I think there's something wrong with the derivation of the angular momentum equation. According to Wikipedia, Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image! Since the velocity crossed with momentum is equals to zero, Please Register or Log in to view the hidden image! However, isn't Please Register or Log in to view the hidden image! which is not equals to the rate of change of radius. This is the part which really bugs me. I hope someone can clarify it.
Torque is the rate of change of angular momentum. The linear analogue is that force is the rate of change of linear momentum.
Yes, torque would be the rate of change of momentum if \(v=\frac{dr}{dt}\) unfortunately, i can't see how is this is so.
Torque is the rate of change of momentum by definition: \(\vec{T}=\frac{d \vec{L}}{dt}\). This is basic physics In introductory physics you get taught that \(\vec{r}\) is the position vector of a particle. Then, by definition, \(\vec{v}=\frac{d \vec{r}}{dt}\). You claimed you took physics, even relativity, no? You have been questioning basic mainstream physics since you joined this forum, what is your problem?
I deliberately changed my tone to sound more confrontational, as i was afraid my thread would soon be buried under. People tend to skip past those mundane looking titles. Anyway, i'm not trying to be contrary, just wanna clear up some confusion. Like if you could give a crystal clear example that would even force all those fools like me to see the light and shut up. for instance, applying kepler's law of equal areas, (vr)sinx is a constant, where r is the distance from the planet to the sun, v is the tangential velocity, and x is the angle between them. So thats v cross r. So i construed r as the radius and \(\frac{dr}{dt}\) as the rate of change of radius. If you consider the planet to be orbiting in the x-y plane, then the x-component of the tangential velocity is the rate of change of displacement in the x-direction. I cant see how the tangential velocity represents the rate of change of position vector, and how the radius represents the position vector. Again, i may be horribly confused, so i hope you can clarify things.
It isn't possible, as evidenced by your continued posting. Have you really taken a physics class? Because your posts show that not to be true. Let's try one last time: Torque is the rate of change of momentum by definition: \(\vec{T}=\frac{d \vec{L}}{dt}\). Also, by definition: \(\vec{L}=\vec{r} X \vec{p}\). Introducing the second expression in the first one and doing the calculations: \(\vec{T}=\vec{r} X \vec{F}\).
aha, but that is only true if (dr/dt) x p is equals to 0. According to wikipedia, (dr/dt) is equals to v, so v x p or v x mv is equal to 0. That part i understand, because any vector crossed with itself is equals to zero. However, i dont understand why (dr/dt) is equals to v. Which i have already explained in post #6. So i hope you can address that point.
\(\frac{dr}{dt}\) is the rate of change of position, so it is exactly the velocity by observation. Really, not a lot more has been done than a relabelling and then splitting v into it's components in spherical coordinates.
but isn't r the radius and not the position? Also, i just realized that even if (dr/dt) x mv = 0, and (dL/dt) = mr x (dv/dt), mr x (dv/dt) is not always equal to torque. They are only equal if the force and tangential acceleration are in the same direction. For a planet orbitting the sun, they are never in the same direction. The magnitude of L = mvr(sinx), where r is the radius of orbit, v the tangential velocity and x the angle between them. So (dL/dt) = mvr(dsinx/dt) + mvsinx(dr/dt) + mrsinx(dv/dt) = 0 we know that gravity cannot exert a torque. But i don't know how to show that mvr(dsinx/dt) + mvsinx(dr/dt) + mrsinx(dv/dt) = torque
We should be careful here and not call the position r, as we have a radial distance from the origin that is already called r: \(\vec{s}\) is the position (strictly, the displacement). In spherical coordinates \( \vec{s} = r_0 \hat{r} + \theta_0 \hat{\theta}\) and I should have written \( \vec{v} = \frac{d}{dt}\vec{s}\) above. I think this is the source of your issues.
So m [d(r x v)/dt] = mv x (ds/dt) + mr x (dv/dt) instead of mv x (dr/dt) + mr x (dv/dt) Looks odd. maybe my notation is a little off. anyway, can mvr(dsinx/dt) + mvsinx(dr/dt) + mrsinx(dv/dt) be shown to equal the torque? where v is the tangential velocity.
so we should do all our calculations in polar coordinates? I'm still not sure about how to show that mvr(dsinx/dt) + mvsinx(dr/dt) + mrsinx(dv/dt) is equal to torque.
Do you understand the difference between \(\vec{L}=\vec{r} X m \vec{v}\) (vector) and \(L=mvr sin x\) (scalar)? Apparently not. This is the source of your errors.
yeah, i know it just represents the magnitude of L. Then τ = Fr(sinθ) too. of course, θ is always zero. it would be good if i could get a good diagram of angular momentum using spherical coordiantes.
No, it isn't. You need to take a class in basic physics, you will never learn from posting nonsense on the internet.
ummm, gravity can't exert a torque right? isn't that why angular momentum is conserved? so isn't θ, the angle between force and radius, always zero? Tach you fool. Please Register or Log in to view the hidden image!
