Target Audience: High School Students, College Freshmen and Sophomores, students preparing for the International Baccalaureate (IB), AP Physics B, AP Physics C, A Level, Singapore/GCE ALevel;Class 11/12 students in India preparing for ISC/ICSE/CBSE and Entrance Examinations like the IITJEE/AIEEE. This might also be helpful in studying topics required by Common Core Physics.
This might also be helpful in studying topics required by Common Core Physics.
A Quick Summary of what we'll study in this chapter and the kind of problems we'll solve(after this intro, there is a comprehensive document with study material as well as solutions to problems.)
CENTROIDS AND CENTRE OF GRAVITY In general when a rigid body lies in a field of force acts on each particle of the body. We equivalently represent the system of forces by single force acting at a specific point. This point is known as centre of gravity. We can extend this concept in many ways and get the various equivalent parameters of a body, which could help us in dealing the situation directly on a rigid body rather than considering each individual particle of the rigid body. Various such parameters include centre of gravity, moment of inertia, centroid , first and second moment of inertias of a line or a rigid body. These parameters simplify the analysis of structures such as beams. Further we will also study the surface area or volume of revolution of a line or area respectively.
CENTRE OF GRAVITYConsider the following lamina. Let’s assume that it has been exposed to gravitational field. Obviously every single element will experience a gravitational force towards the centre of earth. Further let’s assume the body has practical dimensions, then we can easily conclude that all elementary forces will be unidirectional and parallel. Consider G to be the centroid of the irregular lamina. As shown in first figure we can easily represent the net force passing through the single point G. We can also divide the entire region into let’s say n small elements. Let’s say the coordinates to be (x1,y1), (x2,y2), (x3,y3)………. (xn,yn) as shown in figure . Let ΔW1, ΔW2, ΔW3,……., ΔWn be the elementary forces acting on the elementary elements. Clearly, W = ΔW1+ ΔW2+ ΔW3 +…………..+ ΔWn When n tends to infinity ΔW becomes infinitesimally small and can be replaced as dW. Centre of gravity : xc= / , yc= / ( and zc= / in case of a three dimensional body) where x,y are the coordinate of the small element and dw(or ΔW) the elemental force. And we have seen that W.For some type of surfaces of bodies there lies a probability that the centre of gravity may lie outside the body. Secondly centre of gravity represents the entire lamina, therefore we can replace the entire body by the single point with a force acting on it when needed. There is a major difference between centre of mass and centre of gravity of a body. For centre of gravity we integrate with respect to dW whereas for centre of mass we integrate with respect to dm. Mass is a scalar quantity and force a vector quantity. For general practical size objects both of them turn out to be the same as both of them are proportional and the force is unidirected (dW = dm*g) .But when we consider large size objects such as a continent, results would turn out to be different because here the vector nature of dW comes into play.
CENROIDS OF AREAS AND LINES
We have seen one method to find out the centre of gravity, there are other ways too. Let’s consider plate of uniform thickness and a homogenous density. Now weight of small element is directly proportional to its thickness, area and density as: ΔW = ϒ t dA. Where ϒ is the density per unit volume, t is the thickness , dA is the area of the small element. Let’s consider plate of uniform thickness and a homogenous density. Now weight of small element is directly proportional to its thickness, area and density as: ΔW = ϒ t dA. Where ϒ is the density per unit volume, t is the thickness , dA is the area of the small element. So we can replace ΔW with this relationship in the expression we obtained in the prior topic. Therefore we get: Centroid of area : xc= / , yc= / (and zc= / in case of a three dimensional body) Where x,y are the coordinate of the small element and da(or ΔA) the elemental force. Also A (total area of the plate). (xc ,yc,zc) is called the centroid of area of the lamina. If the surface is homogenous we conclude that it is the same as centre of gravity.
There can also arise a case where in crosssectional area is constant and length is variable as in the case of a rope or slender rod. In such cases the situation modifies to: ΔW = ϒ a dl.Where ϒ is the weight per unit length, per unit crosssectional area, A is the area of cross –section, and dl the variable length. So the above results reduce to: Centroid of a line: xl= / , yl= / ( and zl= / in case of a three dimensional body) where x,y are the coordinate of the small element and dl(or ΔL) the elemental force.Also L(total area of the plate).
The coordinate (xl ,yl,zl) is called the centroid of a line. It is important to mention that centroids of line may or may not lie on the line( as shown in diagram above).
FIRST MOMENT OF AREAS AND LINES
First moment of area is defined as: Qy = , Qx = , from the above discussion on the centroid of area it is clear that we can rewrite the expression as : Qy = xc A , Qx = yc A , where A is the total area , and (xc,yc) is the coordinate of the centroid of the given area.
Thus it follows from the above discussion that centroid of a area can be determined by dividing first moment of the area with the area itself. If the first moment of a area with respect to an axis is zero , it indicates that the point lies on that axis itself.[Mx = 0 , implies yc =0 , implies point lies on x axis]. In a similar way we can define the first moment of a line as: Qy = , Qx = , which implies: Qy = xl L , Qx = yl L , If the given line or area is symmetrical finding the centroid becomes easy. x Because of symmetrical nature will always turn out to be zero. Hence Qy = 0. So we can conclude that the first moment about the axis will be zero about the axis of symmetry(y axis in the above example). Further centroid also lies on the axis of symmetry (figure out why?). If a body has more than one axis of symmetry then centroid will lie on the point of intersection of the axes.
CENTROID OF SOME STANDARD GEOMETRIC FIGURESFollowing results are obtained by integration which will be explained later. Results for symmetrical objects like square, circle, cylinder, rectangle, ring etc are omitted. For such cases centroids can be preassumed to be the geometric centre of the body. shape 
 xc  yc  area  Triangular 

