Angular Momentum for Rigid Body

Introduction:

The angular momentum definition of a particle is defined as the moment of linear momentum of the particle.
Let us consider a system of n particles of masses m₁,m₂,m₃..............m_n situated at distances of r₁,r₂,r₃,.......,r_n respectively from the axis of rotation .

Let v₁,v₂,v₃, ....be the linear velocities of the particles respectively, then the linear momentum of the first particle = m₁v₁

Definition of Angular Momentum for Rigid Body:

Since v₁ = r₁`omega`
Linear momentum of the first particle =m₁(r₁`omega` )
The moment of linear momentum of first particle = (m₁r₁`omega` )x r₁
Angular momentum of first particle = m₁r₁²`omega`
Similarly angular momentum of the second particle = m₂r₂²`omega`
and angular momentum of the third particle = m₃r₃²`omega` and so on.
The sum of 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 a rigid body.

Calculating the Angular Momentum for Rigid Body:

`:.` Angular momentum if the rotating rigid body = sum of the angular momenta of all the particles
`rArr` L = m₁r₁²`omega` +m₂r₂²`omega` +m₃r₃²`omega` +.............+m_nr_n²`omega`
`rArr` L = `omega` [ m₁r₁² +m₂r₂²+m₃r₃²+.....+m_nr_n²]
         =`omega[ sum_(i=1)^n m_i r_i^2]`
`rArr` L = `omega` I
where I = `sum_(i=1)^n m_ir_i^2` = moment of inertia of the rotating rigid body about the axis of rotation.

Problem to find the angular momentum of a cylinder:
A solid cylinder of mass 200kg rotates about its axis with angular speed 100 s^-1  .The radius of the cylinder is 0.25m.What is the magnitude of the angular momentum of the cylinder about its axis?
Given data : Mass M = 200 kgs
Angular speed `omega` =100 s^-1
Radius R=0.25m
L= ?
Formulas : I = `(MR^2)/2`
L = I`omega`
Working: I = `(MR^2)/2= (200 xx (0.25)^2)/2 = 6.25 "kg" m^2`
L = I`omega` = 6.25 x 100 = 625 Kg m² s^-1

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