# The rotationnal velocity on a support

• Cri85

## Homework Statement

A support is turning clockwise at Ω around the red axis. A motor on the support is turning at ω, ω is a frame support reference rotationnal velocity. Find the rotationnal velocity of the wheel ω' in the frame support reference.

http://imageshack.com/a/img673/7631/wZhklo.png [Broken]

## The Attempt at a Solution

For me, the rotationnal velocity of the wheel relative to the support is ω'=ω*R1/R2, is it correct ? ω' don't depend of Ω.

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## Answers and Replies

I think you have it right. (if R1 = R2 one expects the omegas to be equal, right ?)

Oh, and it's rotational (or better: angular)

Cri85
I don't understand, could you explain please ?

I don't understand, could you explain please ?
BvU is correcting your spelling of rotational (not rotationnal) and suggesting that 'angular velocity' is the more usual term.

Ok, sorry. But I don't understand his message my calculation is correct only if R1=R2 ?

Ok, sorry. But I don't understand his message my calculation is correct only if R1=R2 ?
No, he is saying your answer is correct. The R1=R2 case is just a sanity check. He is saying that if R1=R2 then you would expect ω'=ω, and that agrees with your answer.
(However, it's not a very strong check since you might have made the error of getting R1 and R2 crossed over. BvU's check doesn't catch that.)

Cri85
Ok, thanks :)

I asked myself about the consequence of the mass of the belt. The Pulley2 is braking on the support. All angular velocities are constant. The sum of torque on the support is 0 if the mass of the belt is 0. But if I take in account the mass of the belt, like radius of pulleys are differents and like Ω is different of 0, the centrifugal force seems to be different.
http://imageshack.com/a/img913/5508/sQ3uiA.png [Broken]

The bigger pulley can be like that:

http://imageshack.com/a/img540/6098/AI8pLJ.png [Broken]

The small pulley is far away the center, so the centrifugal force on the belt depends of Ω and of the position in the support ?

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Yes. But is this still the same exercise ? Because as long as the omegas are constant it has nothing to do with ω'=ω*R1/R2 .

Don't make this too complicated... unless necessary

Where do the brakes come from ? What's the full story ?

Yes. But is this still the same exercise ?
Sorry, no, it's only a question I asked to myself. Maybe I need to ask another question ?

Forget the brake, just the motor, a belt and a disk.

If Fc1 is not equal to Fc2, this could say there is a torque on the support and the support can gives an energy, so it is more difficult for the motor to turn the disk ?

The ΔFc would cause torques on both wheels. With opposite signs, so more or less cancelling (by the belt itself). I don't think the motor would 'notice'.

The support can't receive a torque from the sum of centrifugal forces of the belt ?

Centrifugal forces tend to have a line of action that passes through the axis of rotation, so I would say: no.

(Didn't realize you referred to the central support, the ##\Omega## axis)

Like there is 2 axes of rotation, Ω and the center of each pulley, the velocity of a small part of the belt come from the rotation around Ω and around the center of each pulley, I thought there was a torque on the support. Even the part of the belt in the "line" has 2 velocities.

You've lost me. On several occasions:
2 axes (I see 3 axes)
velocity of a small part of the belt
the part of the belt in the "line" has 2 velocities.

Sorry.

2 axes (I see 3 axes)
yes, there are 3 axes, I spoke about a particular point of the belt that is turning around a pulley and around the Ω axis .

velocity of a small part of the belt
I think for calulate the torque on the support from the centrifugal forces from the belt I need to integrate all the belt ?

the part of the belt in the "line" has 2 velocities.
If I take a point of the belt that is turning around the Pulley1, this point rotates around the pulley but like all the device is turning around Ω axis the velocity is composed of 2 terms, one from the rotation of the pulley: Vp, and other from the rotation around Ω: VΩ. For each point the velocity is composed of 2 terms, no ? I drawn an image:

http://imageshack.com/a/img661/3523/LSAYFE.png [Broken]
The sum of the velocity is Vs but for calculate the centrifugal force I need to take in account the trajectory. Maybe this sort of exercice has been study somewhere ? Or there is a method ?

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You could check out rotating reference frames (and find there's not only a ficititous centrifugal force, but also a coriolis force, dependent on velocity).
But doing such a transformation twice makes life pretty difficult.

Thanks for the link. I will try to calculate the sum of centrifugal force in my last case.

It's easier to draw centrifugal forces in this case :

http://imageshack.com/a/img661/1657/bM8gZu.png [Broken]

The Pulley1 is in the center of rotation, it rotates the belt, it is a crossed belt. I added a Pulley3 and a static belt on it for have the same mass than the Pulley2+belt.

With Ω = 0, the sum of centrifugal forces cancel themselves. But with Ω different of 0, the support seems to have a net force to the left. So why I can't draw my forces like that: red forces are centrifugal forces on the belt on the Pulley2 ?

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Sorry.

yes, there are 3 axes, I spoke about a particular point of the belt that is turning around a pulley and around the Ω axis .

I think for calulate the torque on the support from the centrifugal forces from the belt I need to integrate all the belt ?

If I take a point of the belt that is turning around the Pulley1, this point rotates around the pulley but like all the device is turning around Ω axis the velocity is composed of 2 terms, one from the rotation of the pulley: Vp, and other from the rotation around Ω: VΩ. For each point the velocity is composed of 2 terms, no ? I drawn an image:

http://imageshack.com/a/img661/3523/LSAYFE.png [Broken]
The sum of the velocity is Vs but for calculate the centrifugal force I need to take in account the trajectory. Maybe this sort of exercice has been study somewhere ? Or there is a method ?
The acceleration of the belt provided by each wheel is along the line of centres joining the wheels. Moreover, they are equal and opposite. Hence, no net torque.

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