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Mathematics LibreTexts

7.6: Calculating Centers of Mass and Moments of Inertia

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    17524
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    We have already discussed a few applications of multiple integrals, such as finding areas, volumes, and the average value of a function over a bounded region. In this section we develop computational techniques for finding the center of mass and moments of inertia of several types of physical objects, using double integrals for a lamina (flat plate) and triple integrals for a three-dimensional object with variable density. The density is usually considered to be a constant number when the lamina or the object is homogeneous; that is, the object has uniform density.

    Center of Mass in Two Dimensions

    The center of mass is also known as the center of gravity if the object is in a uniform gravitational field. If the object has uniform density, the center of mass is the geometric center of the object, which is called the centroid. Figure \(\PageIndex{1}\) shows a point \(P\) as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.

    alt
    Figure \(\PageIndex{1}\): A lamina is perfectly balanced on a spindle if the lamina’s center of mass sits on the spindle.

    To find the coordinates of the center of mass \(P(\bar{x},\bar{y})\) of a lamina, we need to find the moment \(M_x\) of the lamina about the \(x\)-axis and the moment \(M_y\) about the \(y\)-axis. We also need to find the mass \(m\) of the lamina. Then

    \[\bar{x} = \dfrac{M_y}{m} \]

    and

    \[\bar{y} = \dfrac{M_x}{m}.\]

    Refer to Moments and Centers of Mass for the definitions and the methods of single integration to find the center of mass of a one-dimensional object (for example, a thin rod). We are going to use a similar idea here except that the object is a two-dimensional lamina and we use a double integral.

    If we allow a constant density function, then \(\bar{x} = \dfrac{M_y}{m}\) and \(\bar{y} = \dfrac{M_x}{m}\) give the centroid of the lamina.

    Suppose that the lamina occupies a region \(R\) in the \(xy\)-plane and let \(\rho (x,y)\) be its density (in units of mass per unit area) at any point \((x,y)\). Hence,

    \[\rho(x,y) = \lim_{\Delta A \rightarrow 0} \dfrac{\Delta m}{\Delta A}\]

    where \(\Delta m\) and \(\Delta A\) are the mass and area of a small rectangle containing the point \((x,y)\) and the limit is taken as the dimensions of the rectangle go to \(0\) (see the following figure).

    alt
    Figure \(\PageIndex{2}\): The density of a lamina at a point is the limit of its mass per area in a small rectangle about the point as the area goes to zero.

    Just as before, we divide the region \(R\) into tiny rectangles \(R_{ij}\) with area \(\Delta A\) and choose \((x_{ij}^*, y_{ij}^*)\) as sample points. Then the mass \(m_{ij}\) of each \(R_{ij}\) is equal to \(\rho (x_{ij}^*, y_{ij}^*) \Delta A\) (Figure \(\PageIndex{2}\)). Let \(k\) and \(l\) be the number of subintervals in \(x\) and \(y\) respectively. Also, note that the shape might not always be rectangular but the limit works anyway, as seen in previous sections.

    alt
    Figure \(\PageIndex{3}\): Subdividing the lamina into tiny rectangles \(R_{ij}\) each containing a sample point \((x_{ij}^*,y_{ij}^*)\).

    Hence, the mass of the lamina is

    \[m =\lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l \rho(x_{ij}^*,y_{ij}^*) \Delta A = \iint_R \rho(x,y) dA.\]

    Let’s see an example now of finding the total mass of a triangular lamina.

    Example \(\PageIndex{1}\): Finding the Total Mass of a Lamina

    Consider a triangular lamina \(R\) with vertices \((0,0), \, (0,3), \, (3,0)\) and with density \(\rho(x,y) = xy \, kg/m^2\). Find the total mass.

    Solution

    A sketch of the region \(R\) is always helpful, as shown in the following figure.

    15.6.1.png
    Figure \(\PageIndex{4}\): A lamina in the \(xy\)-plane with density \(\rho (x,y) = xy\).

    Using the expression developed for mass, we see that

    \[m = \iint_R \, dm = \iint_R \rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} xy \, dy \, dx = \int_{x=0}^{x=3} \left[ \left. x \dfrac{y^2}{2} \right|_{y=0}^{y=3} \right] \, dx = \int_{x=0}^{x=3} \dfrac{1}{2} x (3 - x)^2 dx = \left.\left[ \dfrac{9x^2}{4} - x^3 + \dfrac{x^4}{8} \right]\right|_{x=0}^{x=3} = \dfrac{27}{8}.\]

    The computation is straightforward, giving the answer \(m = \dfrac{27}{8} \, kg\).

