Recent content by DanielA

I'm not sure the best way to upload the MATLAB code I have if anyone wants to see it. I'd prefer to be able to just upload it directly, but doesn't seeme like I can.

Here is the paper again: https://www.mdpi.com/2218-2004/6/2/22?type=check_update&version=2#related_content For a class project I need to calculate the energy levels of atoms using the Hartree Fock method as presented in this paper which essentially brute forces the calculation using finite...

Homework Statement Suppose the potential in a problem of one degree of freedom is linearly dependent upon time such that $$H = \frac{p^2}{2m} - mAtx $$ where A is a constant. Solve the dynamical problem by means of Hamilton's principal function under the initial conditions t = 0, x = 0, ##p =...

I honestly don't know where the difference in factors is. And the class isn't one that would expect us to numerically integrate a function to get an answer. However, I clearly would need to for this one. So I don't know what to do since I'm pretty sure something more fundamental than some...

Wow I jumped the gun on that. The time derivative is of ##\frac{1}{2}\dot \theta^2##. And I messed up the first derivative on the equation of motion. The factor should be ##\frac{6}{5}##. I didn't multiple the ##\frac{5}{12}## by 2 when I took the ##\dot \theta## derivative. Though with how many...

I did a ninja edit on the factors and added an explanation because I realized my ##\dot x## and ##\dot y## were still from origin to ends of the rod and not CoM of the rod. Yeah, I get why the definite integral won't give a constant of integration. I just don't know how to correct my...

Alright, so I think I fixed it. ##T(\theta) = \frac{1}{12}m\ell^2\dot \theta^2 + \frac{1}{2}m((\frac{\dot x}{2})^2 + (\frac{\dot y}{2})^2)## for each rod. There is a similar triangle between the "big" right triangle of the whole rod and the "small" one from where the rod touches the ground to...

It seems for this system, the absolute max possible for two 2-d rods would be 6. 1 rotational and 2 translational for each rod. However, of course, there are constraints on the system that reduces that. The symmetric nature of the system will reduce the 2 rotational DoF (one ##\theta## between...

You're right about the normal force. I didn't consider that. Since the system isn't stable in that position without a string to hold the hinges (it was cut in the problem statement), then does that mean that the force on the point of contact will be split in x and y directions? The gravitational...

Sorry, I wasn't clear, I get what a small angle approximation is and how to use it, but I think he did it with a weird variable substitution. The only requirement for this solution to exist was that the rotation of the hoop was greater than the rotation of the particle around the hoop. He found...

Homework Statement I uploaded the homework question. This is #1. Homework Equations None directly given The Attempt at a Solution My main difficulty with the problem is that I am convinced it is much easier than my classmates make it out to be. This is graduate mechanics so I'm pretty sure...

<Moderator's note: Moved from a technical forum and thus no template.> Technically the homework question is at graduate level, but the area I'm having trouble on I feel is at an undergraduate level. In the question we studied a particle rotating on a vertical hoop that is also rotating about...

I missed that when typing it. I had the mass on my paper. I didn't really think it would be quasiperiodic, as in my study that case was only mentioned for two or more springs. I just haven't encountered a frequency like that and there were no examples of two masses coupled by a spring. I don't...

I guess I misread it. Does the math seem correct though? Or at least the result seems physically sound? It seems like it is to me. Maybe I made a minor math error somewhere, but I think the forms and physical interpretations of the equations are correct now.

Did you type that wrong? I did say the centre of mass can have velocity and kinetic energy associated with it. Anyway, I put my laziness aside and solved the equations I needed to substitute X and x into my Lagrangian (basically I tried to short cut it, confused myself, and created extra work...