Today was a special day because I got to do brunch with my friends. We went to a fancy bakery where they serve the best lattes in town. While waiting for our order, my friends did something cool with their spoons and forks. Let me show you.
Figure 1. Two forks and a toothpick balanced on a glass
At first, I was surprised. If someone has tape then this becomes a simple task. But who carries tape to brunch? Looks like there is a physics hack here.
In fact, one of my friends took advantage of the center of mass of the arrangement of cutlery. By knowing the center of mass, they could do an arrangement such that all the weight of the cutlery is supported by the glass. Here you will learn about density and center of mass and how to calculate each.
Definitions of Density and Center of Mass
Suppose you have a wooden plank of negligible weight, which does not break or flex, resting on top of a fulcrum at half its length.
Figure 2. A plank of negligible weight resting on top of a fulcrum
If you were to put a weight on one of the ends of the plank, then the plank would tilt towards the weight, and the weight would fall down!
Figure 3. A weight placed at one end of the plank, tilting the plank and weight configuration
However, if you place the weight exactly where the fulcrum is, then the plank would remain horizontal, and everything would be balanced.
Figure 4. A weight balanced in the middle of the plank
In order to describe this scenario, you need to know the definition of the moment.
Definition of Moment
The moment with respect to a point is defined as the product of the mass times its distance to that point.
Let \(A\) be an object with mass \(m_A\) located at a relative position \(x_A\) with respect to the origin. The moment of \(A\) is defined as the product of \(m_A\) and its relative position \(x_A\), that is
\[ m_Ax_A.\]
In the above plank example, you found that by placing the weight, which shall be labeled as \(A\), exactly where the fulcrum is located, the system was balanced. This is because the position of the fulcrum represents the origin, so \(x_A=0\) and the moment of \(A\) is given by
\[ \begin{align} m_Ax_A &= m_A(0) \\ &= 0.\end{align}\]
However, if you place the weight elsewhere, the plank will tilt. In this case, \(x_A \neq 0\), so its moment would also be other than zero!
So far, it looks like if the moment is equal to zero, then the system will remain in balance, otherwise, it will start to tilt. Before jumping to any conclusions, you should explore what happens when you add more weight.
First, suppose you place the weight \(A\) on the right end of the plank, and add an identical weight, labeled as \(B\), on the other end of the plank. Since the plank rests on the fulcrum at half of its length \(\ell\), then
\[ x_A = \frac{\ell}{2}\]
and
\[ x_B = -\frac{\ell}{2}.\]
Because the weights weigh the same you can just call them \(m\), so
\[m_A = m\]
and
\[m_B = m.\]
You should expect this system to be balanced, right? Let's find the total moment of the system, which can be found by simply adding the moment of \(A\) and the moment of \(B\), so
\[ \begin{align} m_Ax_A + m_Bx_B &= m\left(\frac{\ell}{2} \right) + m\left( - \frac{\ell}{2} \right) \\ &= m \left( \frac{\ell}{2} - \frac{\ell}{2} \right) \\&= m(0) \\&= 0. \end{align}\]
Figure 5. Two equal weights balanced on a wooden plank of negligible weight on top of a fulcrum
So far so good. Now suppose that one of the weights is heavier. The plank should tilt towards the heavier one, right? If it is the case, then the sum of the moments of \(A\) and \(B\) will not be equal to zero, so you can make a conclusion.
Figure 6. Two different weights on a wooden plank. The plank tilts towards the heavier weight
The weights together form what is called a distribution of mass, and if the system is not tilting in any direction we say that the system is in equilibrium.
If the moments of a distribution of mass add up to zero, then the system is in equilibrium.
Suppose you have two weights, \(m_A=2 \text{ lb} \) and \( m_B=4 \text{ lb}\) which are connected through a thin rod of negligible weight. The rod rests on top of a fulcrum located at \(\frac{2}{3}\) of its length, closer to \(m_A\). Is this distribution of mass in equilibrium?
Solution:
Imagine you just placed the weights, so there is still no tilting on the plank if any.
