Topic 2.1 Extended
Q – Center of mass 4
In the last section you learned how to find the cm of
bodies built of symmetric plates.
In this section you’ll learn a general way to find the
cm of a plate of any shape.
Consider the following generalized plate, which has an
edge cut in the shape of the function y = f(x):
Suppose we lay our plate out flat, and slide rulers to
find xcm and ycm as we did before:
Note
y that xcm is found using a ruler that is
perpendicular to the x-axis.
And ycm
uses a
ruler that
is
perpendicular to the
y-axis.
ycm
0
xcm
L
x
Topic 2.1 Extended
Q – Center of mass 4
y
Note that
0 ≤ x ≤ L, and
0 ≤ y ≤ f(L).
ycm
0
xcm
L
x
Now we need to talk about three types of mass density.
M
Linear mass density λ is mass per unit length: λ =
L
σ = M
Area mass density σ is mass per unit area:
A
M
Volume mass density ρ is mass per unit volume: ρ =
V
this is the one you used in chemistry
mass density
Topic 2.1 Extended
Q – Center of mass 4
y
Note that
0 ≤ x ≤ L, and
0 ≤ y ≤ f(L).
ycm
0
xcm
L
x
The first step in solving asymmetric plates is to find
the area of the plate.
∫
This is given by
L
A =
0
f(x)dx
area of plate
Topic 2.1 Extended
Q – Center of mass 4
y
Note that
0 ≤ x ≤ L, and
0 ≤ y ≤ f(L).
Why did we use σ
instead of λ or ρ?
ycm
0
xcm
L
x
The second step is to find the mass density of the
plate.
This is given by
σ = M
A
=
M
∫
L
f(x)dx
0
mass density
of plate
Topic 2.1 Extended
Q – Center of mass 4
y
Note that
0 ≤ x ≤ L, and
0 ≤ y ≤ f(L).
etc.
x4
x3
x2
x1
0
Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ
m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 m11 m12 m13m14 m15
L
x
f(xi)
x5
Δx
To find xcm we divide our plate into many vertically
oriented rectangles of material having mass Δmi.
Since our rectangles are vertical, all of the mass Δmi
in each rectangle is, on average, the same distance xi
from the origin.
Each rectangle has dimensions Δx by f(xi) so that
Δmi = σΔAi = σf(xi)Δx.
Topic 2.1 Extended
Q – Center of mass 4
y
Note that
0 ≤ x ≤ L, and
0 ≤ y ≤ f(L).
etc.
x5
f(xi)
x4
x3
x2
x1
0
m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 m11 m12 m13m14 m15
L
x
Δx
Now we invoke our xcm formula for discrete masses, and
tailor it slightly with the substitution Δmi = σf(xi)Δx:
15
15
∫
∫
15
1 ∑ σx f(x )Δx
xcm = 1 ∑ Δmixi = 1 ∑ xiΔmi =
i
M i=1 i
M i=1
M i=1
As Δx → 0, Δmi = σf(xi)Δx becomes dm = σf(x)dx
so that
L
L
xcm for
1 x dm
1 σ x f(x)dx
asymmetric plate
xcm =
=
M
M
0
0
Topic 2.1 Extended
Q – Center of mass 4
y
etc…
y1
y2
0
y3
y4
L
Δm7 Note that
Δm6 0 ≤ x ≤ L,
Δm5 0 ≤ y ≤ f(L).
Δm5
Δy
Δm4
Δm3
L - xi
Δm2
Δm1
x
To find ycm we divide our plate into many horizontally
oriented rectangles of material having mass Δmi.
Since our rectangles are horizontal, all of the mass
Δmi in each rectangle is, on average, the same distance
yi from the origin.
Each rectangle has dimensions Δy by L - xi so that
Δmi = σΔAi = σ(L-xi)Δy.
Topic 2.1 Extended
Q – Center of mass 4
y
y1
y2
y3
Δm7
Δm6
Δm5
Δm5
Δm4
Δm3
Δm2
Δm1
y4
L
0
Note that
0 ≤ x ≤ L,
0 ≤ y ≤ f(L).
