TRB BT Assistant - Centre of Gravity and Equilibrium of Rigid Bodies E-Book Study Material

Centre of Gravity and Equilibrium of Rigid Bodies

Centre of Gravity (CG)

The Centre of Gravity (CG) of a rigid body is a unique point where the entire weight of the body can be considered to act. For a uniform gravitational field, this point coincides with the Centre of Mass (CM). Imagine suspending a body from different points; the vertical line of suspension will always pass through the CG.

Concept of Centre of Mass (CM)

The Centre of Mass (CM) is the average position of all the mass that makes up the object. If you were to exert a force on the CM, the object would move without rotating. For a system of discrete particles with masses m₁, m₂, ..., m_n and corresponding position vectors r₁, r₂, ..., r_n, the position vector of the centre of mass (R_CM) is given by:

R_CM = (m₁r₁ + m₂r₂ + ... + m_nr_n) / (m₁ + m₂ + ... + m_n)

In terms of coordinates (x, y, z):

X_CM = (m₁x₁ + m₂x₂ + ... + m_nx_n) / Σm_i

Y_CM = (m₁y₁ + m₂y₂ + ... + m_ny_n) / Σm_i

Z_CM = (m₁z₁ + m₂z₂ + ... + m_nz_n) / Σm_i

For a continuous body, these sums are replaced by integrals:

X_CM = (∫x dm) / M

Y_CM = (∫y dm) / M

Z_CM = (∫z dm) / M where M is the total mass of the body and dm is an infinitesimal mass element.

Centre of Gravity (CG) vs. Centre of Mass (CM)

While often used interchangeably, there's a subtle difference. CG is the point where the resultant gravitational force acts, while CM is the average position of mass. In a uniform gravitational field (like near the Earth's surface), the gravitational force on every particle of the body is parallel and its magnitude depends only on mass. Thus, CG coincides with CM. However, in a non-uniform field (e.g., very large objects like planets or in space), the gravitational field might vary in direction and magnitude across the body, causing CG and CM to differ. For most practical purposes and in competitive exams unless specified otherwise, we assume a uniform gravitational field.

Determination of CG

The location of the CG depends on the shape and mass distribution of the body.

Symmetrical Bodies with Uniform Density: For bodies with regular shapes and uniform density (e.g., sphere, cube, cylinder, ring), the CG coincides with their geometrical center.
Irregular Bodies: For irregularly shaped bodies, the CG can be found experimentally. One common method involves suspending the body from a point, marking the vertical line, then suspending it from another point and marking the vertical line again. The intersection of these lines gives the CG.
Composite Bodies: For a body made up of simpler, known shapes, the CG can be calculated by considering the CG of each component and their respective masses.

Examples of CG for Simple Shapes

Uniform Rod: At its midpoint.
Rectangular Lamina/Plate: At the intersection of its diagonals.
Circular Lamina/Disc: At its center.
Triangle: At the intersection of its medians (centroid).
Sphere (hollow or solid): At its center.
Cone/Pyramid: At a height of 1/4th the height from the base for a solid cone/pyramid. For a hollow cone, it's 1/3rd the height from the base.

Memory Trick: For uniform, symmetrical objects, the CG is almost always at the geometric center. Think of it as the "balancing point."

CG of Composite Bodies (Calculation Method]

Consider a composite body made of two parts with masses m₁ and m₂, and their respective CGs at (x₁, y₁) and (x₂, y₂). The CG of the combined body (X_CG, Y_CG) is given by:

X_CG = (m₁x₁ + m₂x₂) / (m₁ + m₂)

Y_CG = (m₁y₁ + m₂y₂) / (m₁ + m₂)

This can be extended to more than two parts. If a part is removed from a body, its mass is treated as negative.

Example: A uniform rod of mass M and length L has its CG at L/2. If a small segment of mass m is removed from one end, the new CG will shift slightly towards the other end.

Equilibrium of Rigid Bodies

A rigid body is one that does not deform under applied forces. Equilibrium refers to the state of a body where there is no net change in its motion. For a rigid body, this means it is either at rest or moving with constant linear and angular velocity.

Conditions for Equilibrium

For a rigid body to be in equilibrium, two conditions must be satisfied:

Translational Equilibrium: The net external force acting on the body must be zero. This ensures that the centre of mass of the body has zero acceleration (i.e., it is either at rest or moving with constant linear velocity).
ΣF_ext = 0 or ΣF_x = 0, ΣF_y = 0, ΣF_z = 0
Rotational Equilibrium: The net external torque acting on the body about any point must be zero. This ensures that the body has zero angular acceleration (i.e., it is either not rotating or rotating with constant angular velocity).
Στ_ext = 0 or Στ_x = 0, Στ_y = 0, Στ_z = 0

Key Point: Equilibrium means the body is NOT accelerating linearly (velocity is constant) and NOT accelerating angularly (angular velocity is constant). If the body is initially at rest, it will remain at rest.

