# 6: Moments and Centroids

**Mass and Slugs**

Newton's Second Law states that

\[F=ma \]

where \(F\) is the force, \(m\) is the mass, and \(a\) is the acceleration. In the US system, Force is measured in *pounds* and mass is measures in *slugs*.

Example 1

I weigh 165 lbs. What is my mass?

**Solution**

Since weight corresponds with gravitational force, and the acceleration of gravity is 32 ft/sec^{2} we have

\[165 = 32m\]

or

\[m = 5.15 \text{ slugs.}\]

In the metric system,* kg* is a mass unit and *Newtons* is a weight unit.

### Moments and Center of Mass for Discrete Mass Points

Suppose that we have a teeter totter and a 10 kg child is on the left 5 meters from the center of the teeter totter and a 15 kg child is on the right 4 meters from the center of the teeter totter. We define the *moment* as:

\[10(-5) + 15(4) = 10.\]

In general, we define the moment for masses \(m_i\) at the points \(x_i\) to be

\[\text{Moment} = S m_i x_i. \]

If the moment is 0 then we say that the system is in *equilibrium. * Otherwise, let \(x\) be the value such that

\[Sm_i (x_i-x)=0. \]

Then \(x\) is called the *center of mass *of the system.

Theorem

\[x = \dfrac{\text{moment}}{\text{total mass}}\]

**Proof: **

\[\begin{align} \sum m_i (x_i-\bar{x}) &= \sum m_ix_i - \sum m_i \bar{x} \\ &= \sum m_ix_i- \bar{x}\sum m_i \end{align}\]

so that

\[x=\dfrac{\sum m_i}{\sum x_i}. \]

Example 2

Find the center of mass of the teeter totter.

**Solution**

We have

\[\text{moment } =10\]

and

\[\text{total mass}=25 \]

hence the

\[\text{center of mass} = \dfrac{10}{25} = 0.4. \]

We can say that if the center of the teeter totter was 0.4 meters from the current center, then the children would be in balance.

For points in the plane, we can find moments and centers of mass coordinate wise. We define:

Definition

\[m_x = \text{moment about the x axis }= S m_i x_i, \]

\[m_y = \text{moment about the y axis }=S m_i y_i, \]

\[\text{Center of Mass }=\left(\dfrac{m_y}{M}, \dfrac{m_x}{m}\right). \]

Example 3

For the points \((-3,0)\) with mass 4, \((2,2)\) with mass 3, and \((1,-2)\) with mass 1 we have

\[m_x = (4)(-3) + (3)(2) + (1)(1) = -5, \]

\[m_y = (4)(0) + (3)(2) + (1)(-2) = 8, \]

\[\text{Center of Mass} = \left(-\dfrac{5}{8},\dfrac{8}{8}\right) = (-0.625,1). \]

**Center of Mass for a Two Dimensional Plate**

First, we recall that for a region of density \(r\) bounded by \(f(x)\) and \(g(x)\)

\[\text{Mass} = (\text{Density})(\text{Area}) = \rho \int_{a}^{b}\left(f(x)-g(x) \right) dx \]

\[M_x=\rho \int_{a}^{b} \dfrac{f(x)-g(x)}{2}\left( f(x)-g(x) \right) dx \]

\[M_y=\rho\int_{a}^{b} x \left(f(x)-g(x) \right) dx \]

and

\[\text{Center of Mass} = \left(\dfrac{M_y}{M},\dfrac{M_x}{M} \right) \]

Example 4

Find the center of mass for the plate of constant density 2 that is bounded by the curves

\(y = 1 - x\), \(y = 0\) and \(x = 0\).

We have

\[M_x=2\int_{0}^{1} \dfrac{1-x}{2}(1-x) \; dx, \]

\[M_y=2\int_{0}^{1} x(1-x) dx, \]

\[M = 2\int_{0}^{1} (1-x) dx. \]

**Pappus Theorem**

Suppose that we revolve a region around the y-axis. Then the volume of revolution is:

\[V = 2prA \]

where \(A\) is the area of the region and \(r\) is the distance from the centroid (constant density) to the axis of rotation.

Example 5

Suppose that we revolve the 4 x4 frame with width 1 centered about \((6,2)\) about the y-axis. Then we have that the Area is

\[A = 4 + 4 + 2 + 2 = 12.\]

\[ R = 7 \]

so that

\[V = 2p7(12) = 168p.\]

Example 6

Find the volume of the torus formed by revolving the disk

\[(x-11)^2+y^2=4 \]

about the y-axis.

### Contributors

- Larry Green (Lake Tahoe Community College)

- Integrated by Justin Marshall.