Centers of Mass

Introduction

Man and son balancing on a seesaw

Suppose a man and his son balance on a seesaw. If the man weights 200200 lb and his son weighs 5050 lb, where is the balance point on a 1010 ft seesaw? Recall that the magnitude of the moment that they each produce around the balance point is defined as the product of the force they exert and their perpendicular distance from the balance point. In order to balance, these two moments must be equal:

Wfather⋅dfather=Wson⋅dson200x=50(10−x)200x=500−50x250x=500x=2 ft\begin{aligned} W_{father} \cdot d_{father} &= W_{son} \cdot d_{son}\ 200x &= 50 (10-x)\ 200x &= 500 - 50x\ 250x &= 500\ x &= 2\ ft \end{aligned}

Therefore, in order to balance, the balance point should be placed 22 ft from the father's end of the seesaw. This balance point is called the center of mass.

Take a look at the more general case, where x‾\overline{x} is the position of the center of mass:

Balancing diagram

Equate the moments and solve for the location of x‾\overline{x}:

m1(x‾−x1)=m2(x2−x‾)m1x‾+m2x‾=m1x1+m2x2x‾=m1x1+m2x2m1+m2\begin{aligned} m_1 (\overline{x}-x_1) &= m_2 (x_2 - \overline{x})\ m_1 \overline{x} + m_2 \overline{x} &= m_1 x_1 + m_2 x_2\ \ \overline{x} &= \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \end{aligned}

Thus, in the one-dimensional case with discrete masses, x‾=∑mixi∑mi\ans{\overline{x} = \dfrac{\sum m_i x_i}{\sum m_i}}

Non-Discrete Masses

What if, instead of a bar with clear distinct masses resting on it, we have something like the situation below, where we're trying to balance an object, and the center of mass depends on its shape?

Triangle balanced

In this case, instead of using a sum to add up finitely many masses and distances, we'll use an integral (replace each sum with an integral, and replace the discrete masses with a density function ρ(x)ρ(x)):

x‾=∫abxρ(x) dx∫abρ(x) dx=1m∫abxρ(x) dx\ans{\overline{x} = \dfrac{\int_a^b x \rho(x)\ dx}{\int_a^b \rho(x)\ dx} = \dfrac{1}{m} \int_a^b x \rho (x)\ dx}

where mm is the total mass (which can be found by integrating the density function), and ∫abxρ(x) dx\displaystyle\int_a^b x\rho(x)\ dx is called the moment about the y axis.

Two Dimensions

Consider the following figure, where we're looking down on a thin sheet in two dimensions, and we assume that this sheet has varying density given by a density function ρρ.

Thin sheet balancing

Now, there will be two coordinates for the center of mass, or the centroid. These can be found by using similar integrals to what we saw before, but we'll use double integrals this time:

x‾=1m∬Rxρ(x,y) dAy‾=1m∬Ryρ(x,y) dA\ans{\begin{aligned} \overline{x} &= \dfrac{1}{m} \iint_R x \rho(x,y) \ dA\ \ \overline{y} &= \dfrac{1}{m} \iint_R y \rho(x,y) \ dA \end{aligned}}

Consider the region RR bounded by the yy-axis and the curves y=e−x−12y=e^{-x}-\dfrac{1}{2} and y=12−e−xy=\dfrac{1}{2}-e^{-x}. If the density of this sheet is ρ(x,y)=1ρ(x,y)=1, find its center of mass.

