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Spinning disk torque?

  1. Apr 7, 2009 #1
    spinning disk torque???

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

    A uniform solid disk with radius 10 cm has mass 0.7 kg (moment of inertia I = ½MR2). A constant force 12 N is applied as shown. At the instant shown, the angular velocity of the disk is 40 radians/s in the -z direction (where +x is to the right, +y is up, and +z is out of the page, toward you). The length of the string d is 15 cm.


    At this instant, what are the magnitude and direction of the angular momentum about the center of the disk?


    What are the magnitude and direction of the torque on the disk, about the center of mass of the disk?


    3. The attempt at a solution

    i know both of these have a -z direction. for finding the angular momentum i used
    (MR^2/2)W (.7*.1^2/2) * (2(pi)/40) = 5.498e-4
    then to find torque i used RF where F is tension force got this by (M*F) = (.7*12) = 8.4
    then R*(8.4) = .1*8.4 = .84
    these are the wrong answers.
    what did i do wrong??
     
  2. jcsd
  3. Apr 7, 2009 #2
    Re: spinning disk torque???

    Do NOT multiply by the mass (.5kg). So, in short, multiply your final answer by 2 to get the correct one.
     
  4. Apr 7, 2009 #3
    Re: spinning disk torque???


    where did you get the mass of .5kg?? the mass in the problem is .7kg
     
  5. Apr 7, 2009 #4
  6. Apr 7, 2009 #5
    Re: spinning disk torque???

    i did that get the same answer as before. says im wrong

    here is what i did

    L(rot) = Iw
    I = 1/2MR^2 so 1/2*.7*.1^2 = .0035
    w = 2(pi)/T where T = 40 so w = .1570796
    Iw = 5.497787e-4
    do you see anything wrong with this??
     
  7. Apr 7, 2009 #6
    Re: spinning disk torque???

    alrite for part a) L(rot) = Iw
    w=40rad/s (given in problem)
    I=.5*.7*.1^2=.0035
    L=.0035*40=.14
     
  8. Apr 7, 2009 #7
    Re: spinning disk torque???

    OH.. ok
    what i was doing was taking the 40rad/s
    and using w = 2(pi)/T
    so 2(pi)/40
    then solving for that.
     
  9. Apr 7, 2009 #8
    Re: spinning disk torque???

    part b)

    torque=abs(R)*abs(F)*sin(90)
    =R*F*1=.1*12=1.2

    should work
     
  10. Apr 7, 2009 #9
    Re: spinning disk torque???

    Common mistake
     
  11. Apr 7, 2009 #10
    Re: spinning disk torque???

    for part d.) isn't it Lf-Li where Li is part a.) and Lf is I * omega with the given time?
     
  12. Apr 7, 2009 #11
    Re: spinning disk torque???

    for part d) you multiply torque calculated in part c) by the change in time
     
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