Angular velocity, torque and work

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SUMMARY

The discussion focuses on the mechanics of two pulleys, Pulley1 and Pulley2, connected by a belt, with distinct angular velocities influenced by a motor and brake system. Pulley1, with a radius of 3R, rotates at 2w, while Pulley2, with a radius of R, rotates at 4w. Key calculations include the work done by the brake, torque on the supports, and the energy required from the motor, leading to the conclusion that the torque on the supports sums to zero due to equal and opposite forces. The participants emphasize the importance of understanding the relationship between angular velocities and the forces acting on the system.

PREREQUISITES
  • Understanding of angular velocity and torque concepts
  • Familiarity with pulley systems and belt mechanics
  • Knowledge of work and energy principles in physics
  • Ability to analyze forces in mechanical systems
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  • Study the principles of torque in rotating systems
  • Learn about energy transfer in pulley systems
  • Explore the effects of friction in mechanical systems
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Mechanical engineers, physics students, and anyone interested in understanding the dynamics of pulley systems and the calculations involved in torque and work in rotational mechanics.

  • #31
V711 said:
For know the torque on the support I'm looking at all forces to the support not the pulley, could you explain ?
There must be a zero net torque on the pulley, agreed? The four forces you have on the pulley, in the positions shown, will not produce a net zero torque if they are of equal magnitude.
 
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  • #32
Sorry, I changed the last image, it was a little error.

haruspex said:
There must be a zero net torque on the pulley, agreed?
No, because pulleys accelerate more and more like the motor. There is a net torque on the small pulley: F7/F9. It's possible to think a very small mass of the big pulley like that the torque from F1/F2 can be near the same than F5/-F6, no ?
 
  • #33
V711 said:
No, because pulleys accelerate more and more like the motor.
You wrote that the large pulley has no mass. No mass, no torque.
 
  • #34
Just for simplify the study, the big pulley drive the small pulley with the belt and the small pulley has a mass.
 
  • #35
V711 said:
Just for simplify the study, the big pulley drive the small pulley with the belt and the small pulley has a mass.
We're discussing the large pulley. If it has no mass then there is no net torque on it.
 
  • #36
Maybe it's that I don't understand. 2 pulleys are connected with the belt. If I want to turn the big pulley (with a small mass), I need to give a torque, what torque I need to give ? For me, it's not 0, because when I turn the big pulley, the belt move and want to turn the small pulley, if the small pulley has a mass, a force appear on the belt, this force will give F5 to the big pulley, no ? For me it's logical, in the contrary I can turn something without an energy.
 
  • #37
V711 said:
2 pulleys are connected with the belt. If I want to turn the big pulley (with a small mass), I need to give a torque, what torque I need to give ?
The pulley is only an intermediary. There is no net torque on the pulley. The torque exerted by the motor should be exactly canceled by the other forces acting on the pulley. With the magnitudes you claim, that won't happen.
 
  • #38
haruspex said:
There is no net torque on the pulley.
I'm agree. No net torque on the big pulley.

haruspex said:
With the magnitudes you claim, that won't happen.
For me 6*R*F1 = 3*R*F5 so there is no torque on the big pulley. In my last image F1=0.5*F5

I said ok but for me forces are :
|2F1|= |2F2|= |2F3|= |2F4|= |F6|= |F7|= |F8|= |F9| = |F5|, it's possible, no ?
 
Last edited:
  • #39
V711 said:
I'm agree. No net torque on the big pulley.For me 6*R*F1 = 3*R*F5 so there is no torque on the big pulley. In my last image F1=0.5*F5

I said ok but for me forces are :
|2F1|= |2F2|= |2F3|= |2F4|= |F6|= |F7|= |F8|= |F9| = |F5|, it's possible, no ?
Yes, that all looks ok now.
 
  • #40
ok, great ! I would like to know if the forces F6/F8 gives a torque to the support ?
 
  • #41
V711 said:
ok, great ! I would like to know if the forces F6/F8 gives a torque to the support ?
Yes, I said so in post #25.
 
  • #42
ok, and at a time, the motor needs the energy -3FRwt (the motor turn at 2w but it is on a support that turn at w), the support receives -3FRwt (from F3/F4). The small pulley receives +6FRwt and the support receives +2FRwt, the sum is not at 0 so I suppose the small pulley don't receive 6FRwt but 4FRwt, no ?
 
  • #43
V711 said:
ok, and at a time, the motor needs the energy -3FRwt (the motor turn at 2w but it is on a support that turn at w), the support receives -3FRwt (from F3/F4). The small pulley receives +6FRwt and the support receives +2FRwt, the sum is not at 0 so I suppose the small pulley don't receive 6FRwt but 4FRwt, no ?
Let's look at all the torques.
Suppose the motor exerts torque T. I'll not bother with signs, just magnitudes.
Motor:large pulley = T
Axle+belt:large pulley = F7*3R
Motor : support = T
Two axles : support = F6*2R
Support : small pulley = brake torque (B)
Belt+axle : small pulley = F7*R.
What equations do those give you?
 
  • #44
haruspex said:
Support : small pulley = brake torque (B)
Like I drawn my forces in my last image, I supposed the small pulley is not brake, the small pulley has a mass and it is accelerate more and more.
haruspex said:
Motor:large pulley = T
Axle+belt:large pulley = F7*3R
Motor : support = T
Two axles : support = F6*2R
Belt+axle : small pulley = F7*R.

I'm agree with that.

haruspex said:
What equations do those give you?
Sorry, I don't understand your question.
 
  • #45
V711 said:
Like I drawn my forces in my last image, I supposed the small pulley is not brake, the small pulley has a mass and it is accelerate more and more.
Ah, yes. But that means there's an unbalanced torque on the support, so that will accelerate the other way. And nothing will make sense unless you assign a moment of inertia to it.
 
  • #46
haruspex said:
Ah, yes. But that means there's an unbalanced torque on the support, so that will accelerate the other way. And nothing will make sense unless you assign a moment of inertia to it.

Ok, I understood. So with a brake on the small pulley, the support don't receive a torque, correct ?

With a brake, the friction gives the energy 5FRwt (at a time) ?
 
  • #47
V711 said:
Ok, I understood. So with a brake on the small pulley, the support don't receive a torque, correct ?

With a brake, the friction gives the energy 5FRwt (at a time) ?
It depends where the brake is mounted. I would imagine it would be mounted on the support, so the torque of the brake would be transferred to the suppor. If the pulleys are not accelerating then there should be no net torque on the support. Alternatively, the brake may be mounted on something outside entirely.
 
  • #48
haruspex said:
I would imagine it would be mounted on the support
yes, and like the angular velocity of the small pulley is 6w, on the support that angular velocity is 6w-w=5w, correct ? So I thought the work from friction is 5FRwt.
 
  • #49
If I resume:

The support turns at +w (clockwise)
The stator of the motor is fixed on the support so it turns at +w
The rotor turns at +w relatively to the stator, so it turns at +2w in the lab ref
The small pulley turns at +6w (lab ref) so it turns at 5w in the support ref, the small pulley is braking on the support at the radius R
The motor consume the power -3FRw
The brake (small pulley) gives the power +6FRw
So if the support don't receive any torque, this could say the power from brake is lower, could you explain please ?
 

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