Solution

Start by calculating mm, since we'll need it for both formulas. Remember that the mass is the integral of the density function:

m=∬Rρ(x,y) dA=∫0ln⁡2∫1/2−e−xe−x−1/2 dy dx=∫0ln⁡2−2e−x−x∣0ln⁡2=1−ln⁡2\begin{aligned} m &= \iint_R \rho(x,y)\ dA\ &= \int_0^{\ln 2} \int_{1/2-e^{-x}}^{e^{-x-1/2}} \ dy\ dx\ &= \int_0^{\ln 2} -2e^{-x}-x \bigg|_0^{\ln 2}\ &= 1-\ln 2 \end{aligned}

Next, apply the formulas to find the center of mass:

x‾=11−ln⁡2∫0ln⁡2∫1/2−e−xe−x−1/2x dy dx=11−ln⁡2∫0ln⁡22xe−x−x dx=11−ln⁡2[−2xe−x−12x2−2e−x]0ln⁡2=1−ln⁡2−12(ln⁡2)21−ln⁡2≈0.217y‾=11−ln⁡2∫0ln⁡2∫1/2−e−xe−x−1/2y dy dx=11−ln⁡2∫0ln⁡212[(e−x−12)2−(12−e−x)2] dx=12(1−ln⁡2)∫0ln⁡2e−2x−e−x+14−14−e−2x+e−x dx=0\begin{aligned} \overline{x} &= \dfrac{1}{1-\ln 2} \int_0^{\ln 2} \int_{1/2-e^{-x}}^{e^{-x-1/2}} x\ dy\ dx\ &= \dfrac{1}{1-\ln 2} \int_0^{\ln 2} 2xe^{-x} - x\ dx\ &= \dfrac{1}{1-\ln 2} \bigg[ -2xe^{-x}-\dfrac{1}{2}x^2-2e^{-x} \bigg]0^{\ln 2}\ &= \dfrac{1-\ln 2 - \dfrac{1}{2}(\ln 2)^2}{1-\ln 2} \approx 0.217\ \overline{y} &= \dfrac{1}{1-\ln 2} \int_0^{\ln 2} \int{1/2-e^{-x}}^{e^{-x-1/2}} y\ dy\ dx\ &= \dfrac{1}{1-\ln 2} \int_0^{\ln 2} \dfrac{1}{2}\bigg[\left(e^{-x}-\dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}-e^{-x}\right)^2\bigg]\ dx\ &= \dfrac{1}{2(1-\ln 2)} \int_0^{\ln 2} e^{-2x}-e^{-x}+\dfrac{1}{4}-\dfrac{1}{4}-e^{-2x}+e^{-x}\ dx = 0\ \end{aligned}

herefore, the center of mass of this sheet is located at approximately (x‾,y‾)=(0.217,0)\ans{(\overline{x},\overline{y}) = (0.217,0)}

It shouldn't be surprising that the yy component of the center of mass is 00, because this sheet is symmetric about the xx-axis.

Consider the region R={(x,y) : −1≤x≤1,0≤y≤1}R={(x,y)\ :\ -1 \leq x \leq 1, 0 \leq y \leq 1}. If the density of this sheet is ρ(x,y)=2−yρ(x,y)=2−y, find its center of mass.

Solution

Again, start by calculating mm: m=∬Rρ(x,y) dA=∫−11∫012−y dy dx=∫−1132 dx=32x∣−11=3\begin{aligned} m &= \iint_R \rho(x,y)\ dA\ &= \int_{-1}^{1} \int_0^1 2-y \ dy\ dx\ &= \int_{-1}^1 \dfrac{3}{2} \ dx = \dfrac{3}{2}x\bigg|_{-1}^1\ &= 3 \end{aligned}

Next, apply the formulas to find the center of mass:

x‾=13∫−11∫01x(2−y) dy dx=13∫−1132x dx=13(34x2∣−11)=0y‾=13∫−11∫01y(2−y) dy dx=13∫−11y2−13y3∣01 dx=13∫−1123 dx=49\begin{aligned} \overline{x} &= \dfrac{1}{3} \int_{-1}^{1} \int_0^1 x(2-y) \ dy\ dx\ &= \dfrac{1}{3} \int_{-1}^{1} \dfrac{3}{2}x\ dx\ &= \dfrac{1}{3} \left(\dfrac{3}{4}x^2\bigg|{-1}^1\right) = 0\ \overline{y} &= \dfrac{1}{3} \int{-1}^{1} \int_0^1 y(2-y) \ dy\ dx\ &= \dfrac{1}{3} \int_{-1}^{1} y^2-\dfrac{1}{3}y^3\bigg|0^1\ dx\ &= \dfrac{1}{3} \int{-1}^{1} \dfrac{2}{3}\ dx = \dfrac{4}{9}\ \end{aligned}

