Why Is the Total Angular Momentum of a Particle System Constant?

[ElDavidas]

hey everyone, I've been working through a past paper and I'm stuck on a question:

"(a) Consider a system consisting of n distinct particles P1, P2, : : :, Pn, with masses m1, m2, ..., mn and position vectors r1, r2, : : :, rn, relative to the origin O of an inertial frame, respectively.

For each i = 1, 2, : : :, n, suppose that the only forces acting on Pi are internal forces Fij , which always act along the line joining Pi and Pj , for j = 1, 2, : : :, n and j doesn't equal i.

Assume that Fij = ¡Fji for i = 1, 2, : : :, n, j = 1, 2, : : :, n and i doesn't equal j.

Define what is meant by the total angular momentum of the system about O.

Show, as a consequence of Newton's second law, that the total angular momentum of the system about O is constant."

I can define the total angular momentum of the system, it's the 2nd part of the Q I have trouble with.

Thanks

 

[Tide]

Use the defintion of angular momentum:

$$\vec L = \vec r \times \vec p$$

Take the time derivative and sum over all particles noting that

$$\frac {d \vec p}{dt} = \vec F$$

You will find $d\vec L /dt = 0$.

 

[quicknote]

for reaching out! The total angular momentum of the system about O is the sum of the angular momenta of all the particles in the system, taking into account their masses, positions, and velocities. This can be mathematically expressed as:

Q = ∑mi(ri x vi)

where mi is the mass of particle i, ri is its position vector, and vi is its velocity vector.

As for showing that the total angular momentum is constant, we can use Newton's second law, which states that the net force acting on a system is equal to the rate of change of its momentum. In this case, we are dealing with angular momentum, which is a vector quantity, so we need to consider the net torque acting on the system.

Since the only forces acting on the particles are internal forces, the net torque on the system will be zero. This is because the internal forces always act along the line joining the particles, so their torque will always be parallel to the position vector, and therefore their cross product will be zero.

Mathematically, this can be expressed as:

∑τ = ∑(ri x Fij) = 0

Using the fact that Fij = -Fji, we can rewrite this as:

∑τ = ∑(ri x (-Fji)) = -∑(ri x Fji) = 0

Therefore, the total torque on the system is zero, and by Newton's second law, the rate of change of the total angular momentum is also zero. This means that the total angular momentum of the system about O remains constant, as desired.