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Help- speed

  1. Oct 25, 2007 #1

    klm

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    [SOLVED] help- speed

    Mass m_1 on the frictionless table of the figure is connected by a string through a hole in the table to a hanging mass m_2.

    With what speed must m_1 rotate in a circle of radius r if m_2 is to remain hanging at rest?

    [​IMG]

    i know i am suppose to show what i have done, but i really have no idea how to start this problem. any help would be great.

    i know that weight( Fg) and tension are working on M2
    and that Fg, tension, and normal force are working on M1.
     
    Last edited: Oct 26, 2007
  2. jcsd
  3. Oct 25, 2007 #2

    Kurdt

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    Well circular motion seems to be involved. What do you know about that?
     
  4. Oct 25, 2007 #3

    klm

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    umm.. F=ma, for circular motion F=m V^2/r
    and acc points toward the center
     
  5. Oct 25, 2007 #4

    Kurdt

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    So you know the force acting on m1 is the weight of m2. How fast should it rotate so the centripetal force balances the weight?
     
  6. Oct 25, 2007 #5

    klm

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    i dont know? i dont understand
     
  7. Oct 25, 2007 #6

    klm

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    i know that speed=distance/time ...but i dont know how to figure this out
     
  8. Oct 25, 2007 #7
    good.. now that u know this much.. think ahead. Hint: this centripetal force, that u have written, will act on which body.. and would be provided by which force? You are almost there, believe me, just get a pen and notebook.. and work it out! :)
     
  9. Oct 25, 2007 #8
    speed = dist/time .. will incorporate more unknowns (irrelevant for the present problem)!
    Well, u have already got a relation involving speed... u urself posted it above.. think on that equation.
     
  10. Oct 25, 2007 #9

    klm

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    well i think the centripetal force will act on M1 and the force is provided by M2, since it is kind of pulling it down. but i dont understand what i am suppose to do next. i dont know what formula to use..
     
  11. Oct 25, 2007 #10

    klm

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    F=m V^2/r ? so i know the force acting on it is M1..? so square root (Fx r) = v? i dont know?
     
  12. Oct 25, 2007 #11

    Kurdt

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    I practically told you what you have to do. I'll make it clearer. Equate the weight of m2 with the centripetal force.
     
    Last edited: Oct 25, 2007
  13. Oct 25, 2007 #12

    klm

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    i appreciate you trying to help me, but it is pretty obvious i am confused and dont understand what to do
     
  14. Oct 25, 2007 #13
    hmm.. here is a good news: all formula reqd. for this problem have already been quoted in the threads above, u dont need anymore! :)
    and ya, one more thing.. u r right that, it is because of m_2, that m_1 has a tendency to be dragged towards the hole.. but it can help (and this approach generally helps while solving dynamics problems).. if u think in this line: m_2 is being pulled upwards by the tension in the string (and not exactly m_1) and m_1 is being dragged towards the hole by tension in the string (and not exactly m_2)!
     
  15. Oct 25, 2007 #14

    Kurdt

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    You have m1 which is attached to m2 by some string. The weight of m2 (i.e. [itex] W=m_2g[/itex]) is acting on m1 via the string. What you're asked to do is find out how fast you need to rotate m1 so that its centripetal force cancels the weight of the first mass m2.

    Is that any clearer?
     
    Last edited: Oct 25, 2007
  16. Oct 25, 2007 #15
    basically, i m trying to break the whole system in two different (although, related through tension) sub-systems.
     
  17. Oct 25, 2007 #16

    klm

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    okay i think i am understanding. this may be a dumb question but i have to make sure, the centripetal force is acting on m1 correct?
     
  18. Oct 25, 2007 #17

    klm

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    saket- i understand the way you explained it about m2 being pulled up and m1 being pulled down. so does that mean the tension in both are the same then?
     
  19. Oct 25, 2007 #18

    klm

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    so for m2 , would this be correct or necessary : Fnety= T - W = m V^2/r , T-mg=mV^2/r
     
  20. Oct 25, 2007 #19
    centripetal force, by definition, is cocerned with circular motion, and not linear motion. [For linear motion, m.(v^2)/r, as r tends to infinity.. centripetal force tends to zero!]

    Yes, tension in both should be same -- that comes basically from frictionless table.
     
  21. Oct 25, 2007 #20
    i dont mean to interrupt but i need help 2 plz.
     
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