Therefore, the center of mass of this sheet is located at (x‾,y‾)=(0,49)\ans{(\overline{x},\overline{y}) = \left(0,\dfrac{4}{9}\right)}

  1. Consider the region R={(x,y) : 0≤x≤4,0≤y≤2}R={(x,y)\ :\ 0 \leq x \leq 4, 0 \leq y \leq 2}. If the density of this sheet is ρ(x,y)=1+x2ρ(x,y)=1+\dfrac{x}{2}, find its center of mass.

    2. (x‾,y‾)=(73,1)(\overline{x},\overline{y}) = \left(\dfrac{7}{3},1\right)

  2. Find the center of mass of the triangular plate in the first quadrant bounded by x+y=4x+y=4 with ρ(x,y)=1+x+yρ(x,y)=1+x+y.

    3. (x‾,y‾)=(1611,1611)(\overline{x},\overline{y}) = \left(\dfrac{16}{11},\dfrac{16}{11}\right)

Three Dimensions

The center of mass in three dimensions is similar, but now we add a third component, and the integrals become triple integrals. If m=∭Dρ(x,y,z) dV,m=\iiint_D \rho(x,y,z)\ dV, then

x‾=1m∭Dxρ(x,y,z) dVy‾=1m∭Dyρ(x,y,z) dVz‾=1m∭Dzρ(x,y,z) dV\ans{\begin{aligned} \overline{x} &= \dfrac{1}{m} \iiint_D x \rho(x,y,z) \ dV\ \ \overline{y} &= \dfrac{1}{m} \iiint_D y \rho(x,y,z) \ dV\ \ \overline{z} &= \dfrac{1}{m} \iiint_D z \rho(x,y,z) \ dV \end{aligned}}

Consider the cone DD bounded by z=4−x2+y2z=4-\sqrt{x^2+y^2} and z=0z=0. If the density of this cone is ρ(x,y,z)=1ρ(x,y,z)=1, find its center of mass.

Solution

As always, start by calculating mm. Let's try it in rectangular coordinates, but we'll get stuck pretty quickly: m=∭Dρ(x,y,z) dV=∫−44∫−4−x24−x2∫04−x2+y21 dz dy dx=∫−44∫−4−x24−x24−x2+y2 dy dx\begin{aligned} m &= \iiint_D \rho(x,y,z)\ dV\ &= \int_{-4}^{4} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_0^{4-\sqrt{x^2+y^2}} 1\ dz\ dy\ dx\ &= \int_{-4}^{4} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 4-\sqrt{x^2+y^2} \ dy\ dx\ \end{aligned}

We hit a roadblock here, and we notice the x2+y2x^2+y^2 in the expression, which tells us that cylindrical coordinates may work better:

m=∫02π∫04∫04−rr dz dr dθ=∫02π∫044r−r2 dr dθ=∫02π323 dθ=64π3\begin{aligned} m &= \int_0^{2\pi} \int_0^4 \int_0^{4-r} r \ dz\ dr\ d\theta\ &= \int_0^{2\pi} \int_0^4 4r-r^2\ dr\ d\theta\ &= \int_0^{2\pi} \dfrac{32}{3}\ d\theta = \dfrac{64\pi}{3} \end{aligned}

