Recent content by Bullwinckle

This is great! This I understand. I follow your "proof" that... $$\hat{R}=\hat{R}_3(\alpha) \hat{R}_1(\beta) \hat{R}_3(\gamma).$$ is a parameterization of the rotation group SO(3) And I can easily intuit similar proof for the Tait angles (and ditto for the gyroscope) So is that the issue...

Thank you Vanhees, But may I elaborate in the context of your response? How do I KNOW that the Tait or Euler or Gyro are parameterizations of the rotation group? Surely I can rotate about the local 3-axis three times. And I know, obviously, that it will not cover all the rotations. The Tait...

I understand that a gyroscope undergoes precession, nutation, spin. And that the order of the rotations are such that the precession and spin share a common "local axis." I also understand there are, for totally different purposes, Euler angles to model rotations. In this case, the order of the...

Oh, one more thing... Are you saying that books use this: ΔL = mv1 - mv2 Simply to prepare students to solve problems over a finite time difference that is not infinitesimally small? So, here, the Δ is being used for a PRACTICAL purpose?

Thank you

Hello, Sometimes in introductory physics, I see this: After defining P as the momentum, we show: dP/dt = F Then, later, I see this ΔL = mv1 - mv2 So my question is really simple: WHY is there no consistency? Why do we sometimes use the Δ symbol? Is there something about how the...

Say I have a position vector p = e(t) p(t) Where, in 2D, e(t) = (e1(t), e2(t)) and p(t) = (p1(t), p2(t))T And if I conveniently point the FIRST base vector of the frame at the particle, I can use: p(t) = (r1(t), 0)T I want the velocity, so I take v = d(e(t))/dt p(t) + e(t) d(p(t))/dt...

OK, so are you not also stumbling over this: dr = v dt (to progress from the second to third term) Is that not taking this: dr/dt = v And multiplying by dt? Now I feel I am back at square-1 Or, are your words "parametrization" the key. Is this allowed in a parametrization? Is there...

If we take F=ma and multiply both sides by dt, we get Fdt = ma dt And then: Fdt = mdv And then: Impulse = change in momentum. OK; I get that. I get a similar process for Work/Energy multiplying F=ma by ds on both sides as follows Fds = ma ds And using a ds = v dv to get Fds = m v...