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Force Required to stop a Rotating Disc

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Essentially, I am trying to determine the force that must be applied to a rotating disc to that stop that disc from rotating in a certain time period.

    The disc is rotating at 5800 rpm, the disc has a radius of 5 inches, and a thickness of 0.087 inches. The disk is made of titanium (density of 4507 kg/m^3). The time required to stop the disk is 0.005 seconds.


    2. Relevant equations
    K = 1/2*I*ω^2 ; I = 1/2*m*r^2 ; -------> K = 1/2(1/2*m*r^2)*ω^2

    3. The attempt at a solution
    Knowing ω = (2∏) / T , I was able to calculate ω to be ω= 607.375 rad/sec

    Volume of the disc = V = ∏r^2h, which leads to V = 1.1197 E-4 m^3

    Using the density, I found the mass of the disk to be m = 0.50465 kg.

    Plugging all of this into equation above for kinetic energy, I found K to be K = 750.7 joules

    It's been a while since I took a dynamics course. How can I determine the force necessary (I guess I'm kind of assuming a friction-like force such as a break being applied) to stop the disc from spinning in 0.005 seconds? I know I'll need an integral, but I forget the exact formula and steps necessary to finish up this problem.

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 22, 2011 #2

    gneill

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    Staff: Mentor

    Hi rjm3db, welcome to Physics Forums.

    You've done most of the background calculations for the problem. Consider that in order to go from a speed ω to zero in a given time t that there must be an acceleration [itex]\alpha[/itex]. How might you find that?
     
  4. Nov 22, 2011 #3
    Lets see. So, the angular acceleration is equal to α = dω / dt

    So, α = (-607.375 rad / sec) / 0.005 sec

    → α = -121475 rad / sec^2

    Alright, so now I've calculated both the angular acceleration.

    Any other hints to getting me closer to finishing this one up?

    It's been a while since I took a dynamics course, but does it involve the equation a = (alpha x r) + (omega x omega x r) , and later on the equation ƩF = ma ?
     
  5. Nov 22, 2011 #4

    gneill

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    Staff: Mentor

    All you need now is the angular equivalent to F = MA to find the torque, then convert the torque to a force applied (presumably) to the rim of the disk.
     
  6. Nov 22, 2011 #5
    Great, so something along the lines of

    torque = I * α

    torque = r x F → Force = torque / r ?

    I really appreciate all of your help. Thanks a bunch
     
  7. Nov 22, 2011 #6

    gneill

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    Staff: Mentor

    Yup. Looks good.
    Glad to be of help.
     
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