Moments, Center of Mass and Centroid of a Planar Lamina
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how to solve word problems basics of moments and center of mass integration 
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Moment, Center of Mass, Center of Gravity and Centroid of a Planar Lamina
For a Region (Thin Plate, Planar Lamina)
The center of mass a region is the point in which the region will be perfectly balanced horizontally if suspended from that point. [srcPaul's Online Notes] The center of mass is affected by it's shape, it's density and gravity. 
The center of gravity is the same as the center of mass if gravity is the same everywhere in the region. This is so common that many people will use the terms center of mass and center of gravity interchangably. 
The centroid is the same as center of mass if the density is the same everywhere in the region. Again, this is extremely common, so some people just talk about the centroid. 
Before we go on, let's watch a video discussing the center of gravity for a platelike object. We usually call this type of object a planar lamina.
video by Michel vanBiezen 

Here is a related video discussing centroid and center of gravity. His figure actually has a thickness but since it is the same thickness everywhere, it doesn't come into play.
video by Michel vanBiezen 

Okay, now we are ready to discuss the moments of a plane area and a wire or line. Here is a great video.
video by Michel vanBiezen 

Table of Equations
The moment and center of mass equations for a planar lamina are given in Table 1. In this table, ρ is the density.
Table 1  Planar Lamina Moments and Center of Mass Equations 

moment about the xaxis 
\(\displaystyle{M_x=\rho\int_{a}^{b}{(1/2)[f(x)+g(x)][f(x)g(x)]~dx}}\) 
moment about the yaxis 
\(\displaystyle{M_y=\rho\int_{a}^{b}{x[f(x)g(x)]~dx}}\) 
total mass 
\(\displaystyle{m=\rho\int_{a}^{b}{f(x)g(x)~dx}}\) 
center of mass \(\bar{x},\bar{y}\) 
\(\displaystyle{\bar{x}=\frac{M_y}{m}, \bar{y}=\frac{M_x}{m}}\) 
\(f(x) \geq g(x), [a,b]\); \(\rho\) is the planar mass density and is a constant in these equations 
Here is a short video explaining how to derive these equations.
video by David Lippman 

Okay, with all that information you should be able to succesfully work these practice problems.
Practice
Unless otherwise instructed, solve these problems giving your answers in exact, simplified form.
Find the centroid of the region bounded by \( y = \sqrt{x} \), \( y = 0 \) and \( x = 4 \). Also, find the moments \( M_x \) and \( M_y \) for the planar laminar of uniform density ρ.
Problem Statement 

Find the centroid of the region bounded by \( y = \sqrt{x} \), \( y = 0 \) and \( x = 4 \). Also, find the moments \( M_x \) and \( M_y \) for the planar laminar of uniform density ρ.
Final Answer 

\( M_x = 4\rho \), \( M_y = 64\rho/5 \), centroid = \( (12/5, 3/4) \)
Problem Statement
Find the centroid of the region bounded by \( y = \sqrt{x} \), \( y = 0 \) and \( x = 4 \). Also, find the moments \( M_x \) and \( M_y \) for the planar laminar of uniform density ρ.
Solution
video by The Organic Chemistry Tutor 

Final Answer
\( M_x = 4\rho \), \( M_y = 64\rho/5 \), centroid = \( (12/5, 3/4) \)
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Find the centroid of the region bounded by \( y = \sqrt{x} \) and \( y = x \).
Problem Statement
Find the centroid of the region bounded by \( y = \sqrt{x} \) and \( y = x \).
Solution
video by The Organic Chemistry Tutor 

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Find the centroid of the region bounded by \( y = 4x^2 \) and \( y = x+2 \).
Problem Statement
Find the centroid of the region bounded by \( y = 4x^2 \) and \( y = x+2 \).
Solution
video by The Organic Chemistry Tutor 

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Find the centroid of the region bounded by \( y = x^2 \) and \( y = 8x^2 \).
Problem Statement
Find the centroid of the region bounded by \( y = x^2 \) and \( y = 8x^2 \).
Solution
This problem is solved in two consecutive videos.
video by PatrickJMT 

video by PatrickJMT 

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Find the centroid of the region bounded by \( y = 1x^2 \) in the first quadrant.
Problem Statement
Find the centroid of the region bounded by \( y = 1x^2 \) in the first quadrant.
Solution
video by blackpenredpen 

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Find the center of gravity of part of a circle that lies in the first quadrant with radius \(R\).
Problem Statement 

Find the center of gravity of part of a circle that lies in the first quadrant with radius \(R\).
Hint 

Because of the symmetry of the area, \( \bar{x} = \bar{y} \).
Problem Statement
Find the center of gravity of part of a circle that lies in the first quadrant with radius \(R\).
Hint
Because of the symmetry of the area, \( \bar{x} = \bar{y} \).
Solution
video by Michel vanBiezen 

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Find the center of gravity of a semicircle that lies above the xaxis with radius \(R\).
Problem Statement 

