Crossed Belt and Rotational Dynamics: Solving for Angular Velocity and Torque

In summary, the angular velocity of the rotor of the generator relatively from its stator is -11w-w=-12w. There is no torque on the support. The stator of the motor gives a counterclockwise torque -T (T is any value), the stator of the generator gives a counterclockwise torque -T. Axes of the pulley gives a clockwise torque +2T, so the sum of torque is 0 on the support.
  • #1
V711
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0

Homework Statement



http://imageshack.com/a/img908/282/eCETrh.png

The green support turns clockwise at +w around the white fixed axis. The stator of the motor is fixed on the support. The stator of the generator is fixed on the support too. The motor drives the generator with a crossed belt. The rotor of the motor turns clockwise at +10w relatively to its stator (not lab ref). The radius of the pulley1 is the same than the pulley2.

a/ Find the angular velocity of the rotor of the generator in the lab reference
b/ Find the angular velocity of the rotor of the generator relatively from its stator
c/ Is there a torque on the support ?

Homework Equations



--

The Attempt at a Solution



a/

The rotor turns at +10w relatively to its stator and like its stator turns clockwise at +w, the rotor turns clockwise at +11w, is it correct ?

So, the rotor of the generator turns at -11w in the lab ref because the belt is crossed, is it correct ?

b/ Like the stator of the generator turns clockwise at +w, the angular velocity of the rotor relatively from its stator is -11w-w=-12w, is it correct ?

c/ There is no torque on the support. The stator of the motor gives a counterclockwise torque -T (T is any value), the stator of the generator gives a counterclockwise torque -T. Axes of the pulley gives a clockwise torque +2T, so the sum of torque is 0 on the support. Is it correct ?
 
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  • #2
The rotor turns at +10w relatively to its stator and like its stator turns clockwise at +w, the rotor turns clockwise at +11w, is it correct ?
I'd say this is correct for the driving motor. For the generator, I'd only reverse the motion wrt its own stator.
In other words: answer b) first -- and ask yourself if the answer can really depend on the ##\omega## of the support...
 
  • #3
BvU said:
I'd only reverse the motion wrt its own stator.
I don't understand could you explain ?

BvU said:
and ask yourself if the answer can really depend on the ω\omega of the support...
The rotor of the generator is at -11w in the lab ref, is it correct ? it don't depend of the rotation of the support. But the stator of the generator is fixed on the support so it turns at +w, correct ? So why, the angular velocity of the rotor relatively to its stator (not the angulat velocity in the lab ref) is not -12w ?
 
  • #4
Motor and generator turn with the same but opposite ## 10 \omega## wrt to the support.
For the motor this is a given. And it is independent of the angular speed of the support.
If you stand on the support, the motor and the generator have to make the same number of revolutions. Always. ##\Delta \phi_m = - \Delta\phi_g##.

Motor turns with ##\ +10\;\omega\ ## wrt its own stator, ergo idem wrt the support.
Motor and support turn in the same direction, so wrt the lab the ##\omega## add up to ##11\;\omega## .

Generator turns with ##\ -10\;\omega\ ## wrt its own stator, ergo idem wrt the support.
Generator and support turn in the opposite direction, so wrt the lab the ##\omega## add up to ...

If you find this difficult, replace the generator with a motor that runs at ##\ -10\;\omega\ ##.
 
  • #5
BvU said:
Generator turns with −10ω \ -10\;\omega\ wrt its own stator, ergo idem wrt the support.
If there is no support I'm agree because the rotor of the motor is at +10w so the rotor of the generator is at -10w. With the support, the rotor of the motor is at +11w not +10w and I don't understand why the rotor of the generator is not at -11w. Could you explain ?
 
  • #6
Easy to see: stop the motor.
 
  • #7
If I stop the motor, the belt don't move so the angular velocity of the rotor of the generator don't depend of the support, correct ? The rotor of the generator is at -10w. So the relative angular velocity of the rotor of the generator (ref its stator) is -10w-w=-11w ?
 
  • #8
If you stop the motor, both turn with the support. That has an angular speed of ##+1\omega## clockwise according to post #1.

By the way, mathematicians (and therefore physicists too) prefer it if you call anti-clockwise the positive direction of rotation. Matches the right hand coordinate system and makes all the formulas for torque and what have you come out correctly. So better get used to that from the beginning.
 
  • #9
so -9w for the rotor of the generator in the lab ref ?
 
  • #10
Sounds much better to me: in the lab the average is then ##1\omega##. And wrt each other they see ## \pm 20\omega## (depending on which one you sit on :wink:)

[edit] oopsed the 10 to 1 : [ 11 + (-9) ] /2
The other one is [ 11 - (-9) ] or [ 10 - (-10) ] whatever you prefer.
 
  • #11
When the rotor of the motor is at +w, the belt don't move, correct ? So the rotor of the motor is at +w, why in this case the rotor of the generator is not at 0 ?
 
  • #12
V711 said:
When the rotor of the motor is at +w, the belt don't move, correct ? So the rotor of the motor is at +w, why in this case the rotor of the generator is not at 0 ?
I take it you are still talking about the motor not running, stopped, on the handbrake, or whatever. Then both motor and generator turn with the support. At the same angular speed.
 
  • #13
Ok, thanks I understood !
 

Related to Crossed Belt and Rotational Dynamics: Solving for Angular Velocity and Torque

What is the concept of crossed belt and 2 rotations?

Crossed belt and 2 rotations is a mechanical system that involves two rotating pulleys connected by a belt. The belt crosses over itself, creating a figure-eight shape, and the pulleys rotate in opposite directions.

What are the applications of crossed belt and 2 rotations?

Crossed belt and 2 rotations are commonly used in mechanical systems to transmit power and motion between two rotating shafts. They are also used in conveyor systems, bicycle gears, and various other mechanical devices.

How does the direction of rotation affect the speed of the belt in crossed belt and 2 rotations?

The direction of rotation of the pulleys determines the speed of the belt in crossed belt and 2 rotations. When the pulleys rotate in the same direction, the belt moves faster, while rotating in opposite directions slows down the belt.

What are the advantages of using crossed belt and 2 rotations in mechanical systems?

One advantage of crossed belt and 2 rotations is their ability to transmit power between two rotating shafts that are not in line with each other. They also provide a smooth transfer of power and motion, with minimal slipping.

Are there any disadvantages of using crossed belt and 2 rotations?

One disadvantage of crossed belt and 2 rotations is the potential for the belt to slip off the pulleys if they are not properly aligned. They also require regular maintenance to ensure the belt tension is correct and the pulleys are in good condition.

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