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A is the acceleration of the non-inertial frame AS SEEN IN the inertial frame. I switched from R_0 double-dot to A. Sorry for the confusion.

@@brianjackson415 thanks for replying prof. Final exam is in 20 min haha. Say, you reference some videos that seem to exist but also dont seem to be uploaded to TH-cam... like chptr section 9.4/.5 Did you ever record those lectures/ have lectures on later chapters like 11 and 16

@@Twitledum9 I did not manage to make videos for every chapter in Taylor, I'm afraid.

@ no matter, im sure they would have been great. Really enjoyed your (simplified) approach to the material. Really helped me

@@Twitledum9 Happy they're helpful. Good luck on the test.

Woops i see that it was corrected. Thanks prof

u is defined as 1/r. It's NOT a dummy variable.

I had a brain fart... ty for taking the time to respond Prof. I did well on my exam because of your videos. much appreciated!!

@@Twitledum9 Wow, I'm honored that I was able to help that much. Thanks for letting me know.

Good catch. That appears to be a typo, and mu should be squared. Apologies for the confusion.

@@brianjackson415 ahh - Thank you!! I really appreciate your lectures, Professor Jackson! I missed a lot of class and they are definitely a huge aid in helping me prepare for the exam! 👊

Yes. A finite delta phi would give a delta r vector that has some component along r-hat.

@@brianjackson415 Makes sense, thanks for your videos, great resource while going through Taylor

I think the confusion you have here is that you've described two different rotation states. For the first rotation state you've described, you've said the rotation vector has a magnitude of 1 rotation per second ("Let's say it does one full rotation in 1 second"), but for the second state you've described, you're saying the rotation vector has a magnitude of sqrt(3) rotations per second ("all undergo one full rotation in 1 second also"): sqrt((rot about x)^2 + (rot about y)^2 + (rot about z)^2) = sqrt(1 + 1 + 1) = sqrt(3). Those descriptions represent two different rotation states, I think. That's why they don't seem to agree.

@@brianjackson415 Hi Brian, thanks for your reply. Yeah they definitely are different (for one, the projection rotations are ellipses and the rotation around the vector omega is a circle). But I still puzzle over what the connection is between the two types of rotation states (as you call them) since they are related physically. And yeah I could just accept the second rotation state (as you describe) is described mathematically as the square root of the sum of the squares and be done with worrying how the projections' rate of rotation is reconciled with this. But that feels wrong to not explore this and find the relationship between the two rotation states. It also requires me to accept that omega is an actual vector (see next paragraph 🙂). Can you comment on any possible relationship between these relation states. One more thing: accepting the mathematical square root of the sum of the squares relationship obviously means accepting angular velocity vectors as vectors. I can accept this mathematical relationship if I can be convinced that angular velocity is a vector. Normally textbooks just get you to accept that angular velocity is a vector. At the most they just tell you that since infinitesimal rotations commute, d(theta_1)/dt and d(theta_2)/dt etc commute so angular velocity must be a vector. This doesn't quite follow. Such commutativity is necessary but not sufficient for these to be vectors so you're still wondering why exactly angular velocity is a vector and in turn, left unconvinced by the square root of the sum of square relationship. Can you comment on why omega is definitely a vector, given the above. Thanks for your time.

Hi can you comment on this. If you can't comment further on the first bit of my reply, that's ok. But can you explain why angular velocity is regarded as a vector in the context of what I have written in the second half of the reply. Any help would be greatly appreciated. Thanks.

@@markkennedy9767 I'm not sure I could easily explain why the commutativity properties mean that angular momentum has to be a vector. But the main reason we treat angular velocity as a vector is because it has a magnitude (# of spins per second) and a direction (where is the rotation axis pointing). For example, consider the Earth's rotation vector. Small shifts over 26,000 years in the *direction* of the Earth's rotation mean that the Earth's north pole does not always point at the North Star, even though the rate at which the Earth rotates (the length of a day) might not change. In order to completely describe Earth's rotation we need to know both how long a day is and where the rotation axis points. That's a vector.