Δy
L - xi
x
Now we invoke our ycm formula for discrete masses, and
tailor it slightly with the substitution Δmi = σ(L-xi)Δy:
7
7
7
1 ∑ σy (L-x )Δy
ycm = 1 ∑ Δmiyi = 1 ∑ yiΔmi =
i
M i=1 i
M i=1
M i=1
As Δy → 0, Δmi = σ(L-xi)Δy becomes dm = σ(L-x)dy so that
f(L)
f(L)
ycm for
1 y dm
1 σ y (L-x)dy
asymmetric plate
ycm =
=
M
M
0
0
∫
∫
Topic 2.1 Extended
Q – Center of mass 4
y
Note that
0 ≤ x ≤ L,
0 ≤ y ≤ f(L).
ycm
0
xcm
L
x
The formulas, and their derivations, are rather
complicated and hard to memorize. So let’s attack this
problem from a different direction:
Now we define the mass increment dm. We will use dm
in place of the point mass:
Linear mass increment: dm = λdℓ
the mass increment
Area mass increment : dm = σdA
Volume mass increment: dm = ρdV
Topic 2.1 Extended
Q – Center of mass 4
y
 Note that
0 ≤ x ≤ L, and
0 ≤ y ≤ f(L).
It is crucial that
horizontal
rectangles
vertical rectangles
are used for yxcm.
Why?
ycm
0
L
xcm
x
We then rewrite the discrete formulas for xcm and ycm:
n
xcm
= 1 ∑ mixi
M i=1
ycm
= 1 ∑ miyi
M i=1
xcm =
1
M
ycm =
1
M
n
cm, discrete
∫
∫ydm
L
xdm
0
dm
f(L)
0
dm
cm, continuous
Topic 2.1 Extended
Q – Center of mass 4
y
a
Note
0 ≤ x ≤ b,
0 ≤ y ≤ a.
0
b
x
The important thing is that we can use these formulas
correctly. Let’s look at a concrete example:
Suppose we wish to find the center of mass of a right
triangle having a mass M, a base b and an altitude a:
The equation of the line y = f(x) is
y = a x.
b
Topic 2.1 Extended
Q – Center of mass 4
y
dA = ydx
a
dx
0
Note
0 ≤ x ≤ b,
0 ≤ y ≤ a.
To find xcm use vertical rectangles :
xcm =
Then…
1
M
∫xdm
b
0
dm = σdA = σydx
dm,
= ab x
y
b
where dm = σdA.
= σ a xdx
b
x
Topic 2.1 Extended
Q – Center of mass 4
y
a
Then
Note
0 ≤ x ≤ b,
0 ≤ y ≤ a.
0
xcm =
=
=
b
∫xdm
∫xσ ab xdx
σa
x dx
bM ∫
b
1
M
1
M
M
But σ =
A
0
b
0
b
2
0
σa 1 3
= bM 3 b σab3
= 3bM
x
1 3
0
3
and
A
= 1 ba
2
2M
so that σ = ba . Thus
σab2
2Mab2
=
xcm = 3M
3baM
xcm =
2b
3
xcm triangle
Topic 2.1 Extended
Q – Center of mass 4
y
a
Note
0 ≤ x ≤ b,
0 ≤ y ≤ a.
b-x
dA = (b - x) dy
x
Always match the
variables before
dy
integrating.
b
0
x
To find ycm use horizontal rectangles :
ycm =
Then…
1
M
∫ydm
a
0
dm,
y is the
dm = σdA
variable
= σ(b - x)dy
where dm = σdA.
since y = a x, x = b y
b
a
so that
dm = σ(b - b y)dy.
a
Topic 2.1 Extended
Q – Center of mass 4
y
a
Note
0 ≤ x ≤ b,
0 ≤ y ≤ a.
0
Then ycm =
=
=
∫ydm
∫yσ(b - ba y)dy
∫(y - 1a y )dy
1
M
b
a
0
a
1
M 0
a
σb
M 0
2
σb 1 2
1 3
y
y
= M 2
3a
σba2
= 6M
a
0
x
M
But σ =
A
and
A
= 1 ba
2
2M
so that σ = ba . Thus
σba2
2Mba2
=
ycm = 6M
6baM
ycm =
1a
3
ycm triangle
Now you may use triangular plates just like
Topic 2.1
Extended
squares, rectangles,
circles,
etc. to solve
symmetric
sums and
Q –plate
Center
of differences.
mass 4
y
a
a
3
a
3
ycm
a
3
b
0
b
3
To summarize…
xcm =
2b
3
b
3
xcm triangle
xcm
ycm =
x
b
3
1a
3
ycm triangle
The cm of a right triangle is located one third of
the way from the right angle along each side.
