Center of Mass, Gravity, and Volume

The center of mass, center of gravity, and center of volume of an object are all very similar. In fact, they are all the same if the density of the object is constant and gravity is constant along an object.

The center of mass, center of gravity, and center of volume are all represented by a point $ (\bar{x},\bar{y}, \bar{z}) $.

	Center of Mass	Center of Gravity	Center of Volume (Centroid)
Definition	Average location of the mass of an object	Average location of the weight of an object	Geometric center of an object
Equation (Summation Form)	$$ \bar{x} = \frac{\sum \tilde{x}_i m_i}{\sum m_i} $$	$$ \bar{x} = \frac{\sum \tilde{x}_i W_i}{\sum W_i} $$	$$ \bar{x} = \frac{\sum \tilde{x}_i V_i}{\sum V_i} $$
Equation (Integral Form)	$$ \bar{x} = \frac{\int \tilde{x} \,dm}{\int dm} $$	$$ \bar{x} = \frac{\int \tilde{x} \,dW}{\int dW} $$	$$ \bar{x} = \frac{\int \tilde{x} \,dV}{\int dV} $$

When are Centers of Gravity, Mass, and Volume the same? #com

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The center of gravity, center of mass, and center of volume of an object are the same when that object has a constant density (is homogeneous) and gravity is constant along the object.

Centroid

The centroid denotes the geometric center of an object. When an object is made of a homogeneous material, the centroid and the center of mass are at the same point. If the object has an axis of symmetry, the centroid is on the axis of symmetry. In some cases, the centroid may not be on the object.

To solve for the centroid of an object:

Define your coordinate system
Define the infinitesimally small element (usually $dL$, $dA$, or $dV$)
Express the element in terms of the coordinate system you defined
Identify any symmetry
Define the centroid $ (\tilde{x}, \tilde{y}) $of the element
Plug the equations from 3 and 5 into the equations for the centroid

The equation for the centroid of an object is:

Centroid of an Object (3D) #com3D

$$ \bar{x} = \frac{\int{\tilde{x}dV}}{\int{dV}} $$

Centroid of an Object (2D) #com2D

$$ \bar{x} = \frac{\int{\tilde{x}dA}}{\int{dA}} $$

Centroid of an Object (1D) #com1D

$$ \bar{x} = \frac{\int{\tilde{x}dL}}{\int{dL}} $$

with similar equations for $ \bar{y} $.

Example Problem: Centroid of a curved area #com-epc

Solve for $ \bar{x} $
Solve for $ \bar{y} $

Solve for $ \bar{x} $

$$ \bar{x} = \frac{\int \tilde{x} \,dA}{\int dA} = \frac{\int x \ y \,dx}{\int y \,dx} = \frac{\int_{0}^{1} x^3 \,dx}{\int_{0}^{1} x^2 \,dx} = \frac{\frac{x^4}{4} \Big|_{0}^{1}}{\frac{x^3}{3} \Big|_{0}^{1}} = \frac{3}{4} \ m $$
Solve for $ \bar{y} $

$$ \bar{y} = \frac{\int \tilde{y} \,dA}{\int dA} = \frac{\int \frac{y}{2} y \,dx}{\int y \,dx} = \frac{\int_{0}^{1} \frac{x^4}{2} \,dx}{\int_{0}^{1} x^2 \,dx} = \frac{\frac{x^5}{10} \Big|_{0}^{1}}{\frac{x^3}{3} \Big|_{0}^{1}} = \frac{3}{10} \ m $$

Axes of Symmetry

If the shape has an axis of symmetry, the centroid falls on that line of symmetry. Shapes can also have more than one axis. For example, a rectangle has 2 axes, and a circle has infinite number of axes. This applies to composite shapes as well.

Centroids of Basic Shapes

Shape	$$ \boldsymbol{\bar{x}} $$	$$ \boldsymbol{\bar{y}} $$
Rectangle	$$ b/2 $$	$$ h/2 $$
Right Triangle	$$ b/3 $$	$$ h/3 $$
Isoceles Triangle	$$ b/2 $$	$$ h/3 $$
Circle	$$ r $$	$$ r $$
Semicircle	$$ r $$	$$ \frac{4r}{3\pi} $$
Quarter Circle	$$ \frac{4r}{3\pi} $$	$$ \frac{4r}{3\pi} $$

Finding Centroids Using Composite Shapes

We can find the centroid of more complicated shapes by breaking them up into simpler shapes and finding the centroids of those simpler shapes.

To solve for the centroid of our original shape $ (\bar{x}, \bar{y}) $, we take a weighted sum of each of the centroids of our smaller shapes. We will call the centroid locations of the smaller shapes $ \tilde{x} $ and $ \tilde{y} $.

Composite Centroid (x) #compositex

$$ \bar{x} = \frac{\sum \tilde{x}_i A_i}{\sum A_i} $$

Composite Centroid (y) #compositey

$$ \bar{y} = \frac{\sum \tilde{y}_i A_i}{\sum A_i} $$

For these equations, the weighted sum is dependent on the area of each individual shape. This is typically for 2D problems. You can also use length, weight, or mass as your weighing factor depending on what you're being asked to solve for.

Example Problem: Find centroid of a composite shape #com-ccs

Solve for centroid $ (\bar{x} , \bar{y}) $ of the shape below.

First, we notice that there's a line of symmetry along $ x = 2 \ cm $ , so we know that $ \bar{x} = 2 \ cm $. This means that we only need to solve for $ \bar{y} $.

$$ \begin{aligned} (\bar{x}, \bar{y}) &= (2, ?) \\ \bar{y} &= \frac{\sum \tilde{y}_i A_i}{\sum A_i} \end{aligned} $$

We can do this by splitting our shape into composite shapes with known centroid locations. In this case, we will use a rectangle, a semicircle, and an isoceles triangle.

Segment	Area	$ \tilde{x} $	$ \tilde{y} $	$ \tilde{y} A $
1	$ 24 \ cm^2 $	$ 2 \ cm $	$ 3 \ cm $	$ 72 \ cm^3 $
2	$ -4 \ cm^2 $	$ 2 \ cm $	$ \frac{1}{3} (2) = \frac{2}{3} \ cm $	$ - \frac{8}{3} \ cm^3 $
3	$ 2 \pi \ cm^2 $	$ 2 \ cm $	$ \frac{4R}{3 \pi} +6 = \frac{8}{3 \pi} \ cm $	$ \frac{16}{3} + 12 \pi \ cm^3 $

$$ \bar{y} = \frac{\sum \tilde{y}_i A_i}{\sum A_i} = \frac{(72 - \frac{8}{3} + \frac{16}{3} + 12 \pi) \ cm^3}{(24 - 4 +2 \pi \ cm^2)} = 4.27 \ cm $$

$$ (\bar{x} , \bar{y}) = (2, 4.27) \ cm $$

Segment	Area	\( \tilde{x} \)	\( \tilde{y} \)	\( \tilde{y} A \)
1	\( 24 \ cm^2 \)	\( 2 \ cm \)	\( 3 \ cm \)	\( 72 \ cm^3 \)
2	\( -4 \ cm^2 \)	\( 2 \ cm \)	\( \frac{1}{3} (2) = \frac{2}{3} \ cm \)	\( - \frac{8}{3} \ cm^3 \)
3	\( 2 \pi \ cm^2 \)	\( 2 \ cm \)	\( \frac{4R}{3 \pi} +6 = \frac{8}{3 \pi} \ cm \)	\( \frac{16}{3} + 12 \pi \ cm^3 \)