Figure 7. Two weights connected by a rod resting on a fulcrum at \(2/3\) of its length
From here, you can find that
\[ x_A= \frac{1}{3} \ell\]
and
\[ x_B=-\frac{2}{3} \ell, \]
so you can now find the total moment of the system, that is
\[ \begin{align} m_A x_A + m_B x_B &= (2) \left( \frac{1}{3}\ell \right) + (4) \left(- \frac{2}{3}\ell \right) \\ &= \frac{2}{3}\ell - \frac{8}{3}\ell \\&= -2\ell. \end{align} \]
Since \( \ell\) represents a length, it is different from zero, so the total moment of the distribution of mass is not equal to zero. This means that it is not in equilibrium.
Distributions of Mass in Two Dimensions
It is also possible to have a distribution of mass in two dimensions. In this case, you can distinguish between moments about the \(x-\)axis and moments about the \(y-\)axis. Despite finding the position on the \(x-\)axis, in the plank example, you were referring to the moment with respect to the \(y-\)axis, as you were measuring how far each weight was away from the \(y-\)axis. Because of this, the moment with respect to the \(y-\)axis of the weight \(A\) is defined as
\[ m_Ax_A,\]
while its moment with respect to the \(x-\)axis is defined as
\[ m_Ay_A.\]
To find if the system is in equilibrium you have to consider the moments with respect to both axes separately. This means that you should find the sum of moments with respect to the \(x-\)axis, and the sum of moments with respect to the \(y-\)axis.
If the sum of the moments with respect to both axes are both equal to zero, then the system is in equilibrium with respect to the origin.
The positions in the Cartesian plane of three weights are given by
\[ A=(3,2),\]
\[ B=(-2,1),\]
and
\[ C=(0,3).\]
The mass of \(A\) is equal to \(m\), \(B\) weighs twice as much as \(A\), and \(C\) weighs three times as much as \(A\). Is this mass configuration in equilibrium about the origin?
Solution:
To find if the mass distribution is at equilibrium about the origin you need to find if it is in equilibrium about both axes.
Begin by finding if it is in equilibrium about the \(y-\)axis, so you should calculate
\[ m_A x_A + m_B x_B + m_C x_C. \]
Remember that if you want to find the moment about the \(y-\)axis, you need to use the \(x-\)values. Likewise, to find the moment about the \(x-\)axis, you use the \(y-\)values.
You are given that \(m_A=m\). Since \(B\) weighs twice as much and \(C\) three times as much, this means that \( m_B=2m\) and \(m_C=3m\). So
\[ \begin{align} m_A x_A + m_B x_B + m_C x_C &= (m)(3)+(2m)(-2)+(3m)(0) \\ &= 3m-4m+0m \\ &= -m. \end{align}\]
You can stop here, as you just found that the system is not in equilibrium with respect to the \(y-\)axis, which means that it is also not in equilibrium with respect to the origin.
For illustrative purposes you can find the moment about the \(x-\)axis, so
\[ \begin{align} m_A y_A + m_B y_B + m_C y_C &= (m)(2)+(2m)(1)+(3m)(3) \\ &= 2m+2m+9m = 13m. \end{align}\]
This means that the system is not in equilibrium about the \(x-\)axis either.
In these examples, you have been given the mass of the weights along with their positions. While for a few weights this is straightforward, it is possible to have a large amount of weights, so a function describing the distribution of mass is required.
The density is a function that describes the distribution of mass of a system.
Please note that this definition of density is within the context of moment and center of mass.
With all of this setup, you are now ready to look at the definition of center of mass.
The center of mass of a mass distribution is a point such that the sum of the moments with respect to it is equal to zero. The center of mass is also known as the center of gravity.
This means that if you place the fulcrum at the center of mass of a mass distribution, then it will be in balance.
In general, you will be asked to find the center of mass of a mass configuration in two dimensions.
The coordinates \( (x_{CM}, y_{CM}) \) of the center of mass of a mass configuration are given by
\[ x_{CM} = \frac{M_y}{M}\]
and
\[ y_{CM} = \frac{M_x}{M},\]
where \(M_y\) is the sum of the moments with respect to the \(y-\)axis, \(M_x\) is the sum of the moments with respect to the \(x-\)axis, and \(M\) is the total mass of the configuration.
