Many of us have seen performers who spin plates on the ends of sticks. Like the graph showing below. The performers try to keep several of them spinning without allowing any of them to drop. If we look at a single plate (without spinning it), there is a sweet spot on the plate where it balances perfectly on the stick. If we put the stick anywhere other than that sweet spot, the plate does not balance and it falls to the ground. (That is why performers spin the plates; the spin helps keep the plates from falling even if the stick is not exactly in the right place.)
Mathmatically, the sweet spot is called center of mass of the plate. Lucky, we could use the calculus knowlege to find it.
Ler's satrt with finding center of mass in one-dementional case. Thinking about that there are two kids seat on the two side of a seesaw in the playground. If their weight is different, and they all sit at the end of each side of seesaw, the heavier kids will sink down and lighter child is lifted into the air. However, if the heavier kid slides in toward the center, though, the seesaw balances. We can formulate this question into a math question as follwing.
Alone the line, there is mass \(m_1\) at the point \(x_1\) and mass \(m_2\) at the point \(x_2\). Find the center of mass \(\bar{x}\).
According to the Archimedes's Law of Lever: rod will be balanced if and only if \(m_1 d_1 = m_2 d_2\). The \(d_1\) and \(d_2\) is the distance from the center of mass to the mass, such that \(d_1 = \bar{x}-x_1\) and \(d_2 = x_2-\bar{x}\). so we have following :\[m_1(\bar{x}-x_1) = m_2(x_2-\bar{x})\] \[m_1\bar{x}-m_1 x_1 = m_2 x_2-m_2 \bar{x}\] \[(m_1+m_2) \bar{x}= m_1 x_1+m_2 x_2\] \[\bar{x} = \frac{m_1 x_1+m_2 x_2}{(m_1+m_2)}\]
\(m_1 x_1+m_2 x_2\) is called the first moment of the system with respect to the origin, \(m_1+m_2\) is the total mass of the system. In other words, the center of mass of the system is the point at which the total mass of the system could be concentrated without changing the moment.
This idea is not limit to mass of two point. In general, there are \(n\) mass in the system: \(m_1, m_2, \dots , m_n\), the center of mass will be \[\bar{x} = \frac{m_1x_1+ m_2x_2+\dots + m_nx_n}{m_1+m_2+\dots+m_n}\]The moment of the system about the origin which measure the tendency of the system to rotate about the origin will be \[M_0 = m_1x_1+ m_2x_2+\dots + m_nx_n\]
Now, instead of having the mass of a system concentrated at discrete points, we want to know the center of mass for the mass of the system is distributed continuously across a thin plate of material. For mathematical purposes, we assume the object is thin enough for us to treat it as 2-dementional object.
As with system of points mass, in order to find the mass of the thin plate, we need to find the mass of the plate, as well as the moment respect to \(x-aixs\) and \(y-axis\).
Let's use a example to explore how to compute it. Find the center of mass of a thin plate of constant density \(\rho\)covering the region bounded by the parabola \(y=4-x^2\) and bound by \(x-aix\).
The first step in finding the center of mass for this problem is to sketch the region, and in order to help us understand it better, we could also look at the 3-d model. Here is the graph of this region
First, let's calculate the mass of the region
\[m = \int^2_{-2} \rho (4-x^2) dx = \rho \int^2_{-2} 4-x^2 dx = \rho \left[4x-\frac{x^3}{3}\right]\bigg|^2_{-2} = \frac{32}{3} \rho \]
Next, we calculte the moment
\[M_x = \int^2_{-2} \rho \frac{(4-x^2)^2}{2} dx = \frac{\rho}{2} \int^2_{-2} (4-x^2)^2 dx = \frac{\rho}{2} \int^2_{-2} 16-8x^2+x^4 dx = \frac{\rho}{2} \left[16x-\frac{8x^3}{3}+ \frac{x^5}{5}\right]\bigg|^2_{-2} = \frac{256}{15}\rho \]
\[M_y = \int^2_{-2} \rho x (4-x^2) dx = \rho \int^2_{-2}x(4-x^2) dx = \rho \int^2_{-2}4x-x^3 dx= \rho \left[ 2x^2 - \frac{x^4}{4}\right]\bigg|^2_{-2} = 0\]
Then, we have
\[\bar{x} = \frac{M_y}{m} = 0, \ \bar{y} = \frac{M_x}{m} = \frac{256 \rho}{15} \cdot \frac{3}{32 \rho} =\frac{8}{5}\]
Thus, the center of mass of this region is \((0,\frac{8}{5})\)
Note that the \(\bar{x}=0\). Why? Becuase The region is symmetric with respect to the \(y-aixs\), and according to the Symmetry Principle, if a region \(R\) is symmetric about a line \(l\), then the center of mass of R lies on \(l\). Obviously, for this question the center of mass lies on \(y-aixs\), thus the \(\bar{x}=0\).
