[Physics Class Notes] on Angular Momentum About Fixed Axis Pdf for Exam

Angular Momentum 

Angular momentum is the vector product of the angular velocity of a particle and its moment of inertia. If a particle of mass m has linear momentum (p) and position (r) then the angular momentum with respect to its original point O is defined as the product of linear momentum and the change in position. Mathematically 

l = r × p

Derivation of Equation for Angular Momentum

Since l=r × p, when we differentiate it with respect to time we get,

dl/ dt = d (r × p) / dt

Applying the product rule for differentiation,

d (r × p) / dt = (dp/dt)×  r + (dr/dt )×p

Since velocity is the change in position at some time interval, thence, dr/dt = v and p = mv,

Thus, dr/dt ×  p = v × mv

Now since both the  vectors are parallel, their products shall be a zero (o). Now let’s take  dp/dt ×  r,

Since F =  dp/dt

thus ,(dp/dt ) × r = F × r = τ

This means that, d (r×p)/ dt = τ. Since l= r×p, therefore,

dl / dt = τ

Angular Momentum of the Rigid Body Rotating about a Fixed Axis

The angular momentum  what we studied above is on a particle about  any point which states that the rate of change of  total angular momentum wrt time  about a point equals the total net external torque acting on the system about the same point. Thus, the Angular momentum remains conserved when the total external torque is zero. 

But in order to calculate the net rate of change of angular momentum of a rotating object about a fixed axis, we will be learning about the concept of angular momentum of a particle undergoing the rotational motion about a fixed axis.

Now, we shall deal with angular momentum about a fixed axis . 

Thus in order to  study rotational momentum in reference to a rigid body, we consider  it as a vector acting on a system of particles. Since during rotational motion every particle behaves differently, hence we can calculate the angular momentum for a system of many particles.

The angular momentum of any  particle rotating about a fixed axis depends on the net  external torque acting on that body. 

Let us Consider an object rotating about a fixed axis, as shown in the figure. Now consider a particle P in the body that rotates about the axis as shown above. Thus the  total angular momentum for this system is given by,

L = [sum_{i=1}^{N}] ri   X pi

Where, P is the momentum and is equal to mv  and r is the distance of the particle from the axis of rotation.

The contribution of individual particle to the total angular momentum is  given as, l= r×p

Using vector law of addition OP = OC + CP.

So we can write, l = (OC +CP) ×p = (OC×p) +(CP×p)

v = rpw  where rp is the perpendicular distance of  point P from axis of rotation.

Also, the tangential velocity v at the point p is perpendicular to the vector rp. 

Using the right-hand thumb rule, the direction of product CP×v is parallel to the axis of rotation.

Similarly, the product of the vectors OC×V is perpendicular to the axis of rotation.

So, we can write it as: l  = OC × mv + lz                                                      

The component  of angular momentum parallel to the fixed axis of rotation, which is along the z-axis is  Iz 

L = ∑ l = ∑ (lp + lz )

Here Lp is the perpendicular component of momentum can be given as,

Lp = ∑ OC i × mivi

And the parallel component of the momentum is ,

Lz = (∑ miri2 ) ωk’ 

Lz =Iz ωk’

Each  and every particle possessing a velocity vi has a corresponding particle possessing velocity –vi located diametrically opposite on the circle since the object under consideration is generally symmetric about the axis of rotation thus  at a particular perpendicular distance rp,, the total angular momentum due to these particles cancel each other.

For such symmetrical objects, the total momentum of the object is given by,

L = Lz =Iz ωk’

Where, ω is the angular velocity of the body and  gives the direction of the total angular momentum.

and ,I is the moment of inertia of the body.