Recent content by brotherbobby

Yes I agree with the second sentence of yours. It is the first that is more interesting. Let me write it again, despite being repititive. The angular velocity vector of a rotating body ##\vec\omega## is necessarily a constant in space. Thus it would be quite nonsense to write...

Thank you, and a bit silly on my part. My derivation up there in post#1 remains the same even if the the angular velocity vector ##\vec\omega(t)## changed with time. Hence, I suppose it would be ok for us to write that ##\vec\nabla\times\vec v(t)=2\vec\omega(t)##.

Space. So if the angular velocity of the body varies with time, say ##\vec\omega(t)##, I have the following two questions. The first one is more basic and I suppose more important : Can we write ##\vec v(t)=\vec\omega(t)\times\vec r(t)##? Of course the position vector and linear velocity of...

Attempt : I draw the diagram of the problem to the right. A circular plate of radius ##a## rotates with an angular velocity ##\vec\omega = \omega_0\hat k##, assumed constant for the moment. A particle P at rest with the plate lies on the rim along the ##x## axis, as shown. Its velocity is ##\vec...

I don't understand what symmetry really means. I am going by what I suspect.

Text excerpt : For context, I copy and paste the paragraph on "symmetry" from the text, underlining the relevant lines in red. I hope it's readable. Problem situation : This concerns the last three lines from the text excerpt. They say that : "... if a particle of mass ##m_1## travels at a...

As I have found out only recently, and I confess I should have done so a lot earlier, it is best not to think of electricity as the flow of electrons ##(e^{-})## at all. The convention : ##\small{\text{electricity flows from positive to negative}}\; (+ \rightarrow -)## is more than an just an...

Thank you and apologies. A silly error on my part. Something else if I may. There's an agreement to take the conventional flow of current ##I## from positive to negative. In this problem I raised, namely the directiom of force in a current carrying conductor, the agreement becomes crucial. If...

I don't see how. The RHR (Right Hand Rule) mind you. The magnetic field is from left to right. Charge is moving down. Hence force points into the page, not out of it

Positive.

Situation : I have drawn the image to the right. A current ##I## flows "down" along a wire in a magnetic field ##\vec B## directed to the "right". By Flemings's Left Hand Rule (LHR), the wire should experience force out of the page, as shown by the green bullet ##\color{green}\bullet##. However...

I have solved the problem I mentioned in post #1 at the top, which was about the direction of coriolis deflection of a rocket fired "up" from a point in the southern hemisphere. It would be deflected west. Addtionally, I raised two similar cases of the deflection for a body dropped from a height...

Brilliantly explained. Thank you and sorry for the delay in responding.

I should reply to you later when time and the advantage of having done some mathematics first permits. For the case is involved. The particle has acquired a speed (and deflection) due east projected up, after having reached its highest point (P in the northern hemisphere). On its way down, it...

Due to the coriolis force, the particle acquires a velocity due east (along X) on its way up. For a place in the northern hemisphere. Likewise, doesn't it acquire an equal and opposite velocity (along -X towards) west on its way down, that should result in no net horizontal displacement?