Recent content by TSny

The energy ##E = T + V## is conserved in this problem but the angular momentum is not conserved. Energy is conserved since the only forces that do work on the system are the conservative force of gravity and the conservative force of the spring. (There is a force at the point of support, but...

That's correct. Right. No. There can be a constant of the motion that is not associated with an ignorable coordinate. Can you think of such a constant of motion for this problem?

Ok. Thanks. Hopefully the OP will clarify.

Are you making an "ansatz" that the solution inside the sphere will be a uniform field ##B_i## and the solution outside will be the superposition of the external field ##B_0## and the field of a magnetic dipole ##m##? Then, with this assumption, do you want to know how to determine the values...

This is very nice! You used the symbol ##\alpha## for the mass density ##\sigma##. So, your next to last equation can be written as $$\frac{dT}{ds}=\sigma g \sin\phi$$ It is easy to use this equation to derive the result for ##T## as a function of vertical height.

Those answers look right to me. (I guess there is no friction). However, based on what you've written, we can't tell if you thought through it correctly.

Yes, that's the idea. To show it quantitatively, use your approximation ##x = 2l\left(1-\dfrac{9y^2}{8ml^2}\right)##. Show ##\dot x## is of second order in small quantities. Hence, ##\dot x^2## is of fourth order in small quantities. But, you only need to keep quantities to second order in...

Since ##x## varies with time, you might want to show why it’s not necessary to include ##\dot x## in the kinetic energy of the system.

That all looks good to me. [Edit: There's a typographical error in the equation ## m\ddot{y}=-\dfrac{3q^2}{16\pi\epsilon_{\circ}l^3}##. It is missing the factor of ##y## on the right-hand side.]

Here's an attempt at a qualitative explanation. Suppose we have a very long, straight wire with current. Imagine the current is due to motion of both positive and negative charge carriers. The positive charges are shown in red and move upward. The negative charges are blue and move downward...

That looks good. Does the position of the center of mass move during the oscillations? Why or why not? OK. But, since the oscillations are small, you can see that the angle ##\beta## will only deviate a small amount from 90o. So, ##\beta## will never be small. If you let ##\alpha## be the...

If possible, please type your equations using Latex rather than post images of your hand-written work. I'm having difficulty deciphering parts of your diagram and some of the terms in your equations. Using energy is a nice way to approach the problem. Can you find a simple relationship...

Yes, you are right. The textbook solution is wrong where it starts off with ##F_{net} = i_{ind}LBx_2 - i_{ind}LBx_1##. As you noted, it should be ##F_{net} = i_{ind}LB(x_2) - i_{ind}LB(x_1)##, where ##B(x_1)## is the magnetic field at ##x = x_1## and ##B(x_2)## is the magnetic field at ##x...

The answer should be proportional to ##n^2##, not ##n^3##. Show the details of your calculation so we can help you identify any mistakes. The answer that was provided to you has some typographical errors, but the ##n^2## is correct. [EDIT: Nevermind, I was thinking of finding the current in...

Looks good. Check the last equation above. In the parentheses on the left side, the first term should not have ##B##. It should be ##\dfrac{1}{\rho \epsilon_0}##. This will change your expression for ##\alpha## where you have It should be $$\displaystyle \alpha =\frac{1}{\rho \epsilon_0}...