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# Centre of gravity, definition, calculation, difference between COM and COG, practice problems, FAQs

You might have tried balancing a ruler or even a book on your finger. Ever wondered what makes this possible or what so special about the point that lets you balance an entire book on one finger? The reason for this lies behind the concept of centre of gravity.

• Centre of gravity
• Difference between centre of mass and centre of gravity
• Determine the centre of gravity. Calculating the centre of gravity.
• Practice problems
• FAQs

## Centre of gravity

Centre of gravity is an imaginary point.The centre of gravity is a hypothetical point where the whole weight of the object is concentrated, where the whole weight of the object acts. We can also define the centre of gravity as a point about which gravitational torque comes out to be zero. The point of centre of gravity is of great application in predicting the behaviour of bodies.    The centre of gravity finds applications in explaining the behaviour of the bodies when acted upon by gravitational forces.

The image shows the behaviour of a drum under gravitational force. In image A and C, since the drum is supported at the point of centre of gravity, no gravitational torque occurs on the drum. Whereas drum in image B and D, gravitational torque tries to bring the drum to a more stable position.

Centre of gravity is mostly affected by the shape and composition of the objects. In many engineering applications, the concept of centre of gravity plays an important role in the stability of the objects. For example- the lower portion of the ships are generally filled with water or weights are added to keep the ships stable in thunderstorms. Also for racing cars, the centre of gravity is kept as low as possible to provide stability at the higher speeds and prevent their rolling sideways.

## Difference between centre of mass and centre of gravity

 Centre of mass Centre of gravity Point at which the entire mass of the body is supposed to be concentrated. Fixed point at which the entire weight of the body acts. Centre of mass of a body is used to study the motion of the whole body Centre of gravity is generally used to study the stability of a body. Centre of mass depends upon mass distribution. Centre of mass depends upon mass distribution and the gravitational field acting upon it. In case of small and symmetrical bodies or when the gravitational field is uniform, the the centre of mass coincides with the centre of gravity. In case of extended, non-symmetrical bodies and when the gravitational field is non uniform, the centre of gravity does not coincide with the centre of mass.

## Calculating the centre of gravity

Since mass in general is not uniformly distributed in an object, determining the centre of gravity becomes complicated. The general case requires use of calculus which is beyond the scope of the topic. If the mass becomes uniformly distributed, the problem becomes greatly simplified.

For uniformly distributed mass objects, centre of gravity can be found using the following methods:

• When we balance an object with the help of a string or an edge, the point at which the object balances is the centre of gravity.
• One method involves suspending the object using the cord.

Suspending the plate given in the figure by attaching a cord to point A and then by attaching the cord at point C, the centre of gravity of the plate can be located.

Suspending a plate by means of a cord attached to the points A and C, the centre of gravity of the plate can be calculated as follows:

• For a triangular metal plate such as that depicted in the figure, the calculation would involve a summation of the moments of the weights of all the particles that make up the metal plate about point A. By equating this sum to the plate’s weight W, multiplied by the unknown distance from the centre of gravity G to AC, the position of G relative to AC can be determined. The summation of the moments can be obtained easily and precisely by means of integral calculus.

For a triangular metal plate shown in the figure, the centre of gravity is calculated by taking the sum of the moments of the weights of the particles which make up the triangle. This equals to the weight W Multiplying this with the unknown distances from sides AC, CB and AB to the centre of gravity G we get the position of G relative to the sides.

### Centre of gravity of different shape of bodies

 Shape of body Position of CG Thin Uniform Bar middle point of the bar Circular Ring centre of the ring Circular Disk centre of the disk Sphere, Hollow Sphere, Annular Disk at its centre Cubical or Rectangular Block point of intersection of the diagonals Triangular Lamina point of intersection of the medians Square Lamina, Parallelogram and Rectangular Lamina point of intersection of the diagonals Cylinder middle point of the axis

## Practice problems

Q1. An uniform rod of mass 3 kg is hinged at the wall and connected through a light string as shown in the figure. Find the moment due to the weight of the rod about point O.

A. Given, mass of the rod, m = 3 kg

Q2. Four homogeneous bricks, each of length L = 4m, are arranged as shown in the figure. Each bricks is displaced in displaced with respect to the one in contact by L/8 Each brick is placed in contact with one another at a distance of  L/8 from each other. Find the x- coordinate of the centre of gravity relative to origin O.

A.

Length of each brick, L = 4m, Let the mass of each brick be m kg.

Initially all four bricks have their COG at L/2 =2 from the x-axis.

After displacing the bricks, coordinates relative to origin O :

Q3. A square plate of uniform thickness and density is bent along M1M2 till corner C coincides with centre C' as shown in Figures . Determine its centre of gravity.

A. Let w be the uniform weight of the plate per unit area. The entire plate after it is bent can be considered to be made up of three parts.

(1) W1 = weight corresponding to a square plate OACB

= (36w) at location (3,3)

(2) W2 = weight corresponding to overlapped portion M1CM2

= (4.5w) at location (4,4)

(3) - W3 = portion (M1CM2) which is removed

= (- 4.5w) at location (5,3)

Resultant weight, W = W1 + W2 + W3 = 36w + 4.5w + (-4.5w) = 36w

 S. no Wi xi Wixi Yi WiYi 1 W1 = 36w 3 108w 3 108w 2 W2 = 4.5w 4 18w 4 108w 3 W2 = 4.5w 5 - 22.5w 5 - 22.5w ΣW = 36w ΣWixi = 103.5w ΣWixi = 103.5w

Q4. Determine the C.G. of a wire of uniform cross-section bent into shape of a semicircle as shown in figure

A.

Length of the wire, L = na where a = radius of semicircle

## FAQs

Question 1. Why is the centre of gravity important?
Answer. The Centre of gravity vastly simplifies calculations involving gravitation and dynamics to treat the mass of an object as if it is concentrated at one point.

Question 2. How does centre of gravity affect stability?