You found that the mass configuration of the previous example was not in equilibrium about the origin. Now try to find its center of mass.
The positions in the Cartesian plane of three weights are given by
\[ A=(3,2),\]
\[ B=(-2,1),\]
and
\[ C=(0,3).\]
The mass of \(A\) is equal to \(m\), \(B\) weighs twice as much as \(A\), and \(C\) weighs three times as much as \(A\). Find the center of mass of this configuration.
Solution:
You previously found that
\[ M_y = -m \]
and
\[ M_x = 13m, \]
now you only need to find the total mass of the configuration, that is
\[ \begin{align} M &= m_A+m_B+m_C \\ &= m+2m+3m \\ &= 6m. \end{align}\]
Finally, you can find the center of mass, so
\[ \begin{align} x_{CM} &= \frac{M_y}{M} \\ &= \frac{-m}{6m} \\ &= -\frac{1}{6} \end{align}\]
and
\[ \begin{align} y_{CM} &= \frac{M_x}{M} \\ &= \frac{13m}{6m} \\ &= \frac{13}{6}.\end{align}\]
This means that the center of mass of the configuration is located at
\[ \left( -\frac{1}{6}, \frac{13}{6} \right).\]
Density and Center of Mass in Calculus
So far you have been looking at mass configurations that consist of weights that are taken as if they were points. These configurations are called discrete mass configurations because they consist of a finite, countable amount of masses.
However, objects like planks, rods, sheets, and laminas do have mass! These objects can be thought of as if they were continuous mass configurations, so this is where Calculus comes in.
There are two things you can do when dealing with continuous mass configurations:
- You can be given a function that describes the density of an object, in which case you will be tasked with finding its total mass.
- You can be given a function that describes the shape of an object, in which case you will be tasked with finding its center of mass.
For example, suppose you need to find the center of mass of a thin lamina whose shape can be seen as the area below a curve.
Figure 8. A thin lamina that has the shape of the area below a parabola
The area below a curve sounds familiar, right? To address this scenario you will need to use integrals.
Density and Center of Mass using Integrals
You can use integrals to either find the mass of an object given its density or find its center of mass given its shape.
Center of Mass using Integrals
When tasked with finding the center of mass of a lamina described by the area below a curve you will usually be told that it has a constant (or uniform) mass density \( \rho\). In order to do so, you can still use the formulas
\[ x_{CM} = \frac{M_y}{M}\]
and
\[ y_{CM} = \frac{M_x}{M}.\]
The difference is that you have to find \(M_y\) and \(M_x\) using a different approach, namely
\[ M_y =\rho \int_a^b xf(x)\,\mathrm{d}x, \]
\[ M_x = \frac{1}{2} \rho \int_a^b f(x) ^2 \, \mathrm{d}x,\]
and
\[ M = \rho \int_a^b f(x)\,\mathrm{d}x.\]
This is better understood with an example.
Find the center of mass of a lamina with constant density \( \rho\) whose shape is described by the area below
\[ f(x)= 4-x^2\]
in the interval \( [-2,2]\).
Solution:
Here you will need to find the moment with respect to both axes and the total mass of the lamina. For the moment with respect to the \(y-\)axis you need to find
\[ \begin{align} M_y &= \rho \int_a^b x f(x) \, \mathrm{d}x \\ &= \rho \int_{-2}^2 x (4-x^2)\,\mathrm{d}x, \end{align}\]
which you can do with the help of the Power Rule, so
\[ \begin{align} M_y &= \rho \int_{-2}^2 x(4-x^2)\,\mathrm{d}x \\ &= \rho \int_{-2}^2 (4x-x^3)\,\mathrm{d}x \\ &=\rho \left. \left[ 2x^2-\frac{1}{4}x^4\right]\right|_{-2}^2 \\ &= \rho \left[ \left( 2(2)^2-\frac{1}{4}(2)^4 \right) - \left(2(-2)^2-\frac{1}{4}(-2)^4 \right) \right] \\ &= \rho \left( 4-4 \right) \\ &= 0. \end{align}\]
Note that for the above integral you could have also used the fact that it is an odd function integrated over a symmetric interval, so the integral can be expected to be zero.
