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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. MathematicallyÂ

l = r Ã— p

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

dl/ dt = d (r Ã— p) / dt

Applying the product rule for differentiation,

d (r Ã— p) / dt = (dp/dt)Ã—Â r + (dr/dt )Ã—p

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

Thus, dr/dt Ã—Â p = v Ã— mv

Now since both theÂ vectors are parallel, their products shall be a zero (o). Now letâ€™s takeÂ dp/dt Ã—Â r,

Since F =Â dp/dt

thus ,(dp/dt ) Ã— r = F Ã— r = Ï„

This means that, d (rÃ—p)/ dt = Ï„. Since l= rÃ—p, therefore,

dl / dt = Ï„

The angular momentumÂ what we studied above is on a particle aboutÂ any point which states that the rate of change ofÂ total angular momentum wrt timeÂ 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.Â

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 .Â

Thus in order toÂ study rotational momentum in reference to a rigid body, we considerÂ 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 anyÂ particle rotating about a fixed axis depends on the netÂ external torque acting on that body.Â

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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 theÂ total angular momentum for this system is given by,

L = \[\sum_{i=1}^{N}\] riÂ Â X pi

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

The contribution of individual particle to the total angular momentum isÂ given as, l= rÃ—p

Using vector law of addition OP = OC + CP.

So we can write, l = (OC +CP) Ã—p = (OCÃ—p) +(CPÃ—p)

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

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

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

Similarly, the product of the vectors OCÃ—V is perpendicular to the axis of rotation.

So, we can write it as: lÂ = OC Ã— mv + lzÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

The componentÂ of angular momentum parallel to the fixed axis of rotation, which is along the z-axis isÂ IzÂ

L = âˆ‘ l = âˆ‘ (lp + lz )

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

Lp = âˆ‘ OC i Ã— mivi

And the parallel component of the momentum is ,

Lz = (âˆ‘ miri2 ) Ï‰kâ€™Â

Lz =Iz Ï‰kâ€™

EachÂ and every particle possessing a velocity vi has a corresponding particle possessing velocity â€“vi located diametrically opposite on the circle since the object under consideration is generally symmetric about the axis of rotation thusÂ 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 Ï‰kâ€™

Where, Ï‰ is the angular velocity of the body andÂ gives the direction of the total angular momentum.

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

FAQ (Frequently Asked Questions)

Q1. What are Some Common Practical Examples of Angular Momentum?

Ans: Examples of Angular Momentum

1 Ice-skater: When an ice-skater goes for a spin she starts off with her hands and legs far apart from the center of her body.Â

2 Gyroscope : A gyroscope uses the principle of angular momentum to maintain its orientation. It utilities a spinning wheel which has 3 degrees of freedom.Â

Q2Â What are Some Common Examples of Rotation Around a Fixed Axis?

Ans: SomeÂ of the practical common examples of rotation around a fixed axis:

Rotation of Earth around its own axis

Â Â Â Â Â 2.Â AÂ ballet dancer with his hands stretched rotates around his own axis.

Q3. How is Angular Momentum Related to Torque?

Ans: The angular momentumÂ what we studied above is on a particle aboutÂ any point which states that the rate of change ofÂ total angular momentum wrt timeÂ 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.Â

Q4. Can Angular Momentum Convert Linear Momentum?

Ans: Linear momentum and angular momentum cannot be "converted" between each other.Â Linear momentum is not conserved when there is an external force acting on the system but the angular momentum is conserved because there is no external torque acting on the system.

Q5. Why Does the Dancer Stretch Out Her Hands to Slow Down from Spinning?

Ans: When a person is rotating with stretched hands,Â the moment of inertia of the bodyÂ will be more because distribution of mass is far from the axis of rotation.But when theÂ person brings his arms close to body, the moment of inertia of the body decreases because the mass is now distributed close to axis.In this situation no external torque is applied, means angular momentum is conserved thus asÂ I decrease, angular velocity Ï‰ increases.