The reason why I choose this question is that we can not only learn how to find the center of limit, but also know how to use Symmetry Principle for the futher calculas question.
Let's summery those kinds of question in the following theorem. Let R denote a region bounded above by the graph of a continuous function \(f(x)\), below by the x-axis, and on the left and right by the lines \(x=a\) and \(x=b\), respectively. Let \rho denote the density plate. Then we can make the following statements:
1. The mass of the plate is : \[m = \int^a_b \rho f(x) dx\]
2. The moment respect to \(x-aixs\) and \(y-axis\): \[M_x = \int^a_{b} \rho \frac{f(x)}{2} dx \] \[M_y = \int^a_{b} \rho x f(x) dx\]
3. Center of Mass \(\bar{x},\bar{y}\) is : \[\bar{x} = \frac{M_y}{m}, \ \bar{y} = \frac{M_x}{m}\] Author: Yueqi Li (Nicole)
Ler's satrt with finding center of mass in one-dementional case. Thinking about that there are two kids seat on the two side of a seesaw in the playground. If their weight is different, and they all sit at the end of each side of seesaw, the heavier kids will sink down and lighter child is lifted into the air. However, if the heavier kid slides in toward the center, though, the seesaw balances. We can formulate this question into a math question as follwing.
Alone the line, there is mass \(m_1\) at the point \(x_1\) and mass \(m_2\) at the point \(x_2\). Find the center of mass \(\bar{x}\).
This idea is not limit to mass of two point. In general, there are \(n\) mass in the system: \(m_1, m_2, \dots , m_n\), the center of mass will be \[\bar{x} = \frac{m_1x_1+ m_2x_2+\dots + m_nx_n}{m_1+m_2+\dots+m_n}\]The moment of the system about the origin which measure the tendency of the system to rotate about the origin will be \[M_0 = m_1x_1+ m_2x_2+\dots + m_nx_n\]
Now, instead of having the mass of a system concentrated at discrete points, we want to know the center of mass for the mass of the system is distributed continuously across a thin plate of material. For mathematical purposes, we assume the object is thin enough for us to treat it as 2-dementional object.
As with system of points mass, in order to find the mass of the thin plate, we need to find the mass of the plate, as well as the moment respect to \(x-aixs\) and \(y-axis\).
Let's use a example to explore how to compute it. Find the center of mass of a thin plate of constant density \(\rho\)covering the region bounded by the parabola \(y=4-x^2\) and bound by \(x-aix\).
The first step in finding the center of mass for this problem is to sketch the region, and in order to help us understand it better, we could also look at the 3-d model. Here is the graph of this region
The reason why I choose this question is that we can not only learn how to find the center of limit, but also know how to use Symmetry Principle for the futher calculas question.
Let's summery those kinds of question in the following theorem. Let R denote a region bounded above by the graph of a continuous function \(f(x)\), below by the x-axis, and on the left and right by the lines \(x=a\) and \(x=b\), respectively. Let \rho denote the density plate. Then we can make the following statements:
1. The mass of the plate is : \[m = \int^a_b \rho f(x) dx\]
2. The moment respect to \(x-aixs\) and \(y-axis\): \[M_x = \int^a_{b} \rho \frac{f(x)}{2} dx \] \[M_y = \int^a_{b} \rho x f(x) dx\]
3. Center of Mass \(\bar{x},\bar{y}\) is : \[\bar{x} = \frac{M_y}{m}, \ \bar{y} = \frac{M_x}{m}\] Author: Yueqi Li (Nicole)
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