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Topic 2.1 Extended Q – Center of mass 4PowerPoint Presentation

Topic 2.1 Extended Q – Center of mass 4

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Topic 2.1 Extended Q – Center of mass 4

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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.

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

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

∫

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

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

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

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

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

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

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

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

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

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

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

∫

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

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.

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.

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.

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

+

M·

h(x) =

+

M

(multiply top and

bottom by 2H)

+

mx2

MH2

h(x) =

+

2MH

2mx

H

x

Topic 2.1 ExtendedQ – Center of mass 4

(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

H

x

Topic 2.1 ExtendedQ – Center of mass 4

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)

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

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