Topic 2 1 extended q center of mass 4
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y. y = f ( x ). L. 0. y. x. y = f ( x ). y cm. x. 0. L. x cm. 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.

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Topic 2.1 Extended Q – Center of mass 4

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Topic 2 1 extended q center of mass 4

y

y = f (x)

L

0

y

x

y = f (x)

ycm

x

0

L

xcm

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.

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 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.


Topic 2 1 extended q center of mass 41

y

M

A

M

L

M

V

λ =

ρ =

σ =

y = f (x)

L

0

xcm

x

ycm

mass density

Topic 2.1 ExtendedQ – Center of mass 4

Note that

0 ≤x≤L, and

0 ≤ y ≤ f(L).

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

Linear mass densityλismass per unit length:

Area mass densityσismass per unit area:

Volume mass densityρismass per unit volume:

this is the one you used in chemistry


Topic 2 1 extended q center of mass 42

y

y = f (x)

L

0

xcm

x

L

area of plate

f(x)dx

ycm

A =

0

Topic 2.1 ExtendedQ – Center of mass 4

Note that

0 ≤x≤L, and

0 ≤ y ≤ f(L).

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


Topic 2 1 extended q center of mass 43

L

f(x)dx

0

y

M

A

σ =

M

y = f (x)

L

0

xcm

x

mass density

of plate

ycm

=

Topic 2.1 ExtendedQ – Center of mass 4

Note that

0 ≤x≤L, and

0 ≤ y ≤ f(L).

Why did we use σinstead of λorρ?

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


Topic 2 1 extended q center of mass 44

y

x2

y = f (x)

x5

x4

x3

L

0

x1

Δ m14

Δ m3

Δ m5

Δ m7

Δ m9

Δ m11

Δ m2

Δ m4

Δ m13

Δ m6

Δ m8

Δ m15

Δm1

Δ m10

Δ m12

x

Topic 2.1 ExtendedQ – Center of mass 4

Note that

0 ≤x≤L, and

0 ≤ y ≤ f(L).

etc.

f(xi)

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

= σf(xi)Δx.

Δmi = σΔAi


Topic 2 1 extended q center of mass 45

y

x2

y = f (x)

x4

x5

x3

L

0

x1

m14

m3

m5

m7

m9

m11

m2

m4

m13

m6

m8

m15

m1

m10

m12

x

15

1

M

15

Δmixi

xcm=

1

M

σxif(xi)Δx

=

i=1

i=1

xcmfor

asymmetric plate

15

1

M

L

L

xiΔmi

=

1

M

1

M

xdm

σ xf(x)dx

xcm =

=

i=1

0

0

Topic 2.1 ExtendedQ – Center of mass 4

Note that

0 ≤x≤L, and

0 ≤ y ≤ f(L).

etc.

f(xi)

Δx

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

As Δx→ 0,Δmi = σf(xi)Δx becomes

dm = σf(x)dx

so that


Topic 2 1 extended q center of mass 46

y

Δm7

Δm6

Δm5

Δm5

y = f (x)

Δm4

y4

y3

Δm3

L

0

y2

Δm2

y1

Δm1

x

Topic 2.1 ExtendedQ – Center of mass 4

Note that

0 ≤x≤L,

0 ≤ y ≤ f(L).

etc…

Δy

L - xi

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

= σ(L-xi)Δy.

Δmi = σΔAi


Topic 2 1 extended q center of mass 47

y

Δm7

Δm6

Δm5

Δm5

y = f (x)

Δm4

y4

y3

Δm3

L

0

Δm2

y1

Δm1

x

7

1

M

7

Δmiyi

ycm=

1

M

σyi(L-xi)Δy

=

i=1

i=1

ycmfor

asymmetric plate

7

1

M

f(L)

f(L)

yiΔmi

=

1

M

1

M

ydm

σ y(L-x)dy

ycm =

=

i=1

0

0

Topic 2.1 ExtendedQ – Center of mass 4

Note that

0 ≤x≤L,

0 ≤ y ≤ f(L).

Δy

L - xi

y2

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

As Δy→ 0,Δmi = σ(L-xi)Δy becomes

dm = σ(L-x)dy

so that


Topic 2 1 extended q center of mass 48

y

y = f (x)

L

0

xcm

x

the mass increment

ycm

Topic 2.1 ExtendedQ – Center of mass 4

Note that

0 ≤x≤L,

0 ≤ y ≤ f(L).

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

Area mass increment: dm = σdA

Volume mass increment: dm = ρdV


Topic 2 1 extended q center of mass 49

y

y = f (x)

L

0

xcm

x

n

n

L

miyi

mixi

1

M

xdm

xcm=

i=1

i=1

0

ycm

1

M

1

M

f(L)

xcm=

ycm=

1

M

ydm

ycm=

0

cm, discrete

cm, continuous

Topic 2.1 ExtendedQ – Center of mass 4

 Note that

0 ≤x≤L, and

0 ≤ y ≤ f(L).

It is crucial that horizontal rectangles are used for ycm. Why?

