Angular momentum of a particle
The angular momentum in a rotational motion is similar to the linear momentum in translatory motion. The linear momentum of a
particle moving along a straight line is the product of its mass and linear velocity (i.e) Vec p = m. Vec v. The angular momentum of a particle is
defined as the moment of linear momentum of the particle.
Let us consider a particle of mass m moving in the XY plane with a velocity v and linear momentum p = mv at a distance r from the origin (Fig. ).
The angular momentum L of the particle about an axis passing through O perpendicular to XY plane is defined as the cross product of Vec r and Vec p.
(i.e) Vec L = Vec r ? Vec P
Its magnitude is given by L = r p sin θ
Where θ is the angle between Vec r and Vec p and L is along a direction perpendicular to the plane containing Vec r and Vec p .
The unit of angular momentum is kg m2 s?1 and its dimensional formula is, M L2 T?1.
Angular momentum of a rigid body
Let us consider a system of n particles of masses m1, m2 ?.. mn situated at distances r1, r2, ?..rn respectively from the axis of rotation (Fig. ). Let v1,v2, v 3 ?.. be the linear velocities of the particles respectively, then linear momentum of first particle = m1v1.
Since v1= r1ω the linear momentum of first particle = m1(r1 ω)
The moment of linear momentum of first particle
= linear momentum ? perpendicular distance
= (m1r1ω) ? r1
angular momentum of first particle = m1r12ω
angular momentum of second particle = m2r22ω
angular momentum of third particle = m3r32ω and so on.
The sum of the moment of the linear momenta of all the particles of a rotating rigid body taken together about the axis of rotation is known as angular momentum of the rigid body.
Angular momentum of the rotating rigid body = sum of the angular momenta of all the particles.
(i.e) L = m 1r1 2ω+ m2 r2 2ω + m3 r32ω.?. + mn rn 2 ω
L = ω [ m 1r1 2+ m2 r2 2ω + m3 r32.?. + mn rn 2 ]
∴ L = ωI
where I = ∑ mi ri 2 moment of inertia of the rotating rigid body about the axis of rotation.