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Revision: Rotational Motion JEE Main Rotational Motion

Definitions [6]

Definition: Transitional Motion

When a force acts on a stationary rigid body that is free to move, the body starts moving in a straight path in the direction of the applied force. This is called linear or translational motion.

Definition: Rotational Motion

Now consider a body pivoted at a point, i.e., not free to move, and a force is applied on the body at a suitable point, it rotates the body about the axis passing through the pivoted point. This is the turning effect of the force and the motion of the body is called rotational motion.

Define torque and mention its unit.

Torque is defined as the moment of the externally applied force about a point or axis of rotation. The expression for torque is, `vectau = vecr xx vecF`.

Define the centre of gravity of a body.

The point through which the resultant of the weights of all the particles of the body acts is called its centre of gravity.

The centre of gravity is an imaginary location where the body’s whole weight is assumed to be concentrated.

Definition: Centre of Gravity

The Centre of Gravity (c.g.) of a body is the point around which the resultant torque due to the force of gravity on the body is zero.

The centre of gravity (C.G.) of a body is the point about which the algebraic sum of moments of the weights of all the particles constituting the body is zero. The entire weight of the body can be considered to act at this point, howsoever the body is placed.

Definition: Couple

A couple is a pair of forces that satisfy all of the following:

They are equal in magnitude (same strength)
They are opposite in direction
Their lines of action do not coincide (they act along different straight paths)
Because of these properties, a couple produces a pure rotation (turning) effect and no net translation (no sliding) of the object.

Theorems and Laws [2]

State and prove the theorem of the parallel axis about the moment of inertia.

A body's moment of inertia along an axis is equal to the product of two things: Its moment of inertia about a parallel axis through its centre of mass and the product of the body's mass and the square of the distance between the two axes. This is known as the parallel axis theorem.

Proof: Let I_CM represent a body of mass M moment of inertia (MI) about an axis passing through its centre of mass C and let I stand for that body's MI about a parallel axis passing through any point O. Let h represent the separation of the two axis.

Think about the body's minuscule mass element dm at point P. It is perpendicular to the rotation axis through point C and to the parallel axis through point O, with a corresponding perpendicular distance of OP. CP² dm is the MI of the element about the axis through C. As a result, `I_(CM) = int CP^2 dm` is the body's MI about the axis through the CM. In a similar vein, `I = int OP^2 dm` is the body's MI about the parallel axis through O.

Draw PQ perpendicular to OC produced, as shown in the figure. Then, from the figure,

`I = int OP^2 dm`

= `int (OQ^2 + PQ^2) dm`

= `int [(OC + CQ)^2 + PQ^2] dm`

= `int (OC^2 + 2OC.CQ + CQ^2 + PQ^2) dm`

= `int (OC^2 + 2OC.CQ + CP^2)dm` ...(∵ CQ² + PQ² = CP²)

= `int OC^2 dm + int 2OC.CQ dm + int CP^2 dm`

= `OC^2 int dm + 2OC int CQ dm + int CP^2 dm`

Since OC = h is constant and `int dm = M` is the mass of the body,

`I = Mh^2 + 2h int CQ dm + I_(CM)`

The integral `int CQ dm` now yields mass M times a coordinate of the CM with respect to the origin C, based on the concept of the centre of mass. This position and the integral are both zero because C is the CM in and of itself.

∴ I = I_CM + Mh²

This proves the theorem of the parallel axis.

Law: Principle of Moments

Statement:

In equilibrium, the sum of anticlockwise moments equals the sum of clockwise moments about the pivot.

Explanation/Proof:

When several forces act on a pivoted body, they tend to rotate it about an axis passing through the pivot. The resultant moment is obtained by taking the algebraic sum of the moments of all the forces about the pivoted point. By convention, anticlockwise moments are taken as positive and clockwise moments as negative.

A metre rule is suspended at its centre (point O). Two weights W₁ and W₂ are hung on either side at distances l₁ and l₂ using spring balances.

W₁ creates a clockwise moment = W₁ × l₁
W₂ creates an anticlockwise moment = W₂ × l₂

By adjusting the weights or positions, the rule becomes horizontal (in equilibrium).

Conclusion:

At balance, W₁ × l₁ = W₂ × l₂, which confirms the principle of moments.

Key Points

Key Points: Centre of Gravity

The weight of a body acts through a single point called the centre of gravity (C.G.), where the sum of moments of all particles' weights is zero.
The position of the C.G. depends on the shape and mass distribution of the body and changes if the body is deformed.
The C.G. may lie outside the material of the body (e.g., a ring or hollow sphere).
A body balances when supported exactly at its centre of gravity, as seen in a metre rule or square lamina.
The C.G. of an irregular lamina can be found by suspending it from multiple points and tracing the intersection of plumb line paths.

Physics and Measurement Gravitation