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A disk accelerates

  1. Nov 13, 2004 #1
    A .22kg , .22-m-diameter plastic disk is spun on an axle through its center by an electric motor. What torque must the motor supply to take the disk from 0 to 1500 rpm in 5.00 ?

    I used the equations
    T(thau)=alpha/I
    w_f=w_i + alpha*(t_f - t_i)
    I found I to be equal to .021kgm^2
    then found w_f for the first .5s to be:
    (1500rev/1min) * (1min/60s) *(2pi/1rev) = 157.08rad/s
    157.08=0+alpha*(.5s)
    alpha = 31.42rad/s^2
    T= 31.42/.021 = 1496.2Nm

    Am not totally sure of my calculations. I will greatly appreciate any help
     
  2. jcsd
  3. Nov 14, 2004 #2
    I haven't done this in a very long time, but isn't [tex]\tau=I\alpha[/tex]

    And [tex]I=mr^2[/tex]

    Also, is the time 5 seconds? or 0.5 seconds?

    P.S.

    You might want to convert your answer into foot pounds and think about the problem intuitively. Is it really going to take over 1000 ft/lbs of torque to spin a little lightweight plastic disk to 1500 rpm?
     
    Last edited: Nov 14, 2004
  4. Nov 14, 2004 #3

    Yes, you are right [tex]\tau=I\alpha[/tex]
    However, i believe [tex]I=(1/2)mr^2[/tex]
    Also, it is 5seconds not .5s. I made a mistake
    Am supposed to give my answer in Nm
     
    Last edited: Nov 14, 2004
  5. Nov 14, 2004 #4
    You're right about the moment of inertia for a solid disk.

    It would be [tex]I=(1/2)mr^2[/tex] for a solid disk.

    I was thinking of a pendulum because I work with clocks a lot. :biggrin:

    I didn't mean for you to turn your answer in in units of ft-lbs. I simply meant to convert it to ft-lbs to think about it intuitively to check whether or not the answer makes sense. Although that's because I identify with ft-lbs easier. Maybe Nm work better for you intuitively.

    I just thought that your answer is an aweful lot of torque to spin such a small lightweight disk to 1500 rpm (even in a half a second). But if you have 5 seconds to get it spinning 1500 rpm it shouldn't take much torque at all. I would expect a very low amount of torque to be required. That's just my intuitive guess.
     
  6. Dec 11, 2009 #5
    Heres the proper way to compute this answer:

    Torque = Moment of Inertia * Angular Acceleration.
    Moment of Inertia = 1/2 * MR^2. (For a disk)
    Angular Acceleration = change in omega / change in time.

    *Make sure you understand this. If not, please reply or PM me.

    Physics is Phun

    **Sorry about posting in an older thread, but as many students find, mastering physics repeats its questions so im sure we will have students searching through the forums.
     
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