Recent content by jerronimo3000

Ah, I think I understand better. Thank you. How does one calculate the axial force, shear force, and bending moment at each point along the rod?

Could you elaborate on internal moment? I'm not sure I'm grasping what you mean.

In this video, a man applies an angular acceleration to the base of a rod. While the rod rotates, it bends. Why? What force is there that causes the bending, aside from rod's own weight? It seems to me to be the work of a fictitious inertial force. I was always taught that those forces don't...

Oh, duh..thanks, definitely clears that up! Thank you!

If the number of possible values of L is n, and the number of possible values of m is 2*L-1, and there are 2 spin directions.. shouldn't the total number of states be 2*(number of possible L)*(Number of possible m)? But this gives 4n^2 - 2n. I am extremely confused. Thanks for your help!

Aha! That produced 8.14*107 m/s. That looks much better!

Wait, looked through my equations. Relativistic total energy is gamma*m*c^2, and relativistic KE = E - mc^2. So (gamma - 1)*mc^2. Let me try that and see what I get.

I understand that the changes must be equal, but I'm not sure how to equate them mathematically. The potential energy is ΔV*q, right? If 0.5*m*v2 doesn't hold when speeds are near the speed of light then what would the correct equation for kinetic energy be?

In general, kinetic energy is 0.5*m*v2. The energy the particle gains from going through the potential difference, ΔV*q, should equal the particle's kinetic energy after it's gone through the potential difference. Does 0.5*m*v2 hold even at relativistic speeds? If so then I definitely understand...

The kinetic energy comes from its velocity, which comes from the particle being accelerated through the potential difference.

√[ (pc)^2 + (mc^2)^2 ] where p is the momentum of the particle

E is the total energy of the particle. The energy from its mass and then it's kinetic energy as well.

Whoops, I was using the "modern" mass units and Joules for energy. MKS units gives 8.38 * 10^7 m/s. That's 0.28c, which is fast enough to start considering relativistic effects, right?

Well, I guess because we're just coming off of chapters about relativity haha. Also, typically all of the particles we have worked with in class are moving at relativistic speeds. The answer using (V2-V1)*q = 0.5*m*v^2 gives 1.12 * 10^-7. That seems very small.

Homework Statement If a positron (or electron antiparticle) beam is accelerated across a potential of 20 kV, find the final velocity v of the particles. Do this problem TWICE, once using MKS units (J for energy) and a second time using "modern" units (eV for energy). Use the following values...