@@brianjackson415 Hi Brian, thanks for getting back. Yeah, I suppose angular velocity has a magnitude and direction when they are defined this way (which seems like a natural way to define them in fairness). But then I'm back to whether we can be sure that its component vector in the x direction (with its magnitude and direction defined in this way) plus its component in the y direction (with its magnitude and direction defined in this way) plus its component in the z direction (with its magnitude and direction defined in this way) gives back the angular velocity vector (with its magnitude and direction defined in this way). I don't think it's obvious (or trivial) that this follows just by saying we will say that all these objects have a magnitude and direction (as defined in this way). I think what I'm getting at is, it is far easier to see how this would follow if we were dealing with ordinary velocity and its components. Or displacement and its components: I can see how the physical velocity and its components can be defined mathematically (using magnitude and direction) and how we can then add them like vectors. So I can see how it's obvious that ordinary velocity is a vector in this way. It doesn't quite work as simply for angular velocity though when we do the same process as in the last paragraph: when we go from the physical notion of angular velocity to its mathematical definition (using magnitude and direction), there is an extra leap of faith at this point, compared to the ordinary velocity case: for angular velocity, when we go from the physical notion of angular velocity to the mathematical definition of its magnitude and direction, it's not as obvious that this definition of magnitude and direction confers on angular velocity a vector status (as was obvious in the case of ordinary velocity). Does this make sense. So even though we have mapped the physical attributes of angular velocity (the instantaneous angular speed of the object and the direction of the axis around which it rotates) to the mathematical definition of magnitude and direction, i'm still not convinced (as natural as this mapping seems) that this confers on angular velocity vector status. Can you comment on any of this to convince me otherwise. Or failing that, maybe you can comment on how the commutativity of Infinitesimal rotations we talked about previously does actually imply vector status. If you can shed any light on this, it would be brilliant. Thanks.

In my opinion, Rotating frame is rotating with angular velocity. But angular momentum vector can not be parallel with angular velocity vector so (dL)rot comes.

I think it would be a answer. dL/dt (rot) means rate of change of L(angular momentum for fixed reference frame) measured in rotating frame.

I am not. : )

​@@brianjackson415 @3:58 here I have confusion. I think in (dL/dt)rot, L is always zero . Because rotating frame is rotating along with rigid body, so *angular velocity must be zero according to observer in rotating frame* and hence at every instant *angular momentum must also be zero according to observer in rotating frame*....

Happy to help

The functional f doesn't explicitly depends on alpha, but the function capital Y does. So you need to apply chain rule to calculate df/d(alpha). Hope that helps.

Hmm, I don't think so? Why do you think there's a mistake?

@@brianjackson415 at 10:56, when bringing it across the = sign A+B=0 and said A = B. Should be -B. (B=Vter*Tao*ln(1-stuff))

everything else is perfect so great video!

also the - sign didn't matter. He was just showing us that it was transcendental to get us to solve for Ln (1-E).

Good question! We have to be careful what we mean by "time-dependent force" because the force that a particle sees may change (1) because the force at a point in space changes with time AND/OR (2) because the particle is in motion and therefore encounters a force that changes with position over the particle trajectory. I presume you are thinking about the former case (force at a point changes with time). In that case, we can associate a potential energy with that force by taking the line integral at a given instant in time. There's nothing about the line integral calculation that explicitly involves time, so I think would work.

@@brianjackson415 Thanks for the reply. Don't you think taking the line integral of a force at an instant of time would give a potential energy that is independent of time, thus making the force, which is the gradient of that potential energy, time independent(I mean explicitly independent of time)? The force that we were calculating the potential energy for, was explicitly dependent on time (The former case). My confusion is, can we still use the line integral of a force with explicit time dependence(Non-conservative force) to calculate the potential energy for that force, because the line integral won't be equal to the change in potential energy, unlike in the case of conservative forces. Of course we can still calculate the line integral itself, it'd even be equal to the work done by the force, it's just that it won't be equal to the change in potential energy, as far as I know.

@@AkashShaan If the force itself depends on time, then the value of the line integral over a given path will, by definition, depend on time.

Think about what happens in the limit when theta is zero. That represents someone standing, for example, exactly at the North Pole. That person's velocity would be zero. They might be rotating around, but they're not moving at all.

I'm not sure exactly what you're asking, but if you set the gravitational torque equal to (dL/dt)_space, then you can use that equation to calculate the evolution of the top's spin.

@@brianjackson415 Alright sir, thank you indeed.

I'm afraid I don't know what you mean by "Randall".

One of the previous sections in chapter 10. I apologize, but I don't have my textbook with me to give a more specific answer.

@@brianjackson415 please show full derivation

@@a.nelprober4971 A full derivation for what? The kinetic energy of rotation? I direct you to chapter 10 in the textbook.

You can still plot the phase space trajectory of a system (momentum and position), but a time-dependent potential means the energy is not conserved. So you can't use any equations that assume energy conservation. Let me know if that doesn't answer your question.

@@brianjackson828 thank you so much for giving yours expensive time. Yes sir, it is clear now.

The omega vector is the spin vector.