# Angular velocity, torque and work

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1. Jan 23, 2015

### V711

1. The problem statement, all variables and given/known data

Two supports Support1 and Support2 turn at w clockwise around a fixed axis. On the Support1 there is the Pulley1 with a radius 3R, the Pulley1 can turn around an axis fixed on the Support1. On the Support2 there is the Pulley2 with a radius R, the Pulley2 can turn around an axis fixed on the Support2. Pulleys are linking with a belt.

http://imageshack.com/a/img538/2148/WR9V6G.png [Broken]

The Pulley1 turn clockwise at 2w (in lab reference). The Pulley2 turn clockwise at 4w (lab reference). The Pulley1 rotates with a motor fixed on the Support1. The Pulley2 is a brake, it gives the force F on the belt. It is supposed that the angular velocity of the Support1 and the Support2 is always w (external device).

1/ Calculate the work of the brake (energy from friction)
2/ Calculate the torque on the Support2 from the brake, and the work
3/ Give the sum of torque to the Support1 and to the Support2 from the axis of pulleys
4/ Calculate the work of the motor
5/ Calculate the torque on the Support1 from the motor and the work on it

2. Relevant equations

--

3. The attempt at a solution

http://imageshack.com/a/img661/7092/LlOgR2.png [Broken]

1/ The work of the brake is force by distance. The Pulley2 brake on the Support2, the angular velocity of the Pulley2 on the Support2 is $4w-w = 3w$ the energy from friction is $+3FRwt$.

2/ The support receives a clockwise torque from the brake: $FR$, the work is $+FRwt$

3/ The Support1 receives a torque and the Support2 receives the same torque but with a negative value, so the sum of torque is 0, I drawn 3 images for show that:

http://imageshack.com/a/img911/8376/QIgJGA.png [Broken]

http://imageshack.com/a/img673/7281/tzukYR.png [Broken]

http://imageshack.com/a/img540/8715/XVKCky.png [Broken]

I think d1=d2, always, so the sum of torque is 0, because the force F is gave to the Support1 and -F is gave to the Support2 and the position of the Pulley1 is the same the the Pulley2 on their support.

4/ The motor turns at $w$ on the support, so the motor need to give the energy $-3FRwt$, because the radius is $3R$.

5/ The Support1 receives a torque $3FR$, the work on the Support1 is $-3FRwt$ because the Support1 turns at $w$.

Is it correct ?

Last edited by a moderator: May 7, 2017
2. Jan 23, 2015

### haruspex

I'm not sure i understand the set-up. it doesn't say so, but it seems the belt must be slipping on one or other pulley.
Yes.
But you haven't any distances in that calculation. Think about the rate at which the belt must be slipping.

3. Jan 23, 2015

### V711

The brake is not drawn but it is at the distance R on the Pulley2. I calculated the work with length = radius x angle. Angle = 3wt because the angular velocity of the Pulley2 is 3w on the Support2. It is not correct ?

Support1 and Support2 are at +w, the Pulley1 is at +2w and the Pulley2 is at +4w in the lab reference. In the support reference, the Pulley1 is at w and the Pulley2 is at 3w, like the radius is 3R for the Pulley1 and R for the Pulley2 the belt turns, no ? For you, there is a sliding could you explain ?

Each pulley is on a support, each support turns at +w and each pulley turns around itself.

Last edited: Jan 23, 2015
4. Jan 23, 2015

### CWatters

My first impression was that the "supports" are irrelevant. They both turn at the same angular velocity and both pulleys are free to rotate on their respective axis. When you work out the power of a car do you take into account that the earth is rotating?

Caution: I'm in a rush to go out so I might have missed something.

5. Jan 23, 2015

### haruspex

I agree with CWatters, that argument doesn't work. The pulley centres must remain at the same orientation with respect to each other in the lab reference. If there were no motor driving them, the two pulleys would each rotate at -w relative to the supports. The motion of the supports has no relevance for the movement of the belt. If the the pulleys are rotating at +2w and +4w respectively in the lab reference then the belt must slip. It wants to go past one pulley at 6Rw and past the other at 4Rw.

6. Jan 23, 2015

### V711

But like I said in my first post there is a motor on between the Pullet1 and the Support1: "The Pulley1 rotates with a motor fixed on the Support1".

The Pulley1 is forced to turn at 2w, and the Pulley1 drive the Pulley2 with the belt.

But each pulley is on a support and each support turn at +w. I don't understand why the angular velocity in the support frame reference of the Pulley1 is not 2w-w and not 4w-w for the Pulley2, could you explain ?

I don't understand why the belt don't "follow" the pulleys and pulleys are on rotating support. Could you explain ?

7. Jan 23, 2015

### V711

I understood it's ok, thanks

8. Jan 24, 2015

### V711

I asked myself about the torque on the support. With one support (not 2), a big pulley drive a small pulley with a belt like I drawn before but with only one support. If the angular velocity of the big pulley is -2w (clockwise), the angular velocity of the small pulley must be at -6w (clockwise) if I don't want sliding, is it true ?

If the support turn at -w (clockwise). A motor on the support turns the big pulley (no mass in theory) more and more (for accelerate), the small pulley with a mass will accelerate too, the energy is transform in the kinetic energy of the small pulley. The support receive the torque 2FR ? I'm not sure because:

The motor gives the forces F1 and F2 to the big pulley, need the energy -6FRwt at start
The motor gives the forces F3 and F4 to the support, need the energy -6FRwt at start if I want the angular velocity of the support constant
The belt receives the force F5, that force is reported to the axis of the big pulley: F6

The small pulley receives the torque F7/F9, this gives the energy 12FRwt at start
The support receives the torque F6/F8, this gives the energy 2FRwt at start

The sum is not at 0 for the energy where I'm wrong ? There is no torque on the support ?

http://imageshack.com/a/img537/5685/9CpiKN.png [Broken]

Last edited by a moderator: May 7, 2017
9. Jan 24, 2015

### haruspex

I'm not sure what F is in your energy calculations. Is it the same as a particular Fn in the diagram or is it varying according to context?
Here, I'll use F to mean F5 (=F7).
In time t, the large pulley rotates 2wt against a torque of 3RF, so does work 6FRwt.
In time t, the small pulley is rotated 6wt against a torque of RF, so has work 6FRwt done on it.

