Energy of translation compared to the energy of rotation

  • #26
kuruman
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I don't think the OP is asking about kinetic energy.
Judging from the title of this thread, "Energy of translation compared to the energy of rotation", I thought OP is asking that. The title indicates a comparison of how much of the input energy goes into translational and how much into rotational energy. Without knowing the ratio of the masses, this comparison is meaningless.
 
  • #27
JrK
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So the energy needed to rotate the pinion is d2*F1 = 94*1= 94 J. The energy needed to move the pinion in translation is lg*(sin(pi/4)-sin(pi/6))/(pi/12) = 103 J.
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 would like to study with the mass as lower as possible, mass-less is perfect. And the energy-in must be equal to the energy-out.

Even there are external devices to control the position, the sum of energy in that devices are constant, I supposed the external devices perfect, again to simplify the calculations.
 
  • #28
jbriggs444
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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 would like to study with the mass as lower as possible, mass-less is perfect. And the energy-in must be equal to the energy-out.

Even there are external devices to control the position, the sum of energy in that devices are constant, I supposed the external devices perfect, again to simplify the calculations.
Discussion of perpetual motion is not allowed here.
 
  • #29
JrK
139
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Discussion of perpetual motion is not allowed here.
But all is fine, because the circle rotates of near 150° not 135° like I thought. Look the message #21
 

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