Next, find the moment with respect to the \(x-\)axis, so
\[ \begin{align} M_x &= \frac{1}{2} \rho \int_a^b (f(x))^2\,\mathrm{d}x, \\ &= \frac{1}{2}\rho \int_{-2}^2 (4-x^2)^2\,\mathrm{d}x, \end{align} \]
where you will first need to expand the binomial
\[ M_x = \frac{1}{2}\rho \int_{-2}^2 (16-8x^2+x^4)\,\mathrm{d}x.\]
This time you should identify that you are integrating an even function over a symmetric interval, so
\[ \int_{-2}^2 (16-8x^2+x^4)\,\mathrm{d}x = 2 \int_0^2 (16-8x^2+x^4)\,\mathrm{d}x, \]
which means that
\[ M_x = \rho \int_0^2 (16-8x^2+x^4)\,\mathrm{d}x.\]
Now you can use the Power Rule,
\[ \begin{align} M_x &= \rho \left. \left[16x-\frac{8}{3}x^3+\frac{1}{5}x^5 \right] \right|_{0}^2 \\ &= \rho \left[ \left(16(2)-\frac{8}{3}(2)^3+\frac{1}{5}(2)^5 \right) - \left( 16(0)-\frac{8}{3}(0)^3+\frac{1}{5}(0)^5\right)\right] \\ &= \rho \left( \frac{256}{15} - 0\right)\\ &= \frac{256}{15}\rho. \end{align}\]
Finally, find the mass of the lamina, which is given by
\[ \begin{align} M &= \rho \int_a^b f(x)\,\mathrm{d}x \\ &= \rho \int_{-2}^2 (4-x^2) \, \mathrm{d}x \\ & = 2 \rho \int_0^2 (4-x^2)\,\mathrm{d}x \\ &= 2\rho \left. \left( 4x-\frac{1}{3}x^3 \right) \right |_0^2, \end{align}\]
where you have once again used the fact that the integrand is symmetric. Finish the evaluation by substituting \(x=2\) and \(x=0\), that is
\[ \begin{align} M &= 2 \rho \left[ \left( 4(2)-\frac{1}{3}(2)^3\right)-\left( 4(0)-\frac{1}{3}(0)^3\right) \right] \\ &=2\rho \left(8-\frac{8}{3}\right) \\&= \frac{32}{3}\rho. \end{align}\]
Finally, you can find the coordinates of the center of mass by substituting the values you just found, that is
\[ \begin{align} x_{CM} &= \frac{M_y}{M}\ \\ &= \frac{0}{\frac{32}{3} \rho} \\ &=0\end{align},\]
and
\[ \begin{align} y_{CM} &= \frac{M_x}{M}\ \\ &= \frac{\frac{256}{15}\rho}{\frac{32}{3} \rho} \\ &=\frac{8}{5}\end{align}.\]
Figure 9. Lamina with uniform density and its center of mass
Finding the Mass of an Object using Integrals
You might also be tasked with finding the total mass of an object when the density is not constant, that is
\[ \rho = \rho (x).\]
Usually, this will be done in one dimension since doing it in two or three dimensions will require double or triple integrals, which is out of the AP scope.
To find the mass of an object given a function describing its density, you just need to integrate, that is
\[ M = \int_a^b \rho(x) \, \mathrm{d}x.\]
Find the mass \(M\) of a non-uniform density rod of length \(2\) whose density is described by the function
\[ \rho(x) = \frac{1}{4}(x-1)^2,\]
where \(x\) represents the distance away from one end of the rod.
Solution:
Begin by taking a look at the graph of the function describing the density of the rod, which is a parabola.