    Exercise \(\PageIndex{1}\)

    Consider the same region \(R\) as in the previous example, and use the density function \(\rho (x,y) = \sqrt{xy}\). Find the total mass.

    Answer

    \(\dfrac{9\pi}{8} \, kg\)

    Now that we have established the expression for mass, we have the tools we need for calculating moments and centers of mass. The moment \(M_z\) about the \(x\)-axis for \(R\) is the limit of the sums of moments of the regions \(R_{ij}\) about the \(x\)-axis. Hence

    \[M_x = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*)m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*) \rho(x_{ij}^*,y_{ij}^*) \Delta A = \iint_R y\rho (x,y) dA\]

    Similarly, the moment \(M_y\) about the \(y\)-axis for \(R\) is the limit of the sums of moments of the regions \(R_{ij}\) about the \(y\)-axis. Hence

    \[M_x = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (x_{ij}^*)m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*) \rho(x_{ij}^*,y_{ij}^*) \Delta A = \iint_R x\rho (x,y) dA\]

    Example \(\PageIndex{2}\): Finding Moments

    Consider the same triangular lamina \(R\) with vertices \((0,0), \, (0,3), \, (3,0)\) and with density \(\rho (x,y) = xy\). Find the moments \(M_x\) and \(M_y\).

    Solution

    Use double integrals for each moment and compute their values:

    \[M_x = \iint_R y\rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} x y^2 \, dy \, dx = \dfrac{81}{20},\]

    \[M_y = \iint_R x\rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} x^2 y \, dy \, dx = \dfrac{81}{20},\]

    The computation is quite straightforward.

    Exercise \(\PageIndex{2}\)

    Consider the same lamina \(R\) as above and use the density function \(\rho (x,y) = \sqrt{xy}\). Find the moments \(M_x\) and \(M_y\).

    Answer

    \(M_x = \dfrac{81\pi}{64}\) and \(M_y = \dfrac{81\pi}{64}\)

    Finally we are ready to restate the expressions for the center of mass in terms of integrals. We denote the x-coordinate of the center of mass by \(\bar{x}\) and the y-coordinate by \(\bar{y}\). Specifically,

    \[\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x\rho (x,y) dA}{\iint_R \rho (x,y)dA} \]

    and

    \[\bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y\rho (x,y) dA}{\iint_R \rho (x,y)dA}\]

    Example \(\PageIndex{3}\): center of mass

    Again consider the same triangular region \(R\) with vertices \((0,0), \, (0,3), \, (3,0)\) and with density function \(\rho (x,y) = xy\). Find the center of mass.

    Solution

    Using the formulas we developed, we have

    \[\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x\rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{81/20}{27/8} = \dfrac{6}{5},\]

    \[\bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y\rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{81/20}{27/8} = \dfrac{6}{5}.\]

    Therefore, the center of mass is the point \(\left(\dfrac{6}{5},\dfrac{6}{5}\right).\)

    Analysis

    If we choose the density \(\rho(x,y)\) instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid,

    \[x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} = \dfrac{9/2}{9/2} = 1,\]

    \[y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA} = \dfrac{9/2}{9/2} = 1.\]

    Notice that the center of mass \(\left(\dfrac{6}{5},\dfrac{6}{5}\right)\) is not exactly the same as the centroid \((1,1)\) of the triangular region. This is due to the variable density of \(R\) If the density is constant, then we just use \(\rho(x,y) = c\) (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina.

    Exercise \(\PageIndex{3}\)

    Again use the same region \(R\) as above and use the density function \(\rho (x,y) = \sqrt{xy}\). Find the center of mass.

    Answer

    \(\bar{x} = \dfrac{M_y}{m} = \dfrac{81\pi/64}{9\pi/8} = \dfrac{9}{8}\) and \(\bar{y} = \dfrac{M_x}{m} = \dfrac{81\pi}{9\pi/8} = \dfrac{0}{8}\).

    Once again, based on the comments at the end of Example \(\PageIndex{3}\), we have expressions for the centroid of a region on the plane:

    \[x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} \, \text{and} \, y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA}.\]

    We should use these formulas and verify the centroid of the triangular region