1. Home
  2. University physics volume 1
  3. Unit 1. mechanics
  4. Linear momentum and collisions
  5. Center of mass

9.6 Center of mass  (Page 5/13)

<< Chapter < Page
  University physics volume 1     Page 5 / 13
Chapter >> Page >

The total mass of the unit cell is therefore

M = ( 4 ) ( 5.885 × 10 −26 kg ) + ( 4 ) ( 3.816 × 10 −26 kg ) = 3.880 × 10 −25 kg .

From the geometry, the locations are

r → 1 = 0 r → 2 = ( 2.36 × 10 −10 m ) i ^ r → 3 = r 3 x i ^ + r 3 y j ^ = ( 2.36 × 10 −10 m ) i ^ + ( 2.36 × 10 −10 m ) j ^ r → 4 = ( 2.36 × 10 −10 m ) j ^ r → 5 = ( 2.36 × 10 −10 m ) k → r → 6 = r 6 x i ^ + r 6 z k ^ = ( 2.36 × 10 −10 m ) i ^ + ( 2.36 × 10 −10 m ) k ^ r → 7 = r 7 x i ^ + r 7 y j ^ + r 7 z k ^ = ( 2.36 × 10 −10 m ) i ^ + ( 2.36 × 10 −10 m ) j ^ + ( 2.36 × 10 −10 m ) k ^ r → 8 = r 8 y j ^ + r 8 z k ^ = ( 2.36 × 10 −10 m ) j ^ + ( 2.36 × 10 −10 m ) k ^ .

Substituting:

| r → CM, x | = r CM, x 2 + r CM, y 2 + r CM, z 2 = 1 M ∑ j = 1 8 m j ( r x ) j = 1 M ( m 1 r 1 x + m 2 r 2 x + m 3 r 3 x + m 4 r 4 x + m 5 r 5 x + m 6 r 6 x + m 7 r 7 x + m 8 r 8 x ) = 1 3.8804 × 10 −25 kg [ ( 5.885 × 10 −26 kg ) ( 0 m ) + ( 3.816 × 10 −26 kg ) ( 2.36 × 10 −10 m ) + ( 5.885 × 10 −26 kg ) ( 2.36 × 10 −10 m ) + ( 3.816 × 10 −26 kg ) ( 2.36 × 10 −10 m ) + 0 + 0 + ( 3.816 × 10 −26 kg ) ( 2.36 × 10 −10 m ) + 0 ] = 1.18 × 10 −10 m.

Similar calculations give r CM, y = r CM, z = 1.18 × 10 −10 m (you could argue that this must be true, by symmetry, but it’s a good idea to check).

Significance

As it turns out, it was not really necessary to convert the mass from atomic mass units (u) to kilograms, since the units divide out when calculating r CM anyway.

To express r CM in terms of magnitude and direction, first apply the three-dimensional Pythagorean theorem to the vector components:

r CM = r CM, x 2 + r CM, y 2 + r CM, z 2 = ( 1.18 × 10 −10 m ) 3 = 2.044 × 10 −10 m.

Since this is a three-dimensional problem, it takes two angles to specify the direction of r → CM . Let ϕ be the angle in the x , y -plane, measured from the + x -axis, counterclockwise as viewed from above; then:

ϕ = tan −1 ( r CM, y r CM, x ) = 45 ° .

Let θ be the angle in the y , z -plane, measured downward from the + z -axis; this is (not surprisingly):

θ = tan −1 ( R z R y ) = 45 ° .

Thus, the center of mass is at the geometric center of the unit cell. Again, you could argue this on the basis of symmetry.

Check Your Understanding Suppose you have a macroscopic salt crystal (that is, a crystal that is large enough to be visible with your unaided eye). It is made up of a huge number of unit cells. Is the center of mass of this crystal necessarily at the geometric center of the crystal?

On a macroscopic scale, the size of a unit cell is negligible and the crystal mass may be considered to be distributed homogeneously throughout the crystal. Thus,
r → CM = 1 M ∑ j = 1 N m j r → j = 1 M ∑ j = 1 N m r → j = m M ∑ j = 1 N r → j = N m M ∑ j = 1 N r → j N
where we sum over the number N of unit cells in the crystal and m is the mass of a unit cell. Because Nm = M , we can write
r → CM = m M ∑ j = 1 N r → j = N m M ∑ j = 1 N r → j N = 1 N ∑ j = 1 N r → j .
This is the definition of the geometric center of the crystal, so the center of mass is at the same point as the geometric center.

Got questions? Get instant answers now!

Two crucial concepts come out of these examples:

  1. As with all problems, you must define your coordinate system and origin. For center-of-mass calculations, it often makes sense to choose your origin to be located at one of the masses of your system. That choice automatically defines its distance in [link] to be zero. However, you must still include the mass of the object at your origin in your calculation of M , the total mass [link] . In the Earth-moon system example, this means including the mass of Earth. If you hadn’t, you’d have ended up with the center of mass of the system being at the center of the moon, which is clearly wrong.
  2. In the second example (the salt crystal), notice that there is no mass at all at the location of the center of mass. This is an example of what we stated above, that there does not have to be any actual mass at the center of mass of an object.
<< Chapter < Page Page > Chapter >>
Practice Key Terms 4

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 1' conversation and receive update notifications?

Ask