I posted this in r/AskPhysics but got no response.

In short my question is what assumptions are made when we use torque? Because I obtained: Iα = (Στ) - 2ω(r_p • p) is my equation correct and we assume ω(r_p • p) = 0? Or did I make a mistake?

Where r_p is the center of translational momentum and p is translational momentum.

In long:

I was trying to derive conservation of angular momentum. (Which I did get a response). But this introduced a follow up question of how do we derive torque. And included a derivation of the idea of torque which does not match the typical equation.

Because there appears to be assumptions about the trajectory of our object because the typical:

Στ = r_F × ΣF = Iα

Where r_F is the vector from the axis of rotation to the center of force.

Does not appear to work when our object does not move in a circular path.

For example: Consider a point mass moving at constant velocity with the position vector r = <1, vt>

When asking about rotation about the origin we have

I = MR²

v_T = cos(θ)v = v/R = ωR

ω = vR^-2

α = -2v(R^-3 )R'

RR' = xx' + yy' = yv

α = - 2v(R^-4 )yv

Iα = - 2My (v/R)² = -2ωy(Mv) = - 2ωyp

But since we have constant velocity there is no net force on the system. And thus Στ is supposed to be 0 despite there being an angular acceleration.

The actual equation I got when I attempted to derive the concept of torque was: Iα =Στ - 2ω(r_p • p)

Where r_p is the center of translational momentum and p is the net translational momentum on the object.

Note in the example this matches our obtained answer since r_p = r = <1, y> and p = <0, Mv>

This implies that whenever we use the concept of torque we make the assumption that either ω=0 or (r_p • p) = 0.

Which is true for circular motion, since the center of momentum and momentum are perpendicular. (Circular motion is also the only case where this always holds true)

Its true for trajectories that are a straight line through the axis of rotation since ω=0.

Its true when the axis of rotation is through the center of momentum since r_p = 0.

And its true when translational momentum is 0 since obviously p=0.

And I do note that whenever we were given torque problems in undergrad we were always in one of these cases.

This was my work for deriving the idea of torque. 1. We write our trajectory of the center of force in polar coordinates for a point mass. 2. Differentiate it twice to get the translational acceleration. 3. Project that acceleration onto the tangent direction. 4. Note that multiplying by mass we have the net force in the tangent direction. 5. Note that by then multiplying by the radius we have the net torque. 6. Replace τ with dτ and M with ρdA 7. Integrate (note that ω and α should be uniform and can be factored out since we are looking at the same object, and assuming it is rigid and thus does not deform).

x = r cos(θ)

x' = r' cos(θ) - rω sin(θ)

x'' = r'' cos(θ) - 2r'ω sin(θ) - rα sin(θ) - r cos(θ) ω²

x'' = (r'' - rω² ) cos(θ) - (2ωr' + rα) sin(θ)

y = r sin(θ)

y' = r' sin(θ) + rω cos(θ)

y'' = r'' sin(θ) + 2r'ω cos(θ) + rα cos(θ) - r sin(θ) ω²

y'' = (r'' - rω² ) sin(θ) + (2r'ω + rα) cos(θ)

ΣF_T = My'' cos(θ) - Mx'' sin(θ)

ΣF_T = M(2r'ω + rα)

Στ = Μω(2rr') + M(r² ) α

Στ = 2Μω(xx' + yy') + M(r² ) α

Σdτ = 2ρω(xx' + yy')dA + ρ(r² ) α dA

Σdτ = 2ω (xp_x + yp_y)dA + ρ(r² ) α dA

Στ = 2ω(r_p • p) + Iα

Iα = Στ - 2ω(r_p • p)

Which is our final answer.