Topic 2.1 Extended
Q – Center of mass 4
EXERCISE 12*: A can of mass M and height H is filled
with soda of mass m. We punch small holes in the top
and bottom so that the soda drains slowly. As it
drains, the height h of the cm of the can-soda combo
will change.
(a) Where is the cm before the
soda begins to drain?
Since the cm of the can, and the
cm of the soda are located in the
H
center, h = H/2.
H
(b) Where is the cm after the
2
soda has finished draining?
Since the cm of the can is still
located in the center, h = H/2.
Topic 2.1 Extended
Q – Center of mass 4
EXERCISE 12*: A can of mass M and height H is filled
with soda of mass m. We punch small holes in the top
and bottom so that the soda drains slowly. As it
drains, the height h of the cm of the can-soda combo
x
will change.
(c) If x is the height of the
remaining soda at any time, at what
value of x is the cm of the can-soda
combo the lowest?
H
Note that the cm of the soda is at
x/2, and the cm of the can is at H/2.
Note that mass of the can is always
mcan = M.
But the mass of the soda in the can
is proportional to x so that
msoda(x) = mH x.
You can check that the latter works by substituting 0 and H for x.
Topic 2.1 Extended
Q – Center of mass 4
EXERCISE 12*: A can of mass M and height H is filled
with soda of mass m. We punch small holes in the top
and bottom so that the soda drains slowly. As it
drains, the height h of the cm of the can-soda combo
will change.
(c) If x is the height of the
remaining soda at any time, at what
value of x is the cm of the can-soda
combo the lowest?
H
Now we can obtain the cm h(x) of
x
the two-body combo:
h(x) =
h(x) =
M· H2 +
+
M
MH
2MH
2
x
m
·x·
2
H
m
H ·x
+ mx2
+ 2mx
(multiply top and
bottom by 2H)
Topic 2.1 Extended
Q – Center of mass 4
EXERCISE 12*: A can of mass M and height H is filled
with soda of mass m. We punch small holes in the top
and bottom so that the soda drains slowly. As it
drains, the height h of the cm of the can-soda combo
will change.
(c) If x is the height of the
remaining soda at any time, at what
value of x is the cm of the can-soda
combo the lowest?
H
Now we can obtain the cm h(x) of
x
the two-body combo:
(get into product form)
h(x) = (MH
2
+ mx2)·(2MH + 2mx)-1
(take the derivative)
h'(x) = (2mx)·(2MH + 2mx)-1
+ (MH 2 + mx2)(-1)(2MH + 2mx)-2 ·2m
Topic 2.1 Extended
Q – Center of mass 4
EXERCISE 12*: A can of mass M and height H is filled
with soda of mass m. We punch small holes in the top
and bottom so that the soda drains slowly. As it
drains, the height h of the cm of the can-soda combo
will change.
(c) If x is the height of the
remaining soda at any time, at what
value of x is the cm of the can-soda
combo the lowest?
H
Now we can obtain the cm h(x) of
x
the two-body combo:
(set the derivative equal
to zero to minimize)
0 = 2mx(2MH + 2mx)-1
+ (MH 2 + mx2)·(-1)(2MH + 2mx)-2· 2m
2m(MH
2
-1
+ mx2)(2MH + 2mx)-2 = 2mx(2MH + 2mx)-1
(MH
2
+ mx2) = x(2MH + 2mx)
Topic 2.1 Extended
Q – Center of mass 4
EXERCISE 12*: A can of mass M and height H is filled
with soda of mass m. We punch small holes in the top
and bottom so that the soda drains slowly. As it
drains, the height h of the cm of the can-soda combo
will change.
(c) If x is the height of the
remaining soda at any time, at what
value of x is the cm of the can-soda
combo the lowest?
H
Now we can obtain the cm h(x) of
x
the two-body combo:
(MH 2 + mx2) = x(2MH + 2mx)
2 = 2MHx + 2mx2
+
mx
a
b
c
2
2
mx + 2MHx - MH = 0
-2MH ± √4M 2H 2 + 4mMH 2
x =
2m
-2MH ± 2H√M(M + m)
H
= m
=
2m
MH
2
√M(M +m)
- M