Recent content by JrK

No, it is always the same example: a line and a circle, I changed the link in the previous thread and in this thread I added an external force for a short time: F1. I explained the first case to show I understood the sum of forces and the sum of energy but my problem comes from the second case...

Ah, ok. The first case, I apply an horizontal force F1 to the wheel from the ground, the ground receives the force F0. The center of the wheel receives the force F4 (the axle) and the wheel will have a CCW torque on it.

Yes, the center of the wheel can be taken on a rail without any friction. The wheel rotates and moves in translation. I study the geometry of the circle and the line, I try to understand all the movements. Last time, I understood the movement of the wheel: the distance moved by the rotation is...

No, I can translate the wheel just above the ground, at 1 µm (or less) above the ground. I don't want a direct interaction between the ground and the wheel. To have the force F1 on the wheel: I can apply the force from an electromagnet fixed on the ground for example in that case I can have a...

No, I can rotate the wheel around itself before and I can move the wheel in translation after. I don't use the ground to rotate the wheel. Are you sure ? I apply the force F1 from the ground. I use the sum of energy to correct myself when I think alone, and in the first case I have well the sum...

Hi :) 1/ First case A wheel with a mass ##m## and a radius ##r## moves in horizontal translation and rotates around itself. The wheel is just above the ground, doesn't touch it. The wheel rotates CW if the wheel moves in translation to the right. The ground is horizontal and there is no...

But all is fine, because the circle rotates of near 150° not 135° like I thought. Look the message #21

Please, correct that: So the energy needed to rotate the pinion is d2*F1 = 94*1= 94 J. The energy recovered from the pinion in translation is lg*(sin(pi/4)-sin(pi/6))/(pi/12) = 103 J. That calculation is not correct, just approxiative, I corrected in the message to have the true calculation. I...

-- No, I don't think about the kinetics energy, mass-less if possible or the mass as lower as possible. It is just to compare the energy. Sure, in that case all is fine.

Oh, yeah, sorry, the last drawing of the first message was for the notations of the program only but like in the program I can't use ##\alpha## in the program, so I used ##a##. I should have changed the name of the dot. No gravity here, and I think with masses very low. For me, ##a## is a dot...

##a## is a fixed dot on the pinion. Torque about the pivot, you mean A0 ? AT: in fact, I didn't think with a torque on the pinion, because it moves in translation, so I thought there is no energy needed/recover from the rotation of the pinion. The angle of rotation of the pinion is well near...

I need a little help there. From ##p_x = \frac{D-R-Rcos(\alpha)}{\tan(\alpha)}-Rsin(\alpha)## how to calculate the integration without pass from the derivative function ? I used the length by the force (like I could move in translation the rack). But, it is the same result with the torque by...

I think my formula is approximative but not to far from the real result I found with the numerical integration: I directly multiply the total distance moved by the center of the circle by the integrate of ##cos(\alpha)## and I found the mean with the sinus function. But I think it is not perfect...

Yes, but maybe I forgot one work.

I like you AT, you're like a good professor :) I noted ##i## the dot of intersection between the pinion and the rack, ##p## the dot of the center of the circle. I added ##x## and ##y## for easting/norhting and I added ##1## and ##2## to represent the last position ##2## and the new position...