Center of gravity

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This toy uses the principles of center of gravity to keep balance on a finger.

The Center of gravity of a body is that point through which the resultant of the system of parallel forces formed by the weights of all the particles constituting the body passes for all positions of the body. It is denoted as "C.G" or "G". In a uniform gravitational field the center of gravity is identical to the center of mass.[1]

Introduction

Every body is attracted by gravity towards the center of the earth. This force of attraction is proportional to the mass of the body, perpendicular to the surface of the earth, and directed towards the center of the earth. It is known as the weight of the body. For bodies that are small relative to the earth, the constituting parts of the body can be assumed to be at equal distances from the center of the earth, and it can also be assumed that the forces formed by those parts are also parallel to each other.
The resultant of all these parallel forces is the total weight of the body. This resultant force passes through a single point for all positions of the body. That point is called the center of gravity.

Centroid[2]

Different geometrical shapes such as the circle, triangle and rectangle are plane figures having only 2-dimensions. They are also known as laminas. They have only area, but no mass. The center of gravity of these plane figures is called as the Centroid. It is also known as the geometrical center. The method of finding out the centroid of a plane figure is the same as that of finding out the center of gravity of a body. If the lamina is assumed to have uniform mass per unit area, then the centroid is also the center of gravity in a uniform gravitational field.

Methods to calculate center of gravity[3]

1. By geometrical consideration.
2. By moments.
3. By graphical method.

The first two methods are generally used to find out the center of gravity or centroid, as the third method can become tedious.

Center of gravity by geometrical consideration

1. The center of gravity of a circle is its center.
2. The center of gravity of a square,rectangle or a parallelogram is at the points where its diagonals meet each other. It is also the middle point of the length as well as the width.
3. The center of gravity of a triangle is at the point where the medians of the triangle meet.
4. The center of gravity of a right circular Cone is at a distance of ${\displaystyle {\frac {h}{4}}\!}$ from its base.
5. The center of gravity of a hemisphere is at a distance of ${\displaystyle {\frac {3r}{8}}\!}$ from its base.
6. The center of gravity of a segment of a sphere of radius h is at a perpendicular distance of ${\displaystyle {\frac {3}{4}}{\frac {(2r-h)^{2}}{(3r-h)}}\!}$ from the center of the sphere.
7. The center of gravity of a semicircle is at a perpendicular distance of ${\displaystyle {\frac {4r}{3\pi }}\!}$ from its center.
8. The center of gravity of a trapezium with parallel side a and b is at a distance of ${\displaystyle {\frac {2}{3}}{\frac {(b+2a)}{(b+a)}}\!}$ measured from the base b.
9. The center of gravity of a cube of side L is at a distance of ${\displaystyle {\frac {L}{2}}\!}$ from every face.
10. The center of gravity of a Sphere of diameter d is at a distance of ${\displaystyle {\frac {d}{2}}\!}$ from every point.

Center of gravity by Moments

Center of gravity By moments

Consider a body of mass "M" whose center of gravity is required to be found out. Let "g" be the acceleration due to gravity. Then the weight of the body is "Mg". Divide the body into small particles having equal masses, whose center of gravity is known as shown in the figure. Let their weights be m1g1, m2g2, m3g3............,etc., and (X1, Y1), (X2, Y2), (X3,Y3)..............,etc., be the coordinates of their center of gravity from a fixed point "o".

Let "G" be the center of gravity of the whole body then, ${\displaystyle {\bar {X}}}$ and ${\displaystyle {\bar {Y}}}$ are the coordinates of "G" from "o".
From the principle of moments, we know that,
Mg ${\displaystyle {\bar {X}}}$ = m1gX1 + m2gX2 + m3gX3............,

Mg ${\displaystyle {\bar {X}}}$ = g( m1X1 + m2X2 + m3X3............,)

M ${\displaystyle {\bar {X}}}$ = m1X1 + m2X2 + m3X3............,

M ${\displaystyle {\bar {X}}}$ = ${\displaystyle \sum mx}$

but, M = m1 + m2 + m3............ = ${\displaystyle \sum m}$

therefore, ${\displaystyle {\bar {X}}}$ = ${\displaystyle {\frac {\sum mx}{\sum m}}\!}$

Similarly, ${\displaystyle {\bar {Y}}}$ = ${\displaystyle {\frac {\sum my}{\sum m}}\!}$

Axis of reference

The center of gravity of a body is always calculated with reference to some assumed axis known as the axis of reference. The axis of reference for plane figures (laminas) is usually taken as the lowest line touching the lamina which is parallel to the horizontal X axis, for calculating ${\displaystyle {\bar {Y}}}$ the vertical distance of the center of gravity from this axis. Similarly, the line touching the left outermost edge which is parallel to the vertical Y axis is usually used for calculating ${\displaystyle {\bar {X}}}$, the horizontal distance of the center of gravity from this axis.

Centroid of plane figures[4]

Plane geometrical figures such as T-sections, I-sections, L-sections etc., have only areas but no mass. The centroid (center of area) of those figures is found out in the same way as that of solid bodies. The centroid will also be the center of gravity if the lamina has uniform mass per unit area.

Consider a lamina as the above figure,let its area be "A" .Divide the lamina into elemental areas a1, a2, a3............,etc.,. And (X1, Y1), (X2, Y2), (X3,Y3)..............,etc., are the coordinates of their center of areas from the reference axis "Y-Y" .

Let "G" centroid of the whole lamina, and ${\displaystyle {\bar {X}}}$ and ${\displaystyle {\bar {Y}}}$ are the coordinates of "G"(Centroidal distances) from the reference axes "Y-Y" and "X-".
From the principle of moments, we know that,

                 A ${\displaystyle {\bar {X}}}$ =  a1X1 + a2X2 + a3X3............,

                 A ${\displaystyle {\bar {X}}}$ = ${\displaystyle \sum ax}$

        but,     A = a1 + a2 + a3............ =   ${\displaystyle \sum a}$

therefore,      ${\displaystyle {\bar {X}}}$ = ${\displaystyle {\frac {\sum ax}{\sum a}}\!}$

 Similarly,     ${\displaystyle {\bar {Y}}}$ = ${\displaystyle {\frac {\sum ay}{\sum a}}\!}$


References

1. ^ Strength Of Materials - By S.Ramamrutham and R.Narayan , ISBN 81-87433-54-X
2. ^ A Text Book Of Strength Of Materials - By Dr. R.K.Bansal. ISBN 978-81-318-0814-6.
3. ^ A Text Book Of Strength Of Materials - By R.S.Khurmi. ISBN 81-219-2822-2
4. ^ A Text Book Of Engineering Mechanics - By R.S.Khurmi. ISBN 81-219-0651-2