Recent content by domephilis

Here’s the image. The lower left figure is for the second bit. The lower right one is for the torque angle bit. The lowest figure is to figure out how to add angles in 3D. As is the equations on the left side of the board.

My final result was $$\tau = IBR\sin{\theta} \int_{0}^{2\pi} {\frac{\sqrt{1+\sin^2{\theta}\cos^2{\gamma}}}{\sqrt{1+\sin^2{\theta}\cos^2{\gamma}}}d\gamma}$$. I think I am supposed to get a simple answer like $$\tau = \vec{\mu} \times \vec{B}$$ where mu=IA. If I take approximations using the...

Thank you very much for the explanation. It helps a lot.

I see. I didn’t quite get to Lenz’s Law yet. Although I have seen some demonstrations on YouTube to that effect. I had this confusion because in electric fields a charge cannot move on their own (i.e. experience their own field). I thought it might carry over to magnetism. Perhaps this will...

I see my error. The moment of inertia should be taken about the contact point. We can use the parallel axis theorem for that. That does yield the right answer. On your question about the emf generated by the moving rod, my feeling is that unless retardation is significant in this case, the...

Thanks to haruspex, Steve4Physics, and kuruman. I realize my error. If I take the contact point to be static then the torque should be about that point and MoI should be ##\frac{3}{2} MR^2##, hence I’m off by 1/sqrt(3) in the final answer. Regarding the issue of motional emf generated by the...

Torque generated by static friction is IBwR. I wrote it up there. I was wondering it asked that about dw. What’s wrong with just considering the body as a whole? It’s not like there’s a continuum of forces/torques that we need to integrate over. The only torque is from the friction at two...

Also, I am aware that energy conservation should give the correct result. I just don’t know why this way is wrong. Torque is torque.

I can't see what is the problem with my derivation but the answer is incorrect. Please help. We assume here that ##\omega_0 = 0## and ##v_0 = 0##, hence it immediately rolls without slipping without any transitional phase. Hence ##v = \omega R##. Thus, ##v(L) = R\omega(L)##. Since our...

I see. Thank you very much. Have a nice day.

I’m not sure this is right, but here’s my guess. When I pour the hot to the cold, I heat the cold water somewhat and gets warmer water. When I try to reverse it by pouring the water back, that bit back is not the 80C that it used to be and will cool the hotter water. Hence, we end up with a...

That’s my confusion. I added the water in slowly allowing for equilibrium at almost every moment. If I wanted to reverse a step, I can just pour that drop back in. I am mimicking the quasi-static process described in the book for a gas (i.e. slowly take out grains of sand which compresses a...

After re-reading the book, I did figure out what I was supposed to do. Take both waters through a series of reservoirs to bring them down to their final temperature while allowing for a quasi-static process. Thus, $$\Delta S = m_1c \int_{T_1}^{T*} \frac{dT}{T} + m_2c \int_{T_2}^{T*}...

Yes, I agree. I think that’s why the wave equation assumes a different form. As TSny points out, we can still deduce the wave velocity by only considering the boundary points of the pulse which has the special properties that the horizontal velocity (time derivative) is zero and the angle is 0...

Wonderful. I think this exactly what I need to satisfy my little curiosity. Thank you very much for this. I will definitely take some time to digest this.