topic 2.1 extended q – center of mass 4

Topic 2.1 ExtendedQ – 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.

y = f (x)

x

y

0 L

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:

y = f (

x)

0 Lx

y

xcm

ycm

Note that xcm is found using a ruler that is perpendicular to the x-axis.

And ycm uses a ruler that is perpendic-ular to the y-axis.


Note that 0 ≤ x ≤ L, and 0 ≤ y ≤ f(L).

Now we need to talk about three types of mass density.

y = f (

x)

0 Lx

y

xcm

ycm

Linear mass density λ is mass per unit length:

Area mass density σ is mass per unit area:

Volume mass density ρ is mass per unit volume:

λ = ML

σ = MA

ρ = MV

mass densitythis is the one you used in chemistry



The first step in solving asymmetric plates is to find the area of the plate. This is given by

y = f (

x)

0 Lx

y

xcm

ycm

area of plate

∫A =0

Lf(x)dx



The second step is to find the mass density of the plate. This is given by

y = f (

x)

0 Lx

y

xcm

ycm

mass densityof plateσ = M

A ∫0

L

f(x)dx

M=

Why did we use σ instead of λ or ρ?



To find xcm we divide our plate into many vertically oriented rectangles of material having mass Δmi.

y = f (

x)

0 Lx

y

Δm1

Δ m2

Δ m3

Δ m4

Δ m15Δ m5

Δ m6

Δ m7

Δ m8

Δ m9

Δ m10

Δ m11

Δ m12

Δ m13

Δ m14

Since our rectangles are vertical, all of the mass Δmi in each rectangle is, on average, the same distance xi from the origin.

x1

x2

x3

x4

x5

etc.

Each rectangle has dimensions Δx by f(xi) so that Δmi = σΔAi = σf(xi)Δx.

Δx

f(xi)



Now we invoke our xcm formula for discrete masses, and tailor it slightly with the substitution Δmi = σf(xi)Δx:

y = f (

x)

0 Lx

y

xcm = Δmixi∑i=1

151M

m1m2 m3 m4 m15m5 m6 m7 m8 m9 m10

m11 m12 m13m14

x1

x2

x3

x4

x5

etc.

= σxif(xi)Δx∑i=1

151M= xiΔmi∑

i=1

151M

As Δx → 0, Δmi = σf(xi)Δx becomes dm = σf(x)dx so that

1M∫xcm =

0

Lx dm 1

M∫=0

Lσ x f(x)dx

xcm for asymmetric plate

f(xi)

Δx


Note that 0 ≤ x ≤ L, 0 ≤ y ≤ f(L).

To find ycm we divide our plate into many horizontally oriented rectangles of material having mass Δmi.

y = f (

x)

0 Lx

y

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.

L - xi

Δm1

Δm2

Δm3

Δm4

Δm5

Δm5

Δm6

Δm7

etc…Δyy4

y3

y2

y1


Note that0 ≤ x ≤ L,0 ≤ y ≤ f(L).y =

f (x)

0 Lx

y

Δm1

Δm2

Δm3

Δm4

Δm5

Δm5

Δm6

Δm7

Now we invoke our ycm formula for discrete masses, and tailor it slightly with the substitution Δmi = σ(L-xi)Δy:

ycm = Δmiyi∑i=1

71M

= σyi(L-xi)Δy∑i=1

71M= yiΔmi∑

i=1

71M

As Δy → 0, Δmi = σ(L-xi)Δy becomes dm = σ(L-x)dy so that

1M∫ycm =

0

f(L)y dm 1

M∫=0

f(L)σ y (L-x)dy

ycm for asymmetric plate

L - xi

Δyy4

y3

y2

y1


Note that 0 ≤ x ≤ L, 0 ≤ y ≤ f(L).

Now we define the mass increment dm. We will use dm in place of the point mass:

y = f (

x)

0 Lx

y

xcm

ycm

the mass incrementLinear mass increment: dm = λdℓ

Area mass increment : dm = σdA

Volume mass increment: dm = ρdV

The formulas, and their derivations, are rather complicated and hard to memorize. So let’s attack this problem from a different direction:



We then rewrite the discrete formulas for xcm and ycm:

y = f (

x)

0 Lx

y

xcm

ycm

ycm = miyi∑i=1

n1M

xcm = mixi∑i=1

n1M

cm, discrete

xcm = xdm∫0

L1M

ycm = ydm∫0

f(L)1M

dm

dm

cm, continuous

It is crucial that vertical rectangles are used for xcm.