Find the center of gravity of a semicircle that lies above the xaxis with radius \(R\).
Hint 

Because of the symmetry of the area, \( \bar{x} = 0 \).
Problem Statement
Find the center of gravity of a semicircle that lies above the xaxis with radius \(R\).
Hint
Because of the symmetry of the area, \( \bar{x} = 0 \).
Solution
video by Michel vanBiezen 

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Find the center of gravity of a triangle with height \(h\) and width \(b\). Draw the triangle so that the base is on the xaxis and yaxis cuts the triangle in half, going through the third point, which means that the other two sides are equal.
Problem Statement 

Find the center of gravity of a triangle with height \(h\) and width \(b\). Draw the triangle so that the base is on the xaxis and yaxis cuts the triangle in half, going through the third point, which means that the other two sides are equal.
Hint 

Because of the symmetry of the area, \( \bar{x} = 0 \).
Problem Statement
Find the center of gravity of a triangle with height \(h\) and width \(b\). Draw the triangle so that the base is on the xaxis and yaxis cuts the triangle in half, going through the third point, which means that the other two sides are equal.
Hint
Because of the symmetry of the area, \( \bar{x} = 0 \).
Solution
video by Michel vanBiezen 

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Find the center of gravity of the top half of an ellipse, centered at the origin. The equation of the ellipse is \(\displaystyle{ \frac{x^2}{a} + \frac{x^2}{b} = 1 }\) with \(a\) along the xaxis.
Problem Statement
Find the center of gravity of the top half of an ellipse, centered at the origin. The equation of the ellipse is \(\displaystyle{ \frac{x^2}{a} + \frac{x^2}{b} = 1 }\) with \(a\) along the xaxis.
Solution
video by Michel vanBiezen 

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Find the center of gravity of the parabolic area \( y = kx^2 \) with width \(2a\) and height \(h\).
Problem Statement 

Find the center of gravity of the parabolic area \( y = kx^2 \) with width \(2a\) and height \(h\).
Hint 

\( k = h/a^2 \)
Problem Statement
Find the center of gravity of the parabolic area \( y = kx^2 \) with width \(2a\) and height \(h\).
Hint
\( k = h/a^2 \)
Solution
video by Michel vanBiezen 

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Find the center of gravity of the area bounded by \( y = kx^2 \), \( y = 0 \) and \( x = a \).
Problem Statement 

Find the center of gravity of the area bounded by \( y = kx^2 \), \( y = 0 \) and \( x = a \).
Hint 

In this video solution, he labels the height of the area \(h\) and gives his answer in terms of \(h\). However, \(h=ka^2\).
Problem Statement
Find the center of gravity of the area bounded by \( y = kx^2 \), \( y = 0 \) and \( x = a \).
Hint
In this video solution, he labels the height of the area \(h\) and gives his answer in terms of \(h\). However, \(h=ka^2\).
Solution
video by Michel vanBiezen 

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Find the center of gravity of the area bounded by \( y = kx^n \), \( y = 0 \) and \( x = a \).
Problem Statement 

Find the center of gravity of the area bounded by \( y = kx^n \), \( y = 0 \) and \( x = a \).
Hint 

In this video solution, he labels the height of the area \(h\) and gives his answer in terms of \(h\). However, \(h=ka^n\).
Problem Statement
Find the center of gravity of the area bounded by \( y = kx^n \), \( y = 0 \) and \( x = a \).
Hint
In this video solution, he labels the height of the area \(h\) and gives his answer in terms of \(h\). However, \(h=ka^n\).
Solution
This problem is solved in two consecutive videos.
video by Michel vanBiezen 

video by Michel vanBiezen 

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Find the center of gravity of a section of a circle with radius \(R\) and angle \( 2\alpha\). Center the circle so that the origin is at the center and the positive xaxis cuts the sector into two equal sections, \(\alpha\) above the xaxis and \(\alpha\) below the xaxis.
Problem Statement 

Find the center of gravity of a section of a circle with radius \(R\) and angle \( 2\alpha\). Center the circle so that the origin is at the center and the positive xaxis cuts the sector into two equal sections, \(\alpha\) above the xaxis and \(\alpha\) below the xaxis.
Hint 

Because of symmetry, \( \bar{y} = 0 \)
Problem Statement
Find the center of gravity of a section of a circle with radius \(R\) and angle \( 2\alpha\). Center the circle so that the origin is at the center and the positive xaxis cuts the sector into two equal sections, \(\alpha\) above the xaxis and \(\alpha\) below the xaxis.
Hint
Because of symmetry, \( \bar{y} = 0 \)
Solution
video by Michel vanBiezen 

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Find the center of mass of the region bounded by the triangle in the figure.
Problem Statement
Find the center of mass of the region bounded by the triangle in the figure.
Solution
This problem is solved in two consecutive videos.
video by patrickJMTPhysics 

video by patrickJMTPhysics 

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