# Angular Momentum of a Particle

Linear momentum in relation to the linear (or translational) motion of a single particle or of a system of particles (rigid bodies ) is conserved in process like collisions.

For a single particle the linear momentum is
P = mv
For a system of particles, it is
P = MVCM
where M is the total mass of the system and
VCM is the velocity of is centre of mass.

In rotational motion, the corresponding quantity to linear momentum is called angular momentum.

It is the rotational analogue of linear momentum.

Angular momentum is a useful concept in rotational motion, as linear momentum is for translational motion.

# Definition of Angular Momentum

Consider a particle P of mass m whose position vector relative to origin O of an inertial reference frame is r.

Let p be the linear momentum of particle. The angular momentum L of the particle with respect to origin O is defined as the vector product of r and p, i.e.
L = r x p

Angular momentum is a vector. Its magnitude is given by L = rp sin Î¸

where Î¸ is the angle between vectors r and p.

Its dimension are (ML2 T-1).

The units of L are kg m2 s-1.
 Definition of angular momentum of a particle
The direction of L is perpendicular to the plane containing the vectors r and p (which is the plane of this page) and its sense is given by the right-hand thumb rule.

This means that if you were to rotate r to the position of p, through the smaller angle between them with curled fingers of the right hand, the extended thumb would then point in the direction of L. This is shown by the symbol in the figure.
L = (r sin Î¸) p = râŠ¥ p
or as L = r (p sin Î¸) = r pâŠ¥
where râŠ¥ is the component of r at right angles to p and pâŠ¥ is the component of p at right angles to r Fig.
Only the component of p perpendicular to the r (i.e. pâŠ¥ ) contributes to angular momentum. When Î¸ = 0 or 1800, pâŠ¥ = 0 and râŠ¥ = 0, in which case L = 0.

# Relation between Torque and Angular Momentum

From Newton's second law, we have
F = ma
Where p = mv is linear momentum of the particle.
Taking the vector product of r with both sides, we get
But r Ã— F = , the torque about O. Therefore, we have
Now differentiating Eq. (7.18) with respect to t we obtain
The derivative of a vector product is taken in the same way as the derivative of an ordinary product except that the order of the vectors must not be changed. Thus

Therefore
Now (v Ã— v ) is the vector product of two parallel vectors and is, by definition, equal to zero.
Hence

which states that the rate of change of angular momentum of a particle is equal to the torque acting on it.

which states that the rate of change of linear momentum of a particle is equal to the force acting on it.

# Physical Meaning of Angular Momentum

Torque = force Ã— moment arm
angular momentum = linear momentum Ã— moment arm
Torque is a measure of the turning effect of a force and angular momentum is a measure of the turning motion of the object.

# Geometrical Meaning of Angular Momentum

L = r pâŠ¥= r (m vâŠ¥ ) = m r vâŠ¥
where pâŠ¥ and vâŠ¥ are the components of p and v respectively at right angles to r.
Suppose in a small time interval Î” t the particle moves a small distance from P to Q so that its displacement Î”r is
PQ =Î” r=v x Î” t
In this displacement, let Î” Î¸ be the small change in angle Î¸.
 Geometrical meaning of angular momentum
Vector PQ = vÎ” t is resolved into two rectangular components (Ï…^ Î” t) and (Ï…||.Î”t);

vâŠ¥ is called the angular component of velocity of vector v, and v|| the radial component of v.

Only vâŠ¥ is involved in the change Î”Î¸ , so that r Î”Î¸ = vâŠ¥ Î” t, the area Î”A swept out by the radius vector r from O as the particle moves form P to Q in time Î” t is given by
Î”A = =
=
But
Therefore,
=
or L = 2m Î”A/Î” t = 2m Ã— rate of sweeping out of area by radius vector
i.e. L = 2m Ã— a real velocity
Thus, angular momentum = 2 Ã— massÃ— a real velocity. This is the geometrical meaning of angular momentum.