Chapter 09 : Conservation of Momentum

Note: this material fits better with the rotational motion topics covered in the next module, so it will appear on Test 4 instead of Test 3.

Center of Mass : What and why?

The center of mass of an object (or collection of objects) is the mass-weighted average coordinate for the object(s) (or the collection of objects).

Point-Masses
\[ \vec{r}_{cm} = \frac{ \sum\limits_{i=1}^{N} m_i \vec{r}_i }{ \sum\limits_{i=1}^{N} m_i } \]

or simply:
\[ \vec{r}_{cm} = \frac{ \sum\limits_{i=1}^{N} m_i \vec{r}_i }{ M } \]

Breaking the vector equation into components:

where M = Σm_i (the total mass)

Solid Object

\[ \vec{r}_{cm} = \frac{\int \vec{r} dm}{M} \]

Why is the CM concept relevant/important?

Suppose our object is a collection of point-masses (could be ball bearings connected by springs, or maybe a calculator and it's gazillions of 'point-mass' molecules.

Start with definition:

\[ \vec{r}_{cm} = \frac{1}{M} \sum m_i \vec{r}_i \]

Rearrange into:

\[ M \vec{r}_{cm} = \sum m_i \vec{r}_i \]

Differentiate with respect to time:

\[ M \vec{v}_{cm} = \sum m_i \vec{v}_i \]

Differentiate again with respect to time:

\[ M \vec{a}_{cm} = \sum m_i \vec{a}_i \]

Expanding out the sum a bit:

\[ M \vec{a}_{cm} = m_1 \vec{a}_1 + m_2 \vec{a}_2 + \cdots \]

Look at each term in the sum on the right. The first term is actually:
\[ m_1 \vec{a}_1 = \sum \vec{F}_1 \]

That is: it's the sum of all the forces acting on that entity (molecule?)
The right-hand side of that hugely long sum is really:

the sum of all the forces acting on object 1 plus
the sum of all the forces acting on object 2 plus
... and so on

BUT : Most of those forces are INTERNAL forces (between the point-masses or molecules making up the object)

and those all cancel each other out in pairs thanks to Newton's third law (two objects exert "equal and opposite forces" on each other).

We're left with just:

\[ M \vec{a}_{cm} = \sum_{external} \vec{F} \]

THIS IS A BIG DEAL

In words :

If a composite object (book made of gazillions of atoms for example) is tossed into the air,
it's CM will follow the same path a point object of the same mass would have.

Our point-mass model is actually fine for some types of motions of real, extended objects. \( \sum \vec{F} = m\vec{a} \) shows us how the CM of the object will move.

The object (book, etc) can (and almost certainly will) ROTATE about that center of mass point though, and that's what the remainder of the course will focus on: angular motion, and the forces (torques) that cause that rotational motion.

Point-Mass Example : Large mobile

A mobile is constructed with three small steel balls connected with thin (massless) rods:

M₁ = 10 g at x=0, y=0
M₂ = 2 g at x=1 m, y=0
M₃ = 5 g at x=0, y=3 m

Determine the center of mass of the mobile.

Object Mass X coord Y coord m_ix_i m_iy_i

i grams meters meters gram-meter gram-meter

1 10 0 0 0 0

2 2 1 0 2 0

3 5 0 3 0 15

Sum 17 gram --- --- 2 gram-meter 15 gram-meter

Collecting what we need:

\[ X_{cm} = \frac{ \sum ( m_i x_i ) }{\sum m_i } = \frac{ 2~gram \cdot meters }{ 17~grams } = 0.1176~m \]

\[ Y_{cm} = \frac{ \sum ( m_i y_i ) }{\sum m_i } = \frac{ 15~gram \cdot meters }{ 17~grams } = 0.8824~m \]

Solid Object Example : Thin Rod/Plate

Find the center of mass of a thin object of mass M and length L.
(Assume the object has a uniform density throughout.)

Common mass-density symbols:

ρ : mass/volume (kg/m³, grams/cc, etc)
σ : mass/area
λ : mass/length

Method 1 : Do the integral!

\[ x_{cm} = \int x dm \]

Method 2 : Symmetry arguments

Shortcut : CM for common geometric shapes via symmetry

Example : Composite Object

Determine the CM of the thin L-shaped object shown (carpenter's square).
Assume the density/area σ is constant throughout the material.

The 'trick' is to break an object into 'parts' for which we can determine the CM easily (hopefully via symmetry arguments).

This object can be broken into two rectangular parts:

A : a (100 cm) X (4 cm) part along the bottom
B : a (4 cm) X (36 cm) part along the left

where symmetry can be used to find the CM of each part.