Next, apply the formulas to find the center of mass, noting again that x=rcos⁡θx=r \cos θ and y=rsin⁡θy=r \sin θ:

x‾=364π∫02π∫04∫04−rr2cos⁡θ dz dr dθ=364π∫02π∫04(4r2−r3)cos⁡θ dr dθ=364π∫02π(43r3−14r4)∣04cos⁡θ dθ=364π∫02π643cos⁡θ dθ=1π∫02πcos⁡θ dθ=0Similarly, y‾=0z‾=364π∫02π∫04∫04−rzr dz dr dθ=364π∫02π∫0412(4−r)2r dr dθ=3128π∫02π∫0416r−8r2+r3 dr dθ=3128π∫02π643 dθ=1\begin{aligned} \overline{x} &= \dfrac{3}{64\pi} \int_{0}^{2\pi} \int_0^4 \int_0^{4-r} r^2 \cos \theta \ dz\ dr\ d\theta\ &= \dfrac{3}{64\pi} \int_{0}^{2\pi} \int_0^4 (4r^2-r^3) \cos \theta \ dr\ d\theta\ &= \dfrac{3}{64\pi} \int_{0}^{2\pi} \left(\dfrac{4}{3}r^3-\dfrac{1}{4}r^4\right)\bigg|0^4 \cos \theta \ d\theta\ &= \dfrac{3}{64\pi} \int{0}^{2\pi} \dfrac{64}{3} \cos \theta \ d\theta\ &= \dfrac{1}{\pi} \int_{0}^{2\pi} \cos \theta \ d\theta = 0\ \textrm{Similarly, } \overline{y} &= 0\ \overline{z} &= \dfrac{3}{64\pi} \int_{0}^{2\pi} \int_0^4 \int_0^{4-r} zr \ dz\ dr\ d\theta\ &= \dfrac{3}{64\pi} \int_{0}^{2\pi} \int_0^4 \dfrac{1}{2}(4-r)^2r\ dr\ d\theta\ &= \dfrac{3}{128\pi} \int_{0}^{2\pi} \int_0^4 16r-8r^2+r^3\ dr\ d\theta\ &= \dfrac{3}{128\pi} \int_{0}^{2\pi} \dfrac{64}{3}\ d\theta = 1\ \end{aligned}

Therefore, the center of mass of this cone is located at (x‾,y‾,z‾)=(0,0,1)\ans{(\overline{x},\overline{y},\overline{z}) = \left(0,0,1\right)}

Consider the hemisphere DD with radius 44, centered at the origin, and its base on the xyxy plane. If the density of this hemisphere is ρ(ρ,ϕ,θ)=2−ρ4ρ(ρ,ϕ,θ)=2−\dfrac{ρ}{4} (note the abuse of notation, with ρρ meaning both the spherical radius and the density function), find its center of mass.

Solution

Since the body is part of a sphere, we'll use spherical coordinates. If that wasn't enough of a tipoff, the density function was given in terms of spherical coordinates. To use spherical coordinates, we need to recall that x=ρsin⁡ϕcos⁡θx=ρ \sin ϕ \cos θ, y=ρsin⁡ϕsin⁡θy=ρ \sin ϕ \sin θ, and z=ρcos⁡ϕz=ρ \cos ϕ.

m=∭Dρ(ρ,ϕ,θ) ρ2 sin⁡ϕ dρ dϕ dθ=∫02π∫0π/2∫04(2−ρ4)ρ2 sin⁡ϕ dρ dϕ dθ=∫02π∫0π/2803sin⁡ϕ dϕ dθ=∫02π−803cos⁡ϕ∣0π/2 dθ=∫02π803 dθ=160π3\begin{aligned} m &= \iiint_D \rho(\rho,\phi,\theta)\ \rho^2\ \sin\phi\ d\rho\ d\phi\ d\theta\ &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_0^4 \left(2-\dfrac{\rho}{4}\right) \rho^2\ \sin\phi\ d\rho\ d\phi\ d\theta\ &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \dfrac{80}{3} \sin\phi\ d\phi\ d\theta\ &= \int_{0}^{2\pi} -\dfrac{80}{3} \cos\phi\bigg|0^{\pi/2}\ d\theta\ &= \int{0}^{2\pi} \dfrac{80}{3} \ d\theta = \dfrac{160\pi}{3} \end{aligned}

