360
6.3. DENSITY, MASS, AND CENTER OF MASS
(d) Next, suppose that we place four books on the shelf, but of varying weights:
at x 1 = 0 a 2-lb book, at x 2 = 2 a 3-lb book, and x 3 = 4 a 1-lb book, and
at x 4 = 6 a 1-lb book. Use a weighted average of the locations to find x, the
balancing point of the shelf. How does the balancing point in this scenario
compare to that found in (b)?
(e) What happens if we change the location of one of the books? Say that we keep
everything the same in (d), except that x 3 = 5. How does x change?
(f) What happens if we change the weight of one of the books? Say that we keep
everything the same in (d), except that the book at x 3 = 4 now weighs 2 lbs.
How does x change?
(g) Experiment with a couple of different scenarios of your choosing where you
move the location of one of the books to the left, or you decrease the weight of
one of the books.
(h) Write a couple of sentences to explain how adjusting the location of one of the
books or the weight of one of the books affects the location of the balancing
point of the shelf. Think carefully here about how your changes should be
considered relative to the location of the balancing point x of the current
scenario.
⊳
Center of Mass
In Activity 6.8, we saw that the balancing point of a system of point-masses 1 (such as
books on a shelf) is found by taking a weighted average of their respective locations. In
the activity, we were computing the center of mass of a system of masses distributed along
an axis, which is the balancing point of the axis on which the masses rest.
For a collection of n masses m 1 , . . ., m n that are distributed along a single axis at the
locations x 1 , . . ., x n , the center of mass is given by
x =
x 1 m 1 + x 2 m 2 + · · · x n m n
m 1 + m 2 + · · · + m n
.
What if we instead consider a thin bar over which density is distributed continuously?
If the density is constant, it is obvious that the balancing point of the bar is its midpoint.
But if density is not constant, we must compute a weighted average. Let’s say that the
1 In the activity, we actually used weight rather than mass. Since weight is computed by the gravitational
constant times mass, the computations for the balancing point result in the same location regardless of whether
we use weight or mass, since the gravitational constant is present in both the numerator and denominator of
the weighted average.
6.3. DENSITY, MASS, AND CENTER OF MASS
(d) Next, suppose that we place four books on the shelf, but of varying weights:
at x 1 = 0 a 2-lb book, at x 2 = 2 a 3-lb book, and x 3 = 4 a 1-lb book, and
at x 4 = 6 a 1-lb book. Use a weighted average of the locations to find x, the
balancing point of the shelf. How does the balancing point in this scenario
compare to that found in (b)?
(e) What happens if we change the location of one of the books? Say that we keep
everything the same in (d), except that x 3 = 5. How does x change?
(f) What happens if we change the weight of one of the books? Say that we keep
everything the same in (d), except that the book at x 3 = 4 now weighs 2 lbs.
How does x change?
(g) Experiment with a couple of different scenarios of your choosing where you
move the location of one of the books to the left, or you decrease the weight of
one of the books.
(h) Write a couple of sentences to explain how adjusting the location of one of the
books or the weight of one of the books affects the location of the balancing
point of the shelf. Think carefully here about how your changes should be
considered relative to the location of the balancing point x of the current
scenario.
⊳
Center of Mass
In Activity 6.8, we saw that the balancing point of a system of point-masses 1 (such as
books on a shelf) is found by taking a weighted average of their respective locations. In
the activity, we were computing the center of mass of a system of masses distributed along
an axis, which is the balancing point of the axis on which the masses rest.
For a collection of n masses m 1 , . . ., m n that are distributed along a single axis at the
locations x 1 , . . ., x n , the center of mass is given by
x =
x 1 m 1 + x 2 m 2 + · · · x n m n
m 1 + m 2 + · · · + m n
.
What if we instead consider a thin bar over which density is distributed continuously?
If the density is constant, it is obvious that the balancing point of the bar is its midpoint.
But if density is not constant, we must compute a weighted average. Let’s say that the
1 In the activity, we actually used weight rather than mass. Since weight is computed by the gravitational
constant times mass, the computations for the balancing point result in the same location regardless of whether
we use weight or mass, since the gravitational constant is present in both the numerator and denominator of
the weighted average.