The Mass Center

The Mass Center

Alastair McLean

January 5, 2009

The Mass Center

1 The Mass CenterLearning ObjectivesNew SymbolsCenter of Mass MotionDistributed massesSummary

The Mass Center

Copyright

This work is licensed under a Creative CommonsAttribution-Noncommercial-Share Alike 3.0 Unported License.

http://creativecommons.org/licenses/by-nc-sa/3.0/

The Mass Center

Reading Assignment

Knight 12.2

The Mass Center

Learning Objectives

To introduce the concept of the center of mass = CM.

To examine how the CM behaves in 1D collisions.

The Mass Center

New Symbols

Table: New Symbols

xc 1D center or mass m

~rc 3D center of mass m

fi mass fraction

The Mass Center

New Symbols

x x1 2

x0

For a system with two masses, m1 and m2, the mass center is:

xc = f1x1 + f2x2 =m1

Mx1 +

m2

Mx2

whereM =

∑i

mi = m1 + m2

If the fractional masses are f1 = 0.98 and f2 = 0.02 then xc ≈ x1.

If the fractional masses are f1 = 0.02 and f2 = 0.98 then xc ≈ x2.

The Mass Center

New Symbols

Example 1. Find the CM for two point masses located on thex-axis if the masses are equal and show that it lies mid-waybetween the two masses.

x x1 2

x0 xc

The Mass Center

New Symbols

Example 2. Ball 1 and 2 each have mass ≡ m. They are locatedat ±x◦ and they start moving towards the origin at ±v . (a) Derivean expression for xc(t) both before/after the collision. (b) Showthat the relative speeds before/after the collision satisfy:v12 = −u12

The Mass Center

New Symbols

vm m

collision

v

vvt

c

0

-x xoox =0

1 2

The Mass Center

New Symbols

Example 3. (a) Find the speed of the CM vc , before and after thecollision, if m1 = 2m and m2 = m. These objects start equidistantfrom the origin ±x◦ and move towards one another at speed v . (b)Also find the relative speeds before and after the collision.

The Mass Center

New Symbols

v v

2m m

collision

t

x c

c 0x xo-xo

1 2

The Mass Center

New Symbols

The formula we used above for the mass center in 1D can begeneralized to 3D using vector notation:

~rc =∑

i

fi~ri .

1

2m

m

1r

2r

x

y

0

cr

The Mass Center

New Symbols

Example 4. Find the vector that describes the location of the CMwhen the system comprises two point masses.

The Mass Center

New Symbols

Example 5. (a) Find the vector that describes the location of theCM when the system comprises two equal point masses. (b) Showthat the CM lies midway between the two point masses.

The Mass Center

New Symbols

Example 6. Calculate the location of the CM for the compositeobject drawn in the figure. It comprises 4 point masses m, 2m, 3mand 4m oriented in a square of side L, connected my rods ofnegligible mass.

m21

o

4 3

2m

3m4m

xx

oy

y

0

The Mass Center

Center of Mass Motion

Theorem The system momentum ≡ the system mass × CMvelocity.

Proof For a system:

~P =∑

i

mid~ridt

=d

dt

∑i

mi~ri

=d

dtM

(∑i

mi

M~ri

)=

d

dtM~rc = M~vc

The Mass Center

Center of Mass Motion

Theorem The net external force acting on the system ≡ systemmass × CM acceleration.

Proof If we take a time derivative of the above:

d~P

dt= ~F ext = M~ac .

This is N-II for the system.

The Mass Center

Center of Mass Motion

2v

3v

1v1

m

3m

2m

THE SYSTEM

cv

M

CM PICTURE

The CM moves like a single point mass M acted upon by the totalexternal force. It follows that if

~F ext = 0

then ~ac = 0 and the system conspires to keep the CM fromaccelerating. We saw this earlier when we examined CM motionduring collisions.

The Mass Center

Distributed masses

So far, we have considered only point masses. We will nowconsider distributed masses that have constant density.

xx

x

my

0

Si

i

i

∆

∆

Question: How do we find the mass center?

The Mass Center

Distributed masses

For point masses,

xc =∑

i

fixi .

For a continuous object we could break it down into strips locatedat xi with mass ∆mi and width ∆x . Thus fi = ∆mi/M and

(xc)approx =∑

i

fixi .

The Mass Center

Distributed masses

We use a limiting process to get an accurate value for the masscenter:

xc = lim∆mi→0

(xc)approx =

∫dm x

M.

Similar results exist for yc and zc .

The Mass Center

Distributed masses

Example 7. Find the coordinates Pc(xc , yc) of the mass center fora thin sheet with a total mass of M a width of a and a length of b.

x

y

0

a

bM

The Mass Center

Summary

~rc =∑

i

fi~ri . fi = mass fraction =mi

M

In the absence of ~Fext the CM motion, before during and after acollision, is a straight line or point ~ac = 0.

~P = M~vc

d~P

dt= ~F ext = M~ac .