Figure 10. Graph of \( \rho(x)\). The area below the curve represents the mass of each section of the rod
You can note how most of the weight of the rod is located at its ends. To find the total mass you will need to solve the definite integral
\[ M = \int_0^2 \frac{1}{4}(x-1)^2\,\mathrm{d}x.\]
You can expand the binomial, but you will find more straightforward to use substitution with
\[ u=x-1,\]
so
\[ \mathrm{d}u=\mathrm{d}x.\]
This way you can find the indefinite integral
\[ \begin{align} \int \frac{1}{4}(x-1)^2\,\mathrm{d}x &= \int \frac{1}{4}u^2\,\mathrm{d}x \\ &= \frac{1}{4}\left( \frac{1}{3} u^3\right) \\ &= \frac{1}{12}(x-1)^3, \end{align} \]
whereas usual, you do not need to add the integration constant because your goal is the evaluation of a definite integral. Now you can use the Fundamental Theorem of Calculus and evaluate the definite integral, so
\[ \begin{align} M &= \left. \left( \frac{1}{12}(x-1)^3 \right) \right |_0^2 \\ &= \left( \frac{1}{12}(2-1)^3\right) - \left( \frac{1}{12}(0-1)^3 \right) \\ &= \frac{1}{12}-\left(-\frac{1}{12} \right) \\ &= \frac{1}{6}. \end{align}\]
This means that the rod weighs \( \frac{1}{6}\) mass units.
Density and Center of Mass Formulas
Here you can take a look at a compilation of the formulas involving density and center of mass.
The center of mass of a mass configuration has coordinates \( (x_{CM}, y_{CM})\), which are given by
\[ x_{CM} = \frac{M_y}{M}\]
and
\[ y_{CM} = \frac{M_x}{M}.\]
If you are working with a discrete mass configuration, then
\[ M_y = m_1 x_1 + m_2 x_2 + \dots + m_n x_n,\]
\[ M_x = m_1 y_1 + m_2 y_2 + \dots + m_n y_n,\]
and
\[ M= m_1 + m_2 + \dots + m_n.\]
If you are working with a lamina with constant density \(\rho\), then
\[ M_y =\rho \int_a^b xf(x)\,\mathrm{d}x, \]
\[ M_x = \frac{1}{2} \rho \int_a^b f(x) ^2 \, \mathrm{d}x,\]
and
\[ M = \rho \int_a^b f(x)\,\mathrm{d}x,\]
where the shape of the lamina is given by the area between \( f(x) \), \( x=a\), \(x=b\), and the \(x-\)axis.
If you are asked to find the mass of a thin rod with density described by \( \rho(x)\), then
\[ M = \int_a^b \rho(x) \, \mathrm{d}x.\]
Examples of Problems about Density and Center of Mass
Here you can look at more problems involving density and center of mass.
The positions in the Cartesian plane of three weights are given by
\[ A=(-1,2),\]
\[ B=(1,-2),\]
and
\[ C=(2,-1).\]
\(A\) weighs the same as \(B\), while \(C\) weighs the same as the sum of the weights \(A\) and \(B\). Find the center of mass of this configuration.
Solution:
Begin by noting that \(A\) and \(B\) weigh the same, so you can label that value as \(m\). \(C\) weighs the same as the sum of \(A\) and \(B\), so
\[ \begin{align} m_C &= m_A+m_B \\ &= m+m \\ &=2m. \end{align}\]
Now find the moments, starting with the moment with respect to the \(y-\)axis,
\[ \begin{align} M_y &= m_A x_A + m_B x_B + m_C x_C \\ &= (m)(-1)+(m)(1)+(2m)(2) \\ &= -m+m+4m \\ &= 4m, \end{align}\]
then the moment with respect to the \(x-\)axis,
\[ \begin{align} M_x &= m_A y_A + m_B y_B + m_C y_C \\&= (m)(2)+(m)(-2)+2m(-1) \\ &= 2m-2m-2m \\ &=-2m, \end{align} \]
and the total mass of the system,
\[ \begin{align} M &= m_A + m_B + m_C \\ &= m+m+2m \\ &=4m. \end{align} \]
Finally, you can find the coordinates of the center of mass, that is
\[ \begin{align} x_{CM} &= \frac{M_y}{M} \\ &= \frac{4m}{4m} \\ &= 1, \end{align}\]
and
\[\begin{align} y_{CM} &= \frac{M_x}{M} \\ &= \frac{-2m}{4m} \\ &= -\frac{1}{2}.\end{align}\]
This means that the center of mass is located at
\[ \left( 1, -\frac{1}{2} \right).\]
How about the center of mass of a lamina?
