Recent content by TSny

I don't know. A highly reflective metal surface or mirror would not absorb very much. Each ray of light in the beam obeys the law of reflection at the curved surface.

You might experiment with fanning out the beam by reflecting off a smooth cylindrical surface. Here, I used a ceramic mug.

Very good. I'm glad I could help.

Your result ##-\frac 1 2 \lambda v^3## is correct for the rate at which KE is lost to heat at point B. I’m not sure of your thought process in going from ##\Delta KE = - \frac 1 2 \lambda \Delta x v^3## to the result for the rate ##\frac{dKE}{dt}##. But it might be fine. I was just trying to...

Yes Rigidity imposes a condition on the velocities of two points of the rod, not the accelerations.

Good. Ok. This is less of an eyesore if you write it as ##U = U_0 -\lambda ghx##, where ##U_0## is the total potential energy at the moment of release (##x = 0##). The force that the table exerts on the chain does not do any mechanical work since the force doesn't move the chain through any...

As pointed out by @Steve4Physics, mechanical energy is not conserved. However, you can still solve it using energy concepts. (1) Find an expression for the KE of the chain at the instant the upper end of the chain has moved a distance ##x## from point ##A##. Express in terms of ##\lambda##...

Yes, they correctly accounted for gravity. But, they did not account for the change in chemical energy ##U_{chem}## of the battery. Their solution predicts an imaginary value for the final speed of the plate for the case where ##g = 0.## This is a good problem for demonstrating the importance...

Your two methods look correct. I did catch a spot where you have two compensating errors: Check the signs in this calculation. However, your result of ##\frac 1 d## is correct. To see that their answer can't be correct, suppose the experiment is done in zero gravity.

Yes. Nice analogy.

Good. If you wish, you can post your work for ##\nabla \mu## and we can review it.

If you release the block from rest at the position where it starts to submerge, then the only two forces doing work as the block submerges are gravity and buoyancy. But in this case, the object will bob up and down in oscillation. The net work is zero at the instant when the oscillating block...

Is this taken directly from your teacher's PowerPoint slides? Does it represent your teacher's solution to the problem of post #1? Are you trying to understand your teacher's solution? The original problem does not concern spherical shells. So, I'm confused here. For the original problem...

This problem is not the same as the one given in your first post. It appears to be totally unrelated. You mention your teacher's powerpoint. So, I'm assuming the picture above is a PowerPoint slide. If you have a question about this slide, you should start a separate thread and ask specific...

Besides the error pointed out by @anuttarasammyak, check your work for the following: I get an additional term proportional to ##\vec{\beta}##. $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R + \frac 1 {\mu^2} \vec{\beta}$$ Verify that...