Class 11 Physics

Rotational Movement


particles on circular path

Let us take a particle P whose position is r with respect to the origin O. If a force F is acting on particle then moment of force is given as follows:


In other words, moment of force is the vector product of position and force. Magnitude of τ (Greek letter tau) is as follows:

`τ=r F` sin θ

The dimensions of moment of force are ML2T-2, i.e. same as those of work or energy. But while work is a scalar quantity, moment of force (torque) is a vector quantity. SI unit of moment of force is newton meter (Nm). Magnitude of moment of force can be written as follows:

τ = (r sin θ)F = rF

Or, τ = r F sin θ = r F

Where r = r sin θ is the perpendicular distance of the line of action of F from the origin. F is the component of F in the direction perpendicular to r.

If r = 0, F = 0 or θ = 0° or 180° then τ = 0. This explains why you cannot open a door if you apply force on the hinges and not on the door handle.

Angular Momentum of a Particle

Let us consider a particle of mass m and linear momentum p at a position r relative to the origin. Angular momentum of this particle with respect to origin is given as follows:


The magnitude of angular momentum vector is

`l=r p` sin θ

We can also write:

`l=r p_(⊥) = r_(⊥) p`

If the linear momentum is zero or r = 0 or θ = 0° or 180° then angular momentum will be zero.

Equilibrium of a Rigid Body

A rigid body is said to be in mechanical equilibrium:

This means the total force (vector sum of the forces) on the rigid body is zero.

ΣFi = 0 ……………(1)

This equation gives the condition for translational equilibrium of the body.

This also means that the total torque (vector sum of torques) on the rigid body is zero.

Στi = 0 …………..(2)

This equation gives the condition for rotational equilibrium of the body.

Equations (1) and (2) are vector equations. They are equivalent to three scalar equations each. Equation (1) corresponds to following scalar equations.

ΣFix = 0, ΣFiy = 0 and ΣFiz = 0

Where, Fix, Fiy and Fiz are respectively x, y and z components of the forces Fi.

Equation (2) is equivalent to following three scalar equations.

Στix = 0, Στiy = 0 and Στiz = 0

Where, τix, &tauiy and τiz are respectively x, y and z components of the torque τi.

These six equations give six independent conditions to be satisfied for mechanical equilibrium of a rigid body. In a number of cases, all the forces acting on the body are coplanar. Then only three conditions need to be satisfied for mechanical equilibrium. They are as follows:

Partial Equilibrium

A body may be in translational equilibrium and not in rotational equilibrium. Alternately, a body may be in rotational equilibrium and not in translational equilibrium. Following examples illustrate this.

translational equilbrium

Let us take a rod AB of negligible mass. At the two ends of this rod, two parallel forces (both equal in magnitude and along same direction) are applied perpendicular to the rod. C is the midpoint of AB so that CA = CB = a. The moment of forces at both A and B will both be equal in magnitude (aF). So, the net moment of the rod will be zero. Since Σ F ≠ 0, hence the system will be in rotational equilibrium but will not be in translational equilibrium.

rotational equilibrium

Now, let the force at B get reversed so that two forces of equal magnitude are acting in opposite direction. In this case, moments of both the forces are equal but in opposite directions. They will result in rotation of the rod. Now, the body is in translational equilibrium but not in rotational equilibrium.

Couple or Torque: A pair of forces of equal magnitude but in opposite directions with different lines of action is called couple or torque. A couple produces rotation without translation. While opening the lid of a bottle we apply a couple on the lid.

Moment of a couple does not depend on the point about which you take the moments.

moment of couple

Let us assume a couple acting on a rigid body. Forces F and –F are acting respectively at points B and A. Position vectors of these points (with respect to origin) are r1 and r2.

Moment of the couple = Sum of the moments of two forces making the couple

= r1 × (-F) + r2 × F

= (r2 - r1) × F

But r1 + AB = r2

So, AB = r2 - r1

So, moment of couple = AB × F

This means that the moment of couple is independent of origin.

Principle of Moments

An ideal lever is essentially a light (negligible mass) rod pivoted at a point along its length. The pivot point is called the fulcrum. principle of moments

This figure shows a lever AB with fulcrum at O. Two forces F1 and F2 are acting perpendicular to the rod and in opposite directions. R is the reaction of the support at the fulcrum and is directed opposite to the forces. The force F1 is acting at a distance d1 from the fulcrum, while F2 is acting at a distance d2 from the fulcrum.

For translational equilibrium:

`R – F_1- F_2=0`

For rotational equilibrium, the sum of the moments must be zero.


Conventionally, anticlockwise moments are taken to be positive, and clockwise moments are taken to be negative. Here, R acts at the fulcrum itself and has zero moment about the fulcrum.

Load and Load Arm: In case of a lever, the force F1 is generally some weight to be lifted and is called the load. The distance of load d1 from fulcrum is called load arm.

Effort and Effort Arm: Force F2 is the force applied to lift the load and is called effort. Distance d2 of the effort from fulcrum is called effort arm.

Above equation can be written as follows:


Or, Load arm × Load = Effort arm × Effort

This equation gives the principle of moments for a lever.

Above equation can also be written as follows:


This ratio is called mechanical advantage (M.A.)

If the effort arm d2 is larger than the load arm, mechanical advantage is greater than one. If M.A. is greater than one then a small effort can be used to lift a large load.

Center of Gravity

The point where the total gravitational torque on the body is zero is called the center of gravity (CG) of the body.

Let us consider a cardboard in mechanical equilibrium. This object is in mechanical equilibrium because the reaction of the tip (on which the object is resting) is equal and opposite to Mg. It is also in rotational equilibrium and hence is not tilting or falling.

Many particles make up the cardboard and producing forces due to gravity; like m1g, m2g, etc. Torques (due to these forces) are acting on the cardboard. Rotational equilibrium implies that total torque on the body is zero.

If ri is the position vector of the ith particle then the torque about CG, due to the force of gravity on the particle is


Total gravitational torque about CG is zero.


As g is the same for all particles so it is taken out of the summation. Now, as g is non-zero so


This shows that if the sum if zero, the origin must be the center of mass of the body. Thus, the center of gravity of the body coincides with the center of mass in uniform gravity or gravity-free space. This applies for small body because the g does not vary from one point to another in that case.