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Equation of a rotating disk

  1. Apr 24, 2007 #1
    Ive been trying to find formulas that help describe the torque needed to rotate a flat disk on a flat surface when the coefficient of friction between the 2 materials is known.

    The first surface is a glass window (so a vertical plane) the second surface is a rotating disk of wet sponge "cleaing it" (also a vertical plane).

    So im looking for a formula that takes into consideration friction between the surfaces, speed of rotation, radius (or area) of the contact surface and torque required to start the disk spinning.

    Any suggestions?
  2. jcsd
  3. Apr 24, 2007 #2


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    Homework Helper

    If these materials are "normal", then speed of rotation will probably not affect the torque needed to keep it rotating. You're going to need some calculus to solve this problem.

    You basically have a disc that can be divided into infinitesimal rings of area dA proportional to a mass dm. Each ring contributes some frictional force dF applied at a distance r. Integrating over the actual radius of the disc, you will get the total torque required.
  4. Apr 30, 2007 #3
    Thanks for the response but my maths is extremely rusty. I understand the general concept of your response but what would the exact formula look like when complete so that i could use it?
  5. Apr 30, 2007 #4


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    You can make this as easy or complicated as you like. I personally start as easy as possible and then introduce the complications based on how accurate I need to be.

    In this case, why not estimate your frictional force and use the relation

    [tex]\Sigma T = I \alpha[/tex] ?

  6. Apr 30, 2007 #5
    Is that not moment of inertia?

    The idea is to have an equation that works out the torque required to rotated the vertical disk. It must include the coeficient of friction between the surfaces of the disk/window as one of the variables. I assume the torque needed to overcome the friction between the surfaces would be more than the moment of inertia.

    Maybe the total torque would be moment of interia + the frictional force it has to overcome from the 2 materials rubbing?
  7. Apr 30, 2007 #6
    The final formula is:
    [tex]\tau= {2\over 3}F\mu R [/tex]

    [tex] \tau [/tex] is the torque.
    [tex] F [/tex] total force exerted against the glass.
    [tex]\mu [/tex] the coefficient of friction.
    [tex]R [/tex] the radius of the disk.
    As said, it independent of the speed of rotation.
  8. Apr 30, 2007 #7
    Thank you very much thats perfect.

    Could you please take a couple of minutes to explain how you got that formula though please.
  9. Apr 30, 2007 #8
    I gave you the solution because I thought that you are not a physics student but that you need this formula for your job. If this is the case, you do not need the derivation.
    If you are a physics student then do as mezarashi told you (this is, of course, what I did).
  10. Aug 9, 2010 #9
    Sorry for the necro.

    Was working on a similar problem today and devised this integral. After forgetting how to do integrals, I thought my answer was wrong, and then I came here. Seeing the final answer reminded me how to do calculus.

    Here is a picture trying to show some of the math behind it.


    Edit: and sorry for the same step being in there twice.
    Last edited: Aug 9, 2010
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