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
Similarly,
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.
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