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Textbook velocity problem

  1. Mar 12, 2006 #1
    Help please!!

    I found this question in my textbook but the solution is not stated. A force of 2 N vertically down is applied to a vertical disc at a distance 2 cm to the right (horizontally) of the centre of the disc. The disc has a mass of 1.2 kg and a radius of 15 cm. Its initial angular velocity is 2 rad.s-1 anticlockwise.

    How long does it take to come to a (momentary) stop?

    I took a crack at it but is it correct?.
    I=(mr^2)/2, I=[1.2x(15/100)^2]2. Then the value for I is substituted to F=Ia, where 2=value for Ixa, where a is found, and then substitute for a=w/t. So a is substituted=2/t and t is found.

    Is this o.k?
     
  2. jcsd
  3. Mar 12, 2006 #2
    The equation must be [itex]M=Fd=I\ddot{\theta}[/itex]
     
  4. Mar 12, 2006 #3
    The M represents what exactly? and isn't the momentum for a disk about it's center I = ½ mr2?
     
  5. Mar 12, 2006 #4

    Hootenanny

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    I believe the M is turning moment or torque. Your F = Ia equation needs modifing, instead of using F, you must use torque:
    [tex]\tau = I\alpha[/tex]
    This is a version of Newton's second law.
     
  6. Mar 12, 2006 #5
    I see, but is the working correct?
     
  7. Mar 12, 2006 #6
    The first part is allright. After finding a(angular acceleration) you need to
    use the eqn

    [tex] \omega_{f}=\omega_{i}+at [/tex]

    Now solve for t putting [tex]\omega_{f}=0[/tex]

    I recommend you go through the eqns of motion for angular variables.

    Edit:I didn't read ur sol properly.You also need to use [tex]\tau=Ia[/tex]
    and substitute [tex]\tau=rF[/tex]
     
    Last edited: Mar 12, 2006
  8. Mar 12, 2006 #7

    Hootenanny

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    You are quite right arundg, it would be alot simpler using equations of motion, I didn't stop to consider the question, I was simply looking for errors.
     
  9. Mar 12, 2006 #8
    I thank you all for your help. But could you tell me what the T in the equation T=rF stands for. Thanks again!
     
  10. Mar 12, 2006 #9

    Hootenanny

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    [itex]\tau = rF[/itex] is simply torque, force([itex]F[/itex]) multiplied by distance([itex]r[/itex]) from the axis of rotation.
     
    Last edited: Mar 12, 2006
  11. Mar 13, 2006 #10
    Edit:I didn't read ur sol properly.You also need to use [tex]\tau=Ia[/tex]
    and substitute [tex]\tau=rF[/tex][/QUOTE]

    I understand using the [tex]\tau=Ia[/tex] part where we can find [tex]\alpha[/tex](angular acceleration). But why do we need to substitue [tex]\tau=rF[/tex]?. Why can we not just find [tex]\alpha[/tex], and find t so it can be substituted into [tex] \omega_{f}=\omega_{i}+at [/tex]
     
    Last edited: Mar 13, 2006
  12. Mar 13, 2006 #11

    Hootenanny

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    You need to use [itex]\tau = rF[/itex] and put it equal to [itex]\tau = I\alpha[/itex] to find the angular accelertatiion thus;

    [tex]I\alpha = rF[/tex]

    [tex]\alpha = \frac{rF}{I}[/tex]

    Can you go from here?
     
  13. Mar 13, 2006 #12
    got it.thanks
     
  14. Mar 13, 2006 #13

    Hootenanny

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    No problem
     
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