It is crucial that vertical rectangles are used for xcm. Why?

We then rewrite the discrete formulas for xcm and ycm:

dm

dm


Topic 2 1 extended q center of mass 410

a

b

a

b

y = x

y = x.

Topic 2.1 ExtendedQ – Center of mass 4

y

a

Note

0 ≤x≤b,

0 ≤ y ≤ a.

y = f (x)

x

b

0

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


Topic 2 1 extended q center of mass 411

a

b

b

= x

1

M

xdm

xcm=

0

a

b

= σ xdx

a

b

y = x

Topic 2.1 ExtendedQ – Center of mass 4

y

a

Note

0 ≤x≤b,

0 ≤ y ≤ a.

y

dA = ydx

x

dx

b

0

To find xcm use vertical rectangles :

dm,

where dm = σdA.

Then…

dm = σdA

= σydx


Topic 2 1 extended q center of mass 412

M

A

1

3

σ=

- 03

σa

bM

1

3

b3

=

σa

bM

b

x2dx

2M

ba

=

2Mab2

3baM

a

b

=

σ=

b

1

M

xσ xdx

σab2

3M

0

1

2

=

xcm=

A= ba

2b

3

xcm triangle

a

b

0

σab3

3bM

b

1

M

xcm =

y = x

xdm

xcm=

=

0

Topic 2.1 ExtendedQ – Center of mass 4

y

a

Note

0 ≤x≤b,

0 ≤ y ≤ a.

x

b

0

Then

and

But

. Thus

so that


Topic 2 1 extended q center of mass 413

x

y is the variable

a

1

M

ydm

ycm=

0

a

b

b

a

y = x,

x = y

b

a

a

b

dm = σ(b - y)dy.

y = x

Topic 2.1 ExtendedQ – Center of mass 4

y

a

Note

0 ≤x≤b,

0 ≤ y ≤ a.

Always match the variables before integrating.

b - x

dy

dA = (b - x) dy

x

b

0

To find ycm use horizontal rectangles :

dm,

where dm = σdA.

Then…

since

dm = σdA

so that

= σ(b - x)dy


Topic 2 1 extended q center of mass 414

a

b

a

1

M

yσ(b - y)dy

=

1

a

a

σb

M

a

(y - y2)dy

σb

M

1

3a

=

0

- y3

=

0

0

M

A

σ=

1

2

y2

2M

ba

2Mba2

6baM

=

σ=

σba2

6M

1

2

ycm=

A= ba

1a

3

ycm triangle

a

b

σba2

6M

a

1

M

ycm =

y = x

ydm

ycm=

=

0

Topic 2.1 ExtendedQ – Center of mass 4

y

a

Note

0 ≤x≤b,

0 ≤ y ≤ a.

x

b

0

Then

and

But

. Thus

so that


Topic 2 1 extended q center of mass 415

a

3

a

3

a

3

xcm

b

3

b

3

b

3

ycm

1a

3

2b

3

xcm triangle

ycm triangle

xcm =

ycm =

Now you may use triangular plates just like squares, rectangles, circles, etc. to solve symmetric plate sums and differences.

Topic 2.1 ExtendedQ – Center of mass 4

y

a

x

b

0

To summarize…

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 416

H

H

2

Topic 2.1 ExtendedQ – 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 center, h = H/2.

(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.


Topic 2 1 extended q center of mass 417

H

x

But the mass of the soda in the can is proportional to x so that

msoda(x) = x.

m

H

Topic 2.1 ExtendedQ – 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?

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.

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


Topic 2 1 extended q center of mass 418

H

x

m

H

m

H

·x·

·x

Topic 2.1 ExtendedQ – 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?

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

x

2

H

2

+

h(x) =

+

M

(multiply top and

bottom by 2H)

+

mx2

MH2

h(x) =

+

2MH

2mx


Topic 2 1 extended q center of mass 419

H

x

Topic 2.1 ExtendedQ – 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?

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

(get into product form)

h(x) = (MH2 + mx2)·(2MH + 2mx)-1

(take the derivative)

(2mx)

·(2MH + 2mx)-1

h'(x) =

(MH2 + mx2)

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

+

·2m


Topic 2 1 extended q center of mass 420

H

x

Topic 2.1 ExtendedQ – 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?

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

(set the derivative equal

to zero to minimize)

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

2m

+ (MH2 + mx2)·(-1)(2MH + 2mx)-2·

-1

2m(MH2 + mx2)(2MH + 2mx)-2 = 2mx(2MH + 2mx)-1

(MH2 + mx2) = x(2MH + 2mx)


Topic 2 1 extended q center of mass 421

H

a

b

c

x

-2MH ± √4M2H2 + 4mMH 2

x =

2m

-2MH ± 2H√M(M+ m)

H

m

√M(M+m) - M

=

=

2m

Topic 2.1 ExtendedQ – 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?

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

(MH2 + mx2) = x(2MH + 2mx)

MH2 + mx2 = 2MHx + 2mx2

mx2+ 2MHx - MH2 = 0


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