Why?

It is crucial that horizontal rectangles

are used for ycm. Why?


Note0 ≤ x ≤ b, 0 ≤ y ≤ a.

The important thing is that we can use these formulas correctly. Let’s look at a concrete example:

y = f (x)

0 bx

ya

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 = x.ab

y = xab


Note0 ≤ x ≤ b, 0 ≤ y ≤ a.

To find xcm use vertical rectangles : 0 b

x

ya

where dm = σdA.

y = xab

xcm = xdm∫0

b1M dm,

y

dx

= xab

Then…dm = σdA = σydx

dA

= y

dx= σ xdxa

b


Note0 ≤ x ≤ b, 0 ≤ y ≤ a.

Then0 b

x

ya

y = xab

xcm = xdm∫0

b1M

xσ xdxab∫

0

b1M =

x2dx∫0

bσabM =

σabM = b31

3- 031

3σab3

3bM =

But MAσ = 1

2A = baand

so that 2Mbaσ = . Thus

σab2

3Mxcm =2Mab2

3baM=

2b3xcm =

xcm triangle


Note0 ≤ x ≤ b, 0 ≤ y ≤ a.

To find ycm use horizontal rectangles : 0 b

x

ya

where dm = σdA.

y = xab

ycm = ydm∫0

a1M dm,

dy

Then…dm = σdA

= σ(b - x)dy

dA = (b - x) dy

y = x,ab

y is the variable

since x = yba

so that

dm = σ(b - y)dy.ba

b - xx

Always match the variables before

integrating.


Note0 ≤ x ≤ b, 0 ≤ y ≤ a.

Then0 b

x

ya

y = xab

ycm = ydm∫0

a1M

σba2

6M =

But MAσ = 1

2A = baand

so that 2Mbaσ = . Thus

σba2

6Mycm =2Mba2

6baM=

1a3ycm =

ycm triangle

yσ(b - y)dy∫0

a1M =

ba

(y - y2)dy∫0

aσbM =

1a

σbM = y21

2- y31

3a0

a


To summarize…

0 bx

ya

1a3ycm =

ycm triangle2b3xcm =

xcm triangle

xcm

ycm

a3

a3

a3

b3

b3

b3

The cm of a right triangle is located one third of the way from the right angle along each side.

Now you may use triangular plates just like squares, rectangles, circles, etc. to solve

symmetric plate sums and differences.


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 center, h = H/2.

H

(b) Where is the cm after the soda has finished draining?

Since the cm of the can is still located in the center, h = H/2.

H2


(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

x

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.

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) = x.

mH

You can check that the latter works by substituting 0 and H for x.



xNow we can obtain the cm h(x) of the two-body combo:

h(x) =


M·H2 + x2

mH·x·

M + mH·x

h(x) =MH

2 + mx2

2MH + 2mx

(multiply top and bottom by 2H)





h(x) = (MH 2 + mx2)·(2MH + 2mx)-1

h'(x) =+ ·2m

(get into product form)

(take the derivative)(2mx)·(2MH + 2mx)-1

(MH 2 + mx2)(-1)(2MH + 2mx)-2





0 = 2mx(2MH + 2mx)-1

+ (MH 2 + mx2)·(-1)(2MH + 2mx)-2· 2m

(set the derivative equal to zero to minimize)

2m(MH 2 + mx2)(2MH + 2mx)-2 = 2mx(2MH + 2mx)-1

-1

(MH 2 + mx2) = x(2MH + 2mx)





(MH 2 + mx2) = x(2MH + 2mx)

MH 2 + mx2 = 2MHx + 2mx2

mx2 + 2MHx - MH2 = 0

x =-2MH ± √4M

2H 2 + 4mMH 2

2m

=-2MH ± 2H√M(M + m)

2m√M(M +m) - M= H

m

a b c

topic 2.1 extended q – center of mass 4

Documents

mass mi

point mass

area mass density

volume mass density

linear mass density

volume mass increment

linear mass increment

types of mass density