 h/3  bh/2  Semicircular area
Quarter circular area 
 0
4r/3  4r/3
4r/3  */2
 Circular sector 
 2rsin  0  *r2  shape 
 xc  yc  length  Semielliptical area Quarter elliptical area 

4a/3  4b/3
4b/3  *
 Semiparabola area Quarter parabola area 
 0
3a/8  3h/5
3h/5  4ah/3
2ah/3  General spandrel 
 a(n+1)/ (n+2)  h(n+1)/ (4n+2)  ah/n+1  Arc of a circle 
 rsinα/α  0  2αr  Semicircular wire Quarter circular wire 
 0
2r/  2r/
2r/  *

Similar concept of lines and areas can be extended to volume also. The new relationship are: xc= / , yc= / , zc= / .
CENTROIDS OF COMPOSITE AREASWe can end up in situations where the given plate can be broken up into various segments. In such cases we can replace the separate sections by their centre of gravity. One centroid takes care of the entire weight of the section. Further overall centre of gravity can be found out using the same concept we studied before. Xc (W1 + W2 + W3+…..+Wn) = xc1W1 + xc2W2 + xc3W3+…….……..+xcnWn Yc (W1 + W2 + W3+…..+Wn) = yc1W1 + yc2W2 + yc3W3+…….……..+ycnWn Once again if the plate is homogenous and of uniform thickness, centre of gravity turns out to be equal to the centroid of the area. In a similar way we can also define centroid of this composite area by: Xc (A1 + A2 + A3+…..+An) = xc1A1 + xc2A2 + xc3A3+…….……..+xcnAn Yc (A1 + A2 + A3+…..+An) = yc1A1 + yc2A2 + yc3A3+…….……..+ycnAn We can also introduce the concept of negative area. It simply denotes the region where any area is left vacant. We will see its usage in the coming problems.
THEOREMS OF PAPPUSGULDINUSFirstly what is a surface of revolution? A surface of revolution represents the surface generated by rotating a plane curve about an axis. For example a cone can be generated by revolving a semicircle about its diameter. Curved surface of a cone can be generates by revolving a straight line as shown. Theorem 1: The surface area of the revolution is equal to the length of the generating curve times the distance covered by centroid of the curve while the surface is being generated.(case 1 of the diagram, above) Theorem 2: The volume of the body of revolution is equal to the generating area times the distance travelled by the centroid while the body is being generated.(case 2 of the diagram, above). These theorems are very useful when calculating the centroid of a given area. This is because the outputs of the theorems (area, volumes of standard geometries) are already known to us. The only unknown quantity is the location of centroid. OBTAINING CENTROIDS BY INTEGRATIONThe general expression of centroid of a body is given by: xc= / , yc= / dA can be rewritten as dxdy , which turns it into a double integration. However, in most cases this can be simplified to a single integration. We divide the area into thin rectangular strips or sectors. For rectangle it is preknown that its centre of gravity lies at the centre of the rectangle. In such a case dA should be appropriately expressed in terms of coordinates x,y and the differentials. In case of a sector, it is known that the centroid lies at a distance of 2r/3 from the centre. The method of sector should be used when the polar equation of the boundary of the curve is known. And the area dxdy in this case is given by r * rdɵ.
In case of a line, the equations governing the centroid are as follows: xc= / , yc= / . In this case dl = = dx = dy = {in case of polar coordinates} Here's a Quick Look at the kind of Problems which have been solved in the Tutorial document at the end :
Using integration find the centroid of the parabolic area OAB as shown in the figure below. Derive the location of centroid for the following sector. Hence prove the results obtained for a semicircular area.
Find the centroid of the composite area ABCDEF. A circle of radius 0.5 units has been cut out as shown. A triangle and a quarter circle have been cut out in a similar way.
Derive the location of centroid for the following area OAB.
Complete Tutorial Document with Theory, Figures and Solved Problems :
Copy of CENTROIDS AND CENTRE OF GRAVITY
