While riding a two-wheeler bike, you are able to balance the bike. Even if you increase your speed and move fast, you are still able to maintain the balance. Do you know why? The answer to this is Angular Momentum. Spinning the frisbee or the football are the examples of angular momentum. Let us study this in detail.

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## Torque

Torque is the rotational analogue of force. It is also termed as the moment of force and denoted by τ. Suppose you have a rod and you apply the force to the rod from upwards. What will happen is, the rod will start moving downwards. Again if you apply an equal and opposite force from downwards. So as both the time equal and opposite forces are applied, the net force is 0 and the rod will definitely not move.

Again take the same rod and apply equal and opposite forces to the rod, but this time apply the force to the two ends of the rod. You will notice that the rod starts rotating. So it is not true that every time you apply equal and opposite force the net force will be zero.

This is true only when the forces are applied in the same force line. So this kind of force in rotational motion is called as torque. So it is not only the force but how and when the force is applied in rotational motion.

**Browse more Topics under System Of Particles And Rotational Dynamics**

- Introduction to Rotational Dynamics
- Vector Product of Two Vectors
- Centre of Mass
- Motion of Centre of Mass
- Moment of Inertia
- Theorems of Parallel and Perpendicular Axis
- Rolling Motion
- Angular Velocity and Angular Acceleration
- Linear Momentum of System of Particles
- Equilibrium of a Rigid Body
- Angular Momentum in Case of Rotation About a Fixed Axis
- Dynamics of Rotational Motion About a Fixed Axis
- Kinematics of Rotation Motion about a Fixed Axis

### Mathematical Expression of the Torque

Torque is defined as the cross product of the radius vector and the force

τ = r × F = r F Sinθ

- r = position vector of the point on which the force is applied
- F = magnitude of a force
- θ = angle between r and F

Torque is a vector quantity as it is defined by magnitude as well as direction. Its SI unit is Nm. The dimension of the torque is ML²T^{-2}

## Angular Momentum

Torque and angular momentum are closely related to each other. Angular momentum is the rotational analogue of linear momentum **‘p’** and is denoted by **‘l’**. It is a vector product. Angular momentum of the particle is

**l = r × p**

l = r p sinθ, where θ is the angle between r and p.

### Relation between Torque and Angular Momentum

As we know, l = r × p, differentiating on both the sides,

⇒ \( \frac{dl}{dt} \) = \( \frac{d}{dt} \) ( r × p)

⇒ \( \frac{dl}{dt} \) = \( \frac{dr}{dt} \) × p + r × \( \frac{dp}{dt} \)

⇒ \( \frac{dl}{dt} \) = v × p + r × \( \frac{dp}{dt} \)

⇒ \( \frac{dl}{dt} \) = v × mv + r × \( \frac{dp}{dt} \)

since, v = 0

⇒ \( \frac{dl}{dt} \) = r × \( \frac{dp}{dt} \) = r × F = τ

⇒ \( \frac{dl}{dt} \) = τ

So the time rate of change of angular momentum is equal to torque. So finally we can say that rate of change of angular momentum is equal to the torque acting on it. The angular momentum of the system of a particle is denoted by :

L = l_{1 }+ l_{2 }+ …..l_{n}

This can be written as Σ = l_{i}. Now we know that **l = r × p**, so

**l _{i} = r_{i} × p_{i}**

The total angular momentum of the system is** Σ r _{i} × p_{i}**

### Torque for System of Particles

We know that, \( \frac{dl}{dt} \) = τ, a total torque of the system will be the summation of each of the particles.

Σ \( \frac{dl}{dt} \)

τ_{i }= r_{i }× F_{i}

Σ = r_{i }× F_{i }

Torque is external as well as internal.

τ = τ_{ext }+ τ_{int}

τ_{ext }= Σ r_{i} × F_{i}^{ext}

τ_{in = }Σ r_{i} × F_{i}^{int}

So here if internal force becomes zero,

τ_{ext }= \( \frac{dl}{dt} \)

Hence we can say that time rate of the total angular momentum of the system of particles about a point is equal to the sum of the external torques acting on the system taken about the same point.

## Questions For You

Q1. A horizontal flat platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant, a viscous fluid of mass m is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period :

- decreases continuously
- decrease initially and decreases again
- remains unaltered
- increases continuously

Answer: B. The liquid when initially spreads towards the circumference of the platform, its rotational inertia starts increasing as more and more mass is disturbed from the centre. This reduces the angular velocity using conservation of angular momentum. When the fluid finally drops from the platform the rotational inertia decreases which increase the angular velocity.

Q2. Angular momentum is

- Polar vector
- Axial vector
- A scalar
- None of these

Answer: B. Angular momentum is, L = r × m(v). L is the axial vector.

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