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Homework Help: Slowing A Spinning Disc To A Stop

  1. Dec 13, 2009 #1
    1. The problem statement, all variables and given/known data
    I need a quick sanity check here please. (Sorry, my alphas keep coming out as power signs, they're not!).

    To find the force to slow a spinning disc with known moment of inertia (0.0004kgm2) and known angular velocity (120pi rad/s) to a stop in 6s.

    Here, the deceleration ([tex]\alpha[/tex]) is 120pi/6=20 rad/s2. The radius of the disc is 6cm=0.06m


    2. Relevant equations

    Torque = rf (r=radius, f=force)

    Also, torque=I[tex]\alpha[/tex]

    And [tex]\alpha[/tex]=w/t where w=angular velocity (120pi) and t= the time to slow=6s.

    This gives [tex]\alpha[/tex] to be 20pi rad/s2.


    3. The attempt at a solution

    Setting rf=I[tex]\alpha[/tex]

    0.06f=0.0004 x 20pi

    From this, the force is given to be 0.42N.

    Could someone just check if this is right. I think I've done it ok, but its been some time. It kind of looks right to me, but part of me is thinking I've missed something. It seems too easy. Any help greatly appreciated.
     
  2. jcsd
  3. Dec 13, 2009 #2

    Doc Al

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    Staff: Mentor

    Looks good to me. (Assuming the force is applied tangentially.)

    To get your latex to align with the rest of a sentence, use the inline tag (itex) instead of the regular tag (tex): [itex]\alpha[/itex] versus [tex]\alpha[/tex].
     
  4. Dec 13, 2009 #3
    Opps. I forgot to add the lump of 5g of putty 4cm from the centre of the disc. This makes the total moment of inertia (0.004+0.005x0.042.

    Therefore 0.06f=0.000408x20pi

    so f=0.427N

    But, thank you if you think that is right :wink:
     
  5. Dec 13, 2009 #4

    Doc Al

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    Staff: Mentor

    How did the lump of clay get there? If the clay and disk start out rotating together at the given speed, then your answer is fine.
     
  6. Dec 13, 2009 #5
    Yes, the clay was added as an earlier part of the question, hence I originally only had the moment of inertia of the disc (0.0004) and then changed it for the moment of inertia of both the disc and clay (0.004 + 0.005 x 0.0016 where the distance of the clay from the centre is 4cm, and its mass 5g. As it is a single particle on the disc, its own moment of inertia I worked using mr2, unlike 1/2 mr2 for the disc as the disc is a solid rotating object whilst the particle inscribes a hollow track).
     
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