10. Jan 24, 2015

### V711

I drawn forces at the same value but I was not sure. Maybe I need to draw forces like that :

http://imageshack.com/a/img540/8779/7XX9F1.png [Broken]

Ok

In this case, there is no torque on the support ? Why forces F6 and F8 (or F4/F8 in my last image) don't exist ? The small pulley is not brake, it is accelerate more and more.

Last edited by a moderator: May 7, 2017
11. Jan 24, 2015

### haruspex

You described F5 as the force large pulley exerts on belt, but drew it the other way. I'll assume it's the force the belt exerts on the pulley.
So on large pulley you have three forces, F1, F2, F5. If F1, F2 are equal and opposite then F5=0.
I was focussing on the work large pulley does on small pulley via the belt. The torque on the support is not needed for that.

Last edited: Jan 24, 2015
12. Jan 24, 2015

### V711

You're right, sorry I changed (edited) my message because I don't knew if my forces were good or not.

But if there is a torque, this torque work (positive or negative value), my forces on the support are false ?

I don't understand why, could you explain ?

13. Jan 24, 2015

### haruspex

There is a torque there, but it's nothing to do with the work large pulley performs on small pulley via belt.
"Torque on the support" must be from the motor, right? But the side of the motor attached to the support would not move relative to the support, so no work done by that torque. If you mean torque between motor and pulley, yes that does work, but it will be the same as the work the pulley does on the belt.
Yes. Since the belt exerts a net linear attraction between the pulleys, there must be counteracting net linear forces on the pulleys from motor/support. (Whether it's via the motor or directly from the support on large pulley depends on the details of the construction.)

14. Jan 24, 2015

### V711

no, this torque I see it: forces F3 and F4, but with my forces F6/F8 (first image or F4/F8 in the second) I see a torque on the support.

I don't understand why F5=0, could you explain ? F7=0 too ?

http://imageshack.com/a/img913/5298/kJOjH8.png [Broken]

Last edited by a moderator: May 7, 2017
15. Jan 24, 2015

### haruspex

No F5 is not zero. I wrote that if the only forces on the large pulley are F1, F2 and F5, and F1 is equal and opposite to F2, then you'd have to conclude F5=0. Therefore F1 and F2 are not equal and opposite (or there is another force).
I feel that discussing all these different forces on different objects at once is leading to confusion. Please pick one object at a time and list all the forces you believe are acting on it.

16. Jan 25, 2015

### V711

In a general case, why I can't give a torque to a pulley (F1=F2 in value) and have F5 from the belt ?

For me when the motor accelerates more and more (like pulleys):

http://imageshack.com/a/img538/9529/fs36q1.png [Broken]

or forces can be like that ?

http://imageshack.com/a/img537/9495/C8APQU.png [Broken]

The big pulley receives from the motor the forces F1 and F2, equals in value
The green support receives from the motor the forces F3 and F4, equals in value
The big pulley want to pull the belt, the belt gives the force F5 to the big pulley, F5 is reported to the axis : F6 (the support receives F6). The axis gives the forces -F6 to the big pulley, I forgot this force I think ?
The small pulley receives the force F7 from the belt
The axis of the small pulley reveices the force F7 (I noted F8) and gives the force F9 to the pulley, the small pulley has a torque on it, it is accelerated

For me, all these forces are not at 0. For me, if F5=0 => F7=0, the small pulley don't accelerate (or has no mass). With a mass F7 is different of 0 no ?

What's wrong with these forces ?

Last edited by a moderator: May 7, 2017
17. Jan 26, 2015

### haruspex

If those are the only forces acting on the pulley then there is a net linear force. The pulley will accelerate linearly.
Exactly! Or, instead of introducing F6, F2+F5+F1=0, replacing F1=-F2.
Not sure that it matters, but there may be some inconsistency in the way you are think of the physical connection between support, motor and pulley. You can think of the pulleys as mounted on axles directly connected to the support, with the motors only providing torque; or you can think of the pulleys as mounted on the motors and the motors mounted on the support. E.g. you say F2 comes from the motor but -F6 from the support.

So, where are we up to? Is there still a problem?

18. Jan 26, 2015

### V711

You're right it's so logical ! My 2 last images are correct or is it only one ?

In my last message, with the first image, the support receives a torque from forces F6 and F8 ?

Last edited: Jan 26, 2015
19. Jan 26, 2015

### haruspex

Before I answer that, please definehow you want to treat the physical relationship between support, motors and pulleys. Are the pulley axles directly attached to the support, with the motors only supplying torques? If so, it might be better to introduce torque variables for those instead of pair s of forces in arbitrary positions. Or do you want to treat the pulleys as only attached to the motors, and the motors attached to the support?

20. Jan 27, 2015

### V711

There is only one motor, this motor (stator) is fixed to the support. The big pulley is attached to the other part of the motor (rotor). The small pulley has no motor, it is driven by the belt. The axis of the small pulley is fixed to the support. The motor accelerates more and more and all pulleys accelerate more and more, the small pulley has a mass. I would like to undertand how draw all forces and if my image with forces is correct or not.

There is no friction in the axes.