6.3. DENSITY, MASS, AND CENTER OF MASS
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function ρ(x) tells us the density distribution along the bar, measured in g/cm. If we slice
the bar into small sections, this enables us to think of the bar as holding a collection of
adjacent point-masses. For a slice of thickness △x at location x i , note that the mass of the
slice, m i , satisfies m i ≈ ρ(x i )△x.
Taking n slices of the bar, we can approximate its center of mass by
x ≈
x 1 · ρ(x 1 )△x + x 2 · ρ(x 2 )△x + · · · + x n · ρ(x n )△x
ρ(x 1 )△x + ρ(x 2 )△x + · · · + ρ(x n )△x
.
Rewriting the sums in sigma notation, it follows that
x ≈
n
i=1 x i · ρ(x i )△x
n
i=1 ρ(x i )△x
.
(6.4)
Moreover, it is apparent that the greater the number of slices, the more accurate our
estimate of the balancing point will be, and that the sums in Equation (6.4) can be viewed
as Riemann sums. Hence, in the limit as n → ∞, we find that the center of mass is given
by the quotient of two integrals.
For a thin rod of density ρ(x) distributed along an axis from x = a to x = b, the
center of mass of the rod is given by
x =
b
a
x ρ(x) dx
b
a
ρ(x) dx
.
Note particularly that the denominator of x is the mass of the bar, and that this
quotient of integrals is simply the continuous version of the weighted average of locations,
x, along the bar.
Activity 6.9.
Consider a thin bar of length 20 cm whose density is distributed according to the
function ρ(x) = 4 + 0.1x, where x = 0 represents the left end of the bar. Assume that ρ
is measured in g/cm and x is measured in cm.
(a) Find the total mass, M, of the bar.
(b) Without doing any calculations, do you expect the center of mass of the bar to
be equal to 10, less than 10, or greater than 10? Why?
(c) Compute x, the exact center of mass of the bar.
(d) What is the average density of the bar?
(e) Now consider a different density function, given by p(x) = 4e 0.020732x , also for
a bar of length 20 cm whose left end is at x = 0. Plot both ρ(x) and p(x) on
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function ρ(x) tells us the density distribution along the bar, measured in g/cm. If we slice
the bar into small sections, this enables us to think of the bar as holding a collection of
adjacent point-masses. For a slice of thickness △x at location x i , note that the mass of the
slice, m i , satisfies m i ≈ ρ(x i )△x.
Taking n slices of the bar, we can approximate its center of mass by
x ≈
x 1 · ρ(x 1 )△x + x 2 · ρ(x 2 )△x + · · · + x n · ρ(x n )△x
ρ(x 1 )△x + ρ(x 2 )△x + · · · + ρ(x n )△x
.
Rewriting the sums in sigma notation, it follows that
x ≈
n
i=1 x i · ρ(x i )△x
n
i=1 ρ(x i )△x
.
(6.4)
Moreover, it is apparent that the greater the number of slices, the more accurate our
estimate of the balancing point will be, and that the sums in Equation (6.4) can be viewed
as Riemann sums. Hence, in the limit as n → ∞, we find that the center of mass is given
by the quotient of two integrals.
For a thin rod of density ρ(x) distributed along an axis from x = a to x = b, the
center of mass of the rod is given by
x =
b
a
x ρ(x) dx
b
a
ρ(x) dx
.
Note particularly that the denominator of x is the mass of the bar, and that this
quotient of integrals is simply the continuous version of the weighted average of locations,
x, along the bar.
Activity 6.9.
Consider a thin bar of length 20 cm whose density is distributed according to the
function ρ(x) = 4 + 0.1x, where x = 0 represents the left end of the bar. Assume that ρ
is measured in g/cm and x is measured in cm.
(a) Find the total mass, M, of the bar.
(b) Without doing any calculations, do you expect the center of mass of the bar to
be equal to 10, less than 10, or greater than 10? Why?
(c) Compute x, the exact center of mass of the bar.
(d) What is the average density of the bar?
(e) Now consider a different density function, given by p(x) = 4e 0.020732x , also for
a bar of length 20 cm whose left end is at x = 0. Plot both ρ(x) and p(x) on