Types of Equilibrium

The stability of a body in equilibrium depends on the position of its Centre of Gravity (CG) relative to its base of support. There are three main types:

Stable Equilibrium: If a body is slightly displaced from its equilibrium position, it tends to return to its original position. This happens when the CG is at the lowest possible position. When displaced, the weight acting through the CG creates a restoring torque that brings it back.
Example: A cone resting on its base, a pendulum bob hanging freely, a book placed flat on a table. In stable equilibrium, raising the CG increases its potential energy.
Unstable Equilibrium: If a body is slightly displaced from its equilibrium position, it tends to move further away from it. This occurs when the CG is at the highest possible position. Any small displacement causes the weight to create a torque that increases the displacement.
Example: A cone balanced on its apex, a pencil balanced on its tip. In unstable equilibrium, lowering the CG decreases its potential energy.
Neutral Equilibrium: If a body is slightly displaced from its equilibrium position, it remains in the new position. This happens when the CG remains at the same height regardless of the displacement.
Example: A sphere or a cylinder rolling on a flat horizontal surface, a cone resting on its side. The CG's height does not change upon displacement.

Stability & CG Height: A lower CG generally leads to greater stability. Think of a racing car versus a tall, narrow truck – the racing car is much harder to tip over.

Base of Support

The base of support is the area enclosed by the points of contact of the body with the surface on which it rests. For a body to be stable, its CG must remain vertically above its base of support when slightly tilted. If the CG moves beyond the base of support, the body will topple.

Factors Affecting Stability

Height of the CG: Lower CG increases stability.
Size of the Base of Support: A wider base of support increases stability.
Mass of the Body: A heavier body is generally more stable, as a larger torque is required to tip it over.

Illustrative Examples

Example 1: Balancing a Book A book placed flat on a table is in stable equilibrium. If you try to tilt it slightly, it returns to its flat position. If you tilt it beyond a certain point (where the CG is no longer above the base of support), it will topple.

Example 2: A Tilted Ladder A ladder leaning against a wall is in equilibrium if the forces (weight of the ladder, normal forces from the ground and wall, friction) balance. If the angle is too shallow, the friction might not be enough to prevent slipping (translational equilibrium fails). If the ladder is very long and thin, it might be unstable rotationally.

Example 3: A Symmetrical Object on an Inclined Plane A uniform cylinder placed on an inclined plane can be in stable, unstable, or neutral equilibrium depending on how it's placed and the angle of inclination. If it's just sitting there, it's stable as long as it doesn't slide or tip. If it's balanced on its edge, it's unstable. If it's rolling down, it's in neutral equilibrium concerning its rotation.

Torque and Equilibrium

Torque (τ) is the rotational equivalent of force. It is calculated as the product of the force and the perpendicular distance from the pivot point to the line of action of the force (lever arm).

τ = r × F

Magnitude: τ = rF sin(θ), where θ is the angle between r and F.

For rotational equilibrium, the sum of clockwise torques must equal the sum of counter-clockwise torques about any chosen pivot point.

Στ_clockwise = Στ_{counter-clockwise}

Example: A See-Saw Two children of different weights sit on a see-saw. For the see-saw to be balanced (in rotational equilibrium), the torques they produce must be equal. If child A has weight W_A and sits at distance d_A from the pivot, and child B has weight W_B at distance d_B, then for balance:

W_A × d_A = W_B × d_B

This equation highlights that a heavier person needs to sit closer to the pivot (fulcrum) than a lighter person.

Applications in Engineering and Daily Life

The principles of CG and equilibrium are fundamental in many fields:

Structural Engineering: Designing bridges, buildings, and dams ensures they are stable under various loads and environmental conditions. The CG of the structure must be appropriately positioned.
Vehicle Design: The stability of cars, trucks, and aircraft depends heavily on the placement of their CG. Lowering the CG improves handling and reduces the risk of tipping.
Sports: Athletes manipulate their body's CG to maintain balance during complex movements, jumps, and turns.
Everyday Objects: The design of furniture, tools, and even simple objects like cups and lamps considers their CG for stability and ease of use.

Exam Tip: When dealing with equilibrium problems, always choose a pivot point that simplifies calculations, ideally a point where an unknown force acts or where multiple forces intersect. Remember to consider both translational (ΣF=0) and rotational (Στ=0) equilibrium conditions.