Please help finish my deduction for angular momentum

[Cedric Chia]

I'm trying to deduce the angular momentum ( for a rigid body ) on my own, and here is the problem I face.

By introducing the angular momentum of a tiny piece in rigid body (" i ") as :
L_i = r_i × p_i
L_i = r_i × m_i v_i --------------------------------- [ Line 1 ]
L_i = r_i × m_i r_i ω_i

To find the angular momentum of the whole rigid body, summing all the tiny pieces, we have :
∑ L = ∑ r_i m_i r_i ω_i

Since ω is constant everywhere, we then have :
∑ L = ω ∑ m_i r_i²

As ∑ m_i r_i² is the definition of I ( moment of inertia ),
angular momentum of the rigid body thus can be written as :
L = I ω

But what if I want to express it in v ?

Referring to [ Line 1 ] :
L_i = r_i × m_i v_i

To find the angular momentum of the whole rigid body, summing all the tiny pieces, we then have :
∑ L = ∑r_i m_i v_i

What is the v_i and since the velocity of the tiny pieces depend on their position, how can I continue ? Or is the velocity is the same as the edges of the rigid body ?

 

[andrewkirk]

Angular momentum is calculated relative to a reference axis. Change the axis and the angular momentum changes. It is implicitly assumed in your first set of calculations that the reference axis is what the rigid body is rotating around. That axis may or may not pass through part of the body. Given that assumption, the velocity $\vec v_i$ of an infinitesimal piece of the body will always be in a plane that is perpendicular to the axis, and the velocity direction will be perpendicular to the axis. Hence $\vec r_i\times \vec v_i= |\vec v_i|\times |\vec r_i|$.

If we maintain that assumption in your second calc then we can take $v_i$ to be the scalar speed of the infinitesimal piece, and that will be equal to $r_i\omega$.

Note that $\omega$ has no $i$ subscript. If the angular velocity varies between parts of the object, then either the object is not rigid, or it is not rotating around the chosen axis of reference. So the $\omega_i$ items in your calc should just be $\omega$.

 

[Cedric Chia]

Note that ωω\omega has no iii subscript. If the angular velocity varies between parts of the object, then either the object is not rigid, or it is not rotating around the chosen axis of reference. So the ωiωi\omega_i items in your calc should just be ωω\omega.

I did write :
To find the angular momentum of the whole rigid body, summing all the tiny pieces, we have :
∑ L = ∑ ri mi ri ωi

Since ω is constant everywhere, we then have :
∑ L = ω ∑ mi ri2
** Note that the ω here does not have a subscript. **

Let's consider the following scenario :

If the object m₁ traveling linearly with initial velocity u₁ hit a rigid body m₂ ( a uniform rod initially at rest with fixed/pivoted rotation axis at one of it's end ), the object m₁ hit the ends of the rod ( not the one with axis of rotation ).

Using the conservation of momentum, we know that the the rigid body m₂ ( the rod) now have a linear velocity ( which I call it v₂ ).

What is this v₂ mean and where does it act on ?
Why is the angular velocity of the rod ω₂ is simply v₂ / r as we know linear velocity at different position is not the same ?

 

[haruspex]

You can treat the general motion of a rigid body either as a rotation about the instantaneous centre of rotation or as the sum of a linear motion of the mass centre and a rotation about the mass centre.
In your equation using ω, you have taken the first option, so the moment of inertia is not about the mass centre; the parallel axis theorem applies.
In the second view, you still need the ωΣm_ir_i² term for the rotation (the r_i being distances from the mass centre) but you can use mvr for the angular momentum corresponding to the linear motion of the body (r being distance from the axis).
Since v=ωr, the resulting equations are the same.

Not sure if this answers your question, but maybe it helps.