density as a function of position, you first need to be able to find the mass of
such a rod. To do that, one might be tempted to use a method that works only for the special
case of a uniform rod, namely, to try using m = µ L with L being the length of the rod. The
problem with this is, that µ varies along the entire length of the rod. What value would one use
for µ ? One might be tempted to evaluate the given µ at x = L and use that, but that would be
acting as if the linear density were constant at µ = µ(L). It is not. In fact, in the case at hand,
µ (L) is the maximum linear density of the rod, it only has that value at one point on the rod.
What we can do is to say that the infinitesimal amount of mass dm in a segment dx of the rod is
µ dx. Here we are saying that at some position x on the rod, the amount of mass in the
infinitesimal length dx of the rod is the value of µ at that x value, times the infinitesimal length
dx. Here we don’t have to worry about the fact that µ changes with position since the segment
dx is infinitesimally long, meaning, essentially, that it has zero length, so the whole segment is
essentially at one position x and hence the value of µ at that x is good for the whole segment dx.
dm = µ( x) dx (22-2)
1
44
Chapter 22 Center of Mass, Moment of Inertia
y
dx
(L, 0)
•
z
x
x
dm = µ dx
Now this is true for any value of x, but it just covers an infinitesimal segment of the rod at x. To
get the mass of the whole rod, we need to add up all such contributions to the mass.
Of course, since each dm corresponds to an infinitesimal length of the rod, we will have an
infinite number of terms in the sum of all the dm’s. An infinite sum of infinitesimal terms, is an
integral.
L
∫ dm = ∫ µ(x)dx (22-3)
0
where the values of x have to run from 0 to L to cover the length of the rod, hence the limits on
the right. Now the mathematicians have provided us with a rich set of algorithms for evaluating
integrals, and indeed we will have to reach into that toolbox to evaluate the integral on the right,
but to evaluate the integral on the left, we cannot, should not, and will not turn to such an
algorithm. Instead, we use common sense and our conceptual understanding of what the integral
on the left means. In the context of the problem at hand, ∫dm means “the sum of all the
infinitesimal bits of mass making up the rod.” Now, if you add up all the infinitesimal bits of
mass making up the rod, you get the mass of the rod. So ∫dm is just the mass of the rod, which
we will call m. Equation 22-3 then becomes
L
m = ∫ µ(x) dx (22-4)
0
kg
Replacing µ
µ = .
(x) with the given expression for the linear density
2
0 650
x which I choose
3
m
kg
to write as
2
µ = bx with b being defined by b ≡ 0 650
.
we obtain
3
m
L
m = ∫ bx2dx
0
Factoring out the constant yields
1
45
Chapter 22 Center of Mass, Moment of Inertia
L
m = b ∫ x2dx
0
When integrating the variable of integration raised to a power all we have to do is increase the
power by one and divide by the new power. This gives
L
x3
m = b
3 0
Evaluating this at the lower and upper limits yields
3
L
0 3
m = b
−
3
3
3
L
0 3
m = b
−
3
3
3
bL
m =
3
kg
The value of L is given as 0.890 m and we defined b to be the constant 0 650
.
in the given
3
m
kg
expression for µ,
2
µ = 0 650
.
x , so
3
m
kg
0.650
(0.
m
890
)3
m3
m =
3
m = 0 1527
.
kg
That’s a value that will come in handy when we calculate the position of the center of mass.
Now, when we calculated the center of mass of a set of discrete particles (where a discrete
particle is one that is by itself, as opposed, for instance, to being part of a rigid body) we just
carried out a weighted sum in which each term was the position of a particle times its weighting
factor and the weighting factor was that fraction, of the total mass, represented by the mass of the
particle. We carry out a similar procedure for a continuous distribution of mass such as that
which makes up the rod in question. Let’s start by writing one single term of the sum. We’ll
consider an infinitesimal length dx of the rod at a position x along the length of the rod. The
position, as just stated, is x, and the weighting factor is that fraction of the total mass m of the rod
1
46
Chapter 22 Center of Mass, Moment of Inertia
that the mass dm of the infinitesimal length dx represents. That means the weighting factor is
dm , so, a term in our weighted sum of positions looks like:
m