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Impule & Momentum

  1. May 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Note Diagram.

    If they are released from rest, determine the time required for block A to attain a speed of 2 m/s. Neglect the mass of the ropes


    2. Relevant equations



    3. The attempt at a solution

    I = mk2 = (30)(.252) = 1.875 kg m2

    X Direction

    mv + [tex]\int[/tex] Fdt = mv2

    0 + Oxt = 0

    Y Direction

    mv + [tex]\int[/tex] Fdt = mv2

    0 + Oyt - 30(9.81)t = 0

    Oy = 294.3

    Moments

    Iw1 + [tex]\int[/tex]MOdt = Iw2

    0 + 25(.3) - 10