How does that work? Start with the definition:
\[ x_{cm} = \frac{ \sum ( m_i x_i ) }{M} \hspace{2em} so \hspace{2em} \sum ( m_i x_i ) = M x_{cm} \]

\[ But: \hspace{2em} \sum ( m_i x_i ) = \sum_A (m_i x_i) + \sum_B (m_i x_i) \]

\[ So: \hspace{2em} M x_{cm} = M_A x_{cm,A} + M_B x_{cm,B} \]

\[ Finally: \hspace{2em} x_{cm} = \frac{M_A x_{cm,A} + M_B x_{cm,B}}{M} \]

In words:

Break the object into parts we can easily find the CM of.
Replace each part with a point located at the CM of that part
Use the point-mass equations to find the overall CM for the object.

Why is the CM concept relevant/important?
Suppose our object is a collection of point-masses (could be ball bearings connected by springs, or maybe a calculator and it's gazillions of 'point-mass' molecules.
Start with definition:	\[ \vec{r}_{cm} = \frac{1}{M} \sum m_i \vec{r}_i \]
Rearrange into:	\[ M \vec{r}_{cm} = \sum m_i \vec{r}_i \]
Differentiate with respect to time:	\[ M \vec{v}_{cm} = \sum m_i \vec{v}_i \]
Differentiate again with respect to time:	\[ M \vec{a}_{cm} = \sum m_i \vec{a}_i \]
Expanding out the sum a bit:	\[ M \vec{a}_{cm} = m_1 \vec{a}_1 + m_2 \vec{a}_2 + \cdots \]
Look at each term in the sum on the right. The first term is actually: \[ m_1 \vec{a}_1 = \sum \vec{F}_1 \] That is: it's the sum of all the forces acting on that entity (molecule?) The right-hand side of that hugely long sum is really: the sum of all the forces acting on object 1 plus the sum of all the forces acting on object 2 plus ... and so on
BUT : Most of those forces are INTERNAL forces (between the point-masses or molecules making up the object) and those all cancel each other out in pairs thanks to Newton's third law (two objects exert "equal and opposite forces" on each other).
We're left with just:	\[ M \vec{a}_{cm} = \sum_{external} \vec{F} \]
THIS IS A BIG DEAL
In words : If a composite object (book made of gazillions of atoms for example) is tossed into the air, it's CM will follow the same path a point object of the same mass would have. Our point-mass model is actually fine for some types of motions of real, extended objects. \( \sum \vec{F} = m\vec{a} \) shows us how the CM of the object will move. The object (book, etc) can (and almost certainly will) ROTATE about that center of mass point though, and that's what the remainder of the course will focus on: angular motion, and the forces (torques) that cause that rotational motion.

Object	Mass	X coord	Y coord	m_ix_i	m_iy_i
i	grams	meters	meters	gram-meter	gram-meter
1	10	0	0	0	0
2	2	1	0	2	0
3	5	0	3	0	15
Sum	17 gram	---	---	2 gram-meter	15 gram-meter
Collecting what we need: \[ X_{cm} = \frac{ \sum ( m_i x_i ) }{\sum m_i } = \frac{ 2~gram \cdot meters }{ 17~grams } = 0.1176~m \] \[ Y_{cm} = \frac{ \sum ( m_i y_i ) }{\sum m_i } = \frac{ 15~gram \cdot meters }{ 17~grams } = 0.8824~m \]

Solid Object Example : Thin Rod/Plate
Find the center of mass of a thin object of mass M and length L. (Assume the object has a uniform density throughout.) Common mass-density symbols: ρ : mass/volume (kg/m³, grams/cc, etc) σ : mass/area λ : mass/length
Method 1 : Do the integral!	\[ x_{cm} = \int x dm \]
Method 2 : Symmetry arguments

Example : Composite Object
Determine the CM of the thin L-shaped object shown (carpenter's square). Assume the density/area σ is constant throughout the material. The 'trick' is to break an object into 'parts' for which we can determine the CM easily (hopefully via symmetry arguments). This object can be broken into two rectangular parts: A : a (100 cm) X (4 cm) part along the bottom B : a (4 cm) X (36 cm) part along the left where symmetry can be used to find the CM of each part.
How does that work? Start with the definition: \[ x_{cm} = \frac{ \sum ( m_i x_i ) }{M} \hspace{2em} so \hspace{2em} \sum ( m_i x_i ) = M x_{cm} \] \[ But: \hspace{2em} \sum ( m_i x_i ) = \sum_A (m_i x_i) + \sum_B (m_i x_i) \] \[ So: \hspace{2em} M x_{cm} = M_A x_{cm,A} + M_B x_{cm,B} \] \[ Finally: \hspace{2em} x_{cm} = \frac{M_A x_{cm,A} + M_B x_{cm,B}}{M} \]
In words: Break the object into parts we can easily find the CM of. Replace each part with a point located at the CM of that part Use the point-mass equations to find the overall CM for the object.