Stopping a disk with a disk

[pendodecahedron]

Not sure whether this is the right forum but here goes.

I have two disks of the same material and same radius. One is rotating and accelerating the other is stationary.

The stationary one is spring loaded which when released is pushed into the spinning disk to stop it spinning.

Is there some way to analytically derive the linear force needed to be applied to the stationary disk to stop the spinning disk ?

Thanks

 

[Skogsgurra]

Sounds like a very typical exam or school question. Is it?

Gunnar Englund

http://www.gke.org

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[Skogsgurra]

What happened to my answer/question? Is this a school thing?

Otherwise, it is a simple matter of applying law of friction to infinite radii, find the friction force for each element, multiply by r and integrate over r0 to r1, where r0 is inner radius and r1 is outer radius.

But I really do not see any practical or engineering need for such an excersise. So, I ask: What are you going to use it for?

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[pendodecahedron]

Hello shogsgurra

It is one of three different prototype choices for a safety mechanism on a rehabilitation robot. The others being a cenrifugal clutch and a disk caliper set up. Unfortunatly space and wieght are issues so although the disk on disk set up is not conventional it is looking like the most light weight. Well its looking like the most light wieght that all depends on the kind of linear force actuation that is needed

I have managed to experimentally collect acceleration data, but am getting a headache trying to equate the linear force to the generated torque.

I know that the torque is equal to the angular acceleration mulitplied by the angular moment of inertia and that the stopping force is related to the linear spring force time the kinetic coeffiecant of friction. But for some reason I cant get my head around linking the two together!

Thanks for your time,

Nic

 

[Skogsgurra]

Ok, I see. Couldn't imagine that it was for such a purpose. I am sitting here waiting for a prototype transformer. So I have some time left for this. See picture below.

Use SI units to get answer in newtonmeters.

Example: R0=0.05 m R1=0.15 m P=10 N k=0.4

T=.4x100x6.28x(0.003375-0.000125)/3=0.27 Nm.

You seem to need much more spring force than my 100 N to get any usable torque.



Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Skogsgurra]

Sorry, read P=100 N above.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Skogsgurra]

Hey! There is something wrong here!

I get the unit Nm^3 for torque. That cannot be right. Should be Nm.

Hmm...

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Skogsgurra]

OK. Got it.

P is total force from spring. You must divide by total area before using it in the formula.

So, your torque should be 0.27/0.0628=4.3 Nm. Which sounds better.

Better check this before using it in a design.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Skogsgurra]

Still haven't got that transformer. Did a test with a cardboard model in the meantime. Seems to be right if P is N/m^2.

So, if you input P=(spring force)/(disk area) in the formula in my first post, you will get torque in Nm.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[pendodecahedron]

Thankyou very much.

So P is the pressure, yes that makes sense.

Thanks again was very good of you to help me.

Nic

 

[Skogsgurra]

You are welcome. I had fun doing this.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Tmoose]

If you want the disks to make full contact they may have to be very thick. That's why clutches and disk brakes squeeze their disks from both sides at once.

 

[Skogsgurra]

Yes, that is how I tested it. Two flat surfaces pressing the disk from both sides. Stacking several (thin) disks between the corresponding number of stationary disks makes torque add up. Five disks will give you ten times more torque than one rotating disk being pressed against a stationary one.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...