You should check this, but it turns out that x‾=y‾=0\overline{x}=\overline{y}=0, which shouldn't be surprising, since the hemisphere is centered at the origin.

z‾=3160π∫02π∫0π/2∫04ρ cos⁡ϕ (2−ρ4)ρ2 sin⁡ϕ dρ dϕ dθ=3160π∫02π∫0π/2∫04(ρ3−18ρ4) sin⁡2ϕ dρ dϕ dθ=3160π∫02π∫0π/21925sin⁡2ϕ dϕ dθ=1850π∫02πcos⁡2ϕ∣0π/2 dθ=1850π∫02π2 dθ=3625\begin{aligned} \overline{z} &= \dfrac{3}{160\pi} \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_0^4 \rho\ \cos\phi\ \left(2-\dfrac{\rho}{4}\right) \rho^2\ \sin\phi\ d\rho\ d\phi\ d\theta\ &= \dfrac{3}{160\pi} \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_0^4 \left(\rho^3-\dfrac{1}{8}\rho^4\right)\ \sin 2\phi\ d\rho\ d\phi\ d\theta\ &= \dfrac{3}{160\pi} \int_{0}^{2\pi} \int_{0}^{\pi/2} \dfrac{192}{5} \sin 2\phi\ d\phi\ d\theta\ &= \dfrac{18}{50\pi} \int_{0}^{2\pi} \cos 2\phi\bigg|0^{\pi/2}\ d\theta\ &= \dfrac{18}{50\pi} \int{0}^{2\pi} 2\ d\theta = \dfrac{36}{25} \end{aligned}

Therefore, the center of mass of this hemisphere is located at (x‾,y‾,z‾)=(0,0,3625)\ans{(\overline{x},\overline{y},\overline{z}) = \left(0,0,\dfrac{36}{25}\right)}

  1. Find the center of mass of the tetrahedron in the first octant bounded by z=1−x−yz=1−x−y and the coordinate planes if the density function is ρ(x,y,z)=1ρ(x,y,z)=1.

    2. (x‾,y‾,z‾)=(14,14,14)(\overline{x},\overline{y},\overline{z}) = \left(\dfrac{1}{4},\dfrac{1}{4},\dfrac{1}{4}\right)

  2. Consider the rectangular prism D={(x,y,z):0≤x≤4,0≤y≤1,0≤z≤1}D={(x,y,z) : 0≤x≤4,0≤y≤1,0≤z≤1}. If the density of this prism is ρ(x,y,z)=1+x2ρ(x,y,z)=1+\dfrac{x}{2}, find its center of mass.

    3. (x‾,y‾,z‾)=(73,12,12)(\overline{x},\overline{y},\overline{z}) = \left(\dfrac{7}{3},\dfrac{1}{2},\dfrac{1}{2}\right)

  3. Find the center of mass of the solid bounded by the upper half of the sphere ρ=6ρ=6 and z=0z=0 with ρ(ρ,ϕ,θ)=1+ρ4ρ(ρ,ϕ,θ)=1+\dfrac{ρ}{4}.

    4. (x‾,y‾,z‾)=(0,0,19885)(\overline{x},\overline{y},\overline{z}) = \left(0,0,\dfrac{198}{85}\right)

  4. Find the center of mass of the prism formed by z=xz=x, x=1x=1, y=4y=4, and the coordinate planes, if ρ(x,y,z)=2+yρ(x,y,z)=2+y.

    5. (x‾,y‾,z‾)=(23,73,13)(\overline{x},\overline{y},\overline{z}) = \left(\dfrac{2}{3},\dfrac{7}{3},\dfrac{1}{3}\right)