Find the center of mass of a lamina with constant density \( \rho\) whose shape is described by the area below
\[ f(x)= 2x\]
in the interval \( [0,3]\).
Solution:
As usual, you will need to find both moments and the mass of the lamina. For the moment with respect to the \(y-\)axis you will need to calculate
\[ \begin{align} M_y &= \rho \int_a^b x f(x)\,\mathrm{d}x \\ &= \rho \int_0^3 x (2x)\,\mathrm{d}x \\ &= \rho \int_0^3 2x^2\,\mathrm{d}x, \end{align} \]
which you can find with the help of the Power Rule, that is
\[ \begin{align} M_y &= \rho \int_0^3 2x^2 \, \mathrm{d}x \\ &= \rho \left. \left( \frac{2}{3}x^3\right)\right|_0^3 \\ &= \rho \left( \frac{2}{3}(3)^3-\frac{2}{3}(0)^3 \right) \\ &= \frac{54}{3}\rho \\ &= 18\rho. \end{align}\]
Next, find the moment with respect to the \(x-\)axis by using the respective formula, so
\[\begin{align} M_x &= \frac{1}{2}\rho \int_a^b (f(x))^2\,\mathrm{d}x \\ &= \frac{1}{2}\rho \int_0^3 (2x)^2 \,\mathrm{d}x \\ &= \frac{1}{2}\rho\int_0^3 4x^2\,\mathrm{d}x \\ &= \rho \int_0^3 2x^2\,\mathrm{d}x.\end{align}\]
Note that this is the same integral you just found before, so
\[M_x = 18\rho\]
as well. Now find the mass of the lamina by using
\[ \begin{align} M &= \rho \int_a^b f(x)\,\mathrm{d}x \\ &= \rho \int_0^3 2x\,\mathrm{d}x, \end{align}\]
doing so will give you
\[ \begin{align} M &= \rho \left.\left( x^2 \right) \right|_0^3 \\ &= \rho( (3)^2-(0)^2 ) \\&= 9\rho. \end{align}\]
Finally, you can use these values to find the coordinates of the center of mass, that is
\[ \begin{align} x_{CM} &= \frac{M_y}{M} \\ &= \frac{18\rho}{9\rho} \\ &= 2\end{align}\]
and
\[ \begin{align} y_{CM} &= \frac{M_x}{M} \\ &= \frac{18\rho}{9\rho} \\ &= 2.\end{align}\]
This means that the center of mass is located at
\[ (2,2).\]
Density and Center of Mass - Key takeaways
- The moment with respect to a point is defined as the product of the mass times its distance to that point.
- If the sum of the moments with respect to a point is equal to zero, then the system is in equilibrium with respect to that point.
- The distribution of mass is a system made of many bodies that have weight.
- The distribution of mass can be either discrete or continuous.
- The density is a function that describes the distribution of mass of a system.
- The center of mass of a mass distribution is a point such that the sum of the moments with respect to it is equal to zero.
- To find the center of mass \( (x_{CM},y_{CM}) \) of a mass distribution you use the formulas\[x_{CM} = \frac{M_y}{M}\]and\[y_{CM} = \frac{M_x}{M}.\]
- If you have a discrete mass distribution then\[ M_y = m_1 x_1 + m_2 x_2 + \dots + m_n x_n, \] \[ M_x = m_1 y_1 + m_2 y_2 + \dots + m_n y_n,\]and\[ M = m_1 + m_2 + \dots + m_n.\]
- If you have a thin lamina of uniform density \( \rho\) whose shape is described by \( f(x)\), then\[ M_y =\rho \int_a^b xf(x)\,\mathrm{d}x, \] \[ M_x = \frac{1}{2} \rho \int_a^b f(x) ^2 \, \mathrm{d}x,\]and\[ M = \rho \int_a^b f(x)\,\mathrm{d}x.\]
- If you are given a non-uniform density in the form of \( \rho(x)\) then you can find the total mass of the object with an integral\[ M = \int_a^b \rho(x)\,\mathrm{d}x.\]
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