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Circular motion of an air puck

  1. Dec 18, 2008 #1
    1. The problem statement, all variables and given/known data
    An air puck of mass 0.029 kg is tied to a string and allowed to revolve in a circle of radius 1.6m on a frictionless horizontal surface. The other end of the string passes through a hole
    in the center of the surface, and a mass of 1.4kg is tied to it, as shown in the figure. The
    suspended mass remains in equilibrium while the puck revolves on the surface. The acceleration of gravity is 9.81 m/s2 .
    What is the linear speed of the puck? answer in m/s.
    I got what the forces were- i drew a freebody diagram and had tension and gravity pointing down, with centrifigal acc pointing down as well.
    I got tension to be 13.734N and gravity to be .28449N.
    I plugged them into Fnet= ma
    so T+Fg= m(centrif acc)
    thing is, I'm not sure what to use for mass..is it the .029, the combined mass of the two or the mass of the weight on the bottom?
     
  2. jcsd
  3. Dec 18, 2008 #2
    Centripetal acceleration=v^2/r
    T=(1.4 kg block)*g
    T=(mass of small block)(centripetal acceleration) or T=(.029kg)v^2/r

    Therefore, (1.4kg)*g=(.029Kg)v^2/r

    I think you can take it from there.

    With respect to deciding which masses go where: Think about which object*acceleration is supplying the force. The lower object is not spinning and is therefore not contributing to the "spinning force"
     
  4. Dec 18, 2008 #3
    Also, I just noticed: If the object is falling near the surface of earth the acceleration due to gravity is 9.81m/s^2. Or maybe I didn't understand part of your question.
     
  5. Dec 18, 2008 #4
    Actually the more I look at this,

    There are two free body diagrams to draw. The first one is of the block rotating around the table. The second is of the block underneath the table.

    From that you will get two equations. Put them together as I did and the rest is algebra.
     
  6. Dec 18, 2008 #5
    hmm i set them equal to each other and i got velocity to be 31.27982714m/s. However, when I submit that, it says I'm wrong...
     
  7. Dec 18, 2008 #6
    That's not what I got for velocity.
    (1.4kg)(9.81m/s^2)=(.029Kg)(v^2)/(1.6m)
     
  8. Dec 18, 2008 #7
    ah faulty algebra on my part... Thanks for your help :)
     
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