Recent content by haruspex

To figure out the velocity at a given point in the string you need to find the tension there. The element at that position does not care how the tension is created. Write an expression for ##T(x)##.

Any conservation laws that you can apply?

Shouldn't that denominator quadratic be raised to the power of 3/2? Don't forget to apply an approximation for small r. And as @kuruman notes, it is the field normal to the axis that is interesting.

It's just Pythagoras. To avoid confusion, I will take the point of interest as being at z' from the solenoid, instead of at x. In (x,y,z) coordinates, where the negative z axis is the axis of the cylinder, you have a fixed point at (0,r,z'). Your solenoid element is at ##(R\cos(\phi)...

Surely ##\theta## is a function of ##\phi##.

I thought this question was familiar, even in that the official answer is wrong. https://www.physicsforums.com/threads/how-is-work-calculated-on-a-conveyor-belt-with-friction.1061325/

If there were a component of the field parallel to the surface of the conductor, what would the charges there do?

In part a, nothing is rotating yet, so Euler's equation reduces to ##\vec\tau=I\dot{\vec\omega}##. Part b is solved most simply using energy. If you mean the generalisation of being at angle ##\theta##, that should work, but it still seems like overkill.

Yes. Which Euler equations? https://en.wikipedia.org/wiki/List_of_things_named_after_Leonhard_Euler

Yes, it starts horizontal (part a) then swings down around the pivot to become vertical (part b), and continues, presumably.

… and part b specifies "when the rod is vertical after rotating 90 degrees". As I read the question, part a asks for the angular acceleration when released from horizontal and part b asks for the angular velocity at the vertical. Neither requires you to consider an arbitrary angle in between.

The two forces, gravity and the support force, define a plane. The rod and masses are all in that plane.

I didn’t read that in detail because it was evidently much more complicated than it needed to be. You should be able to write down the MoIs about the pivot straight off: ##ML^2/3+m(L/3)^2+2m(2L/3)^2##. Similarly the torque about the pivot: ##(ML/2+mL/3+2m(2L/3))g##.

That's rarely the easiest way. Better to calculate the moments of inertia separately then add them.

I agree with your electrostatic force. You need to calculate q, though. For the magnetic force, you could divide each strip into narrower strips and find the horizontal components of the pairwise attractions.