# Centre of Gravity and Mass

## Centre of Gravity

The centre of gravity (CG) of a body is a point where the weight of the body acts and total gravitational torque on the body is zero. In a sphere, it is at its centre. By trial and error, we can locate a point G on a cardboard, where it is gravity of the card board. The tip of pencil at G provides the normal reaction R to the total weight mg of cardboard. The cardboard is in translational equilibrium, as R = mg.

Forces of gravity like m_{1}g, m_{2}g, …. Etc. act on individual particles of the card board. They make up the torques on the cardboard. For its particle of mass m_{i} , the force of gravity is m_{1}g. If \(\vec r_i\) is the position vector of the particle from C.G. of the cardboard, then the torque about the C.G. is

$$\vec \tau _i = \vec r_1 \times m_i \vec g \dots (i) $$

As C.G. of the cardboard is located that the total torque on it to forces of gravity on all the particles is zero, so

$$ \vec \tau = \sum _{i=1}^n \vec \tau _i = \sum _{i=1}^n \vec r_i \times m_i \vec g = 0 \dots (ii) $$

As g is non-zero and same for all particles of the body, so equation (ii) can be written as

$$\sum _{i=1}^n m_i \vec r_i = 0 \dots (iii) $$

This is the condition, when centre of mass of the body lies at the origin. As the position vectors \(\vec r _1 \) and taken with respect to the C.G., therefore the centre of gravity of the body coincides with the centre of mass of the body. However, if the body is so extended that \(\vec g\) varies from part of the body, then the centre of gravity shall not coincide with the centre of mass of the body.

**Shape of a body and the position of Its Centre of Gravity**

The centre of the gravity of a body depends on the shape and size of the body. The C.G. of a body may not necessarily be within the body. For example, the C.G. of the ring does not lie in the material of the ring; it is at its geometric centre. The position of centre of gravity changes with change in shape of a body. An iron bar has C.G. at its middle point. But when the bar is bent into a circular ring, the C.G. is not within the material. It is its geometric centres.

S.N |
Shape of body |
Position of C.G. |

1. |
Thin uniform bar |
Middle point of the bar. |

2. |
Circular ring |
Centre of the ring. |

3. |
Circular disc |
Centre of disc. |

4. |
Sphere, hollow sphere and annular disc |
At centre. |

5. |
Cubical or rectangular block |
Point of intersection of the diagonals. |

6. |
Triangular lamina |
Point of intersection of the medians. |

7. |
Square lamina, parallelogram and rectangular lamina |
Point of intersection of the diagonals. |

8. |
Cylinder |
Middle point of the axis. |

9. |
Cone or pyramid |
On the line joining the apex to the centre of the base at the distance equal to ¼ of the length of this line from the base. |

## Centre of Mass

A rigid body consists of a large number of particles and the mass of the body is the sum of mass of the individual particles. We can consider a point in the body such that whole mass of the body is concentrated in it and the motion of the point is same as the motion of a particle of same mass as that of body when the same force is applied on it. That point is called centre of mass.

### Calculation of Centre of Mass

Let us consider a system of n particles m_{1}, m_{2}, m_{3}, …., m_{n} with coordinate position (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), …., (x_{n}, y_{n}) respectively as shown in the figure. Let (x, y) be the co-ordinate of centre of mass of the body. For the various particles, we have F_{1} = m_{1}a_{1}, F_{2} = m_{2}a_{2}, F_{1} = m_{3}a_{3} and so on. Similarly, F_{1}, F_{2}, F_{3} …. Also produce acceleration a_{1}, a_{2}, a_{3} ….. respectively. Total force acting on the body,

$$\begin{align*} \therefore F_1 + F_2 + F_3 + \dots &= m_1a_1 + m_2a_2 + m_3a_3 + \dots \\ \text {or,} \: \sum F &= m_1 \frac {d^2x_1}{dt^2} + m_2 \frac {d^2x_2}{dt^2} + m_3 \frac {d^2x_3}{dt^2} + \dots \\ \text {or,} \: \sum F &= \frac {d^2}{dt^2} (m_1x_1 + m_2x_2 + m_3x_3 + \dots ) \\ \text{or,} \frac {\sum F}{\sum m} &= \frac {d^2}{dt^2} \left (\frac {m_1x_1 + m_2x_2 + m_3x_3 + \dots }{m_1 + m_2 + m_3 + \dots } \right ) \\ \text {From this relation, we can say that x-coordinate of centre of mass is} \\ x &= (\frac {m_1x_1 + m_2x_2 + m_3x_3 + \dots }{m_1 + m_2 + m_3 + \dots } = \frac {\sum mx}{M} \\ \text {Similarly, its y-coordinate is } \\ y &= \frac {m_1y_1 + m_2y_2 + m_3y_3 + \dots }{m_1 + m_2 + m_3 + \dots } = \frac {\sum my}{M} \\ \text {So, centre of mass} (x,y) &= \left [\frac {m_1x_1 + m_2x_2 + m_3x_3 + \dots }{M} , \frac {m_1y_1 + m_2y_2 + m_3y_3 + \dots } {M} \right ]\\ where M &= m_1 + m_2 + m_3 + \dots , \end{align*}$$

**Centre of Mass of two Bodies**

Suppose two bodies of masses m_{1} and m_{2} as shown in the figure. Let the masses be connected by a rigid rod and their centre of mass, C. Let x_{1} be the distance of m_{1} from C and x_{2} that of m_{2} from C. Then we have,

$$\begin{align*} m_1x_1 &= m_2x_2 \\ \text {or} \: x_1 &= x_2 \frac {m_2}{m_1} \\ \text {or} \: x_2 &= x_1 \frac {m_1}{m_2} \\ \end{align*}$$

So greater the mass of the body, nearer is their C.G. and so on.

### Equilibrium of Concurrent Forces

A number of forces acting at a point are called concurrent force. If the resultant of the concurrent forces acting on a body is zero, the body is in equilibrium. When concurrent force are in equilibrium, then the state of the body will not change i.e. if the body is at rest it remains at rest and if in uniform motion, continue to move with uniform velocity on straight line.

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