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# derivation of angular momentum?

## 1 Answers

8 years ago

Dear Samuel,

Consider a single particle of mass m travelling with a velocity v a radiusr from an axis, as shown below. Figure %: A single particle moving with respect to an axis, O

The angular momentum of the single particle, then, is defined as:

 l = rmv sinθ

Notice that this equation is equivalent to l = rp sinθ , where p is the linear momentum of the particle: a particle does not need to move in a circular path to possess angular momentum. However, when calculating angular momentum, only the component of the velocity moving tangentially to the axis of rotation is considered (explaining the presence of sinθ in the equation). Another important aspect of this equation is that the angular momentum is measured relative to the origin chosen. This choice is arbitrary, and our origin can be chosen to correspond to the most convenient calculation.

Because angular momentum is the cross product of position and linear momentum, the angular momentum formula is expressed in vector notation as:

 l = r×p

This equation provides the direction of the angular momentum vector: it always points perpendicular to the plane of motion of the particle.

#### Angular Momentum and Net Torque.

It is possible to derive a statement relating angular momentum and net torque. Unfortunately, the derivation requires quite a bit of calculus, so we will simply revert to the linear analogue. Recall that: F = . In a similar way, τ = A net torque changes a particle''s angular momentum in the same way that a net force changes a particle''s linear momentum.

In circumstances of rotational motion, however, we usually deal with rigid bodies. In such cases the definition of the angular momentum of a single particle is of little use. Thus we extend our definitions to systems of particles.

#### Angular Momentum of Systems of Particles

Consider a rigid body rotating about an axis. Each particle in the body moves in a circular path, implying that the angle between the velocity of the particle and the radius of the particle is 90 o . If there are n particles, we find the total angular momentum of the body by summing the individual angular moments:

L = l 1 + l 2 + ... l n

Now we express each l in terms of the particle''s mass, radius and velocity:

L = r 1 m 1 v 1 + r 2 m 2 v 2 + ... r n m n v n

We now substitute σ for v using the equation v = σr :

L = m 1 r 1 2 σ 1 + m 2 r 2 2 σ 2 + ... m n r n 2 σ n

However, in a rigid body, each particle moves with the same angular velocity. Thus:

 L = ( mr 2)σ = Iσ

Here we have a concise equation for the angular momentum of a rigid body. Note the similarity to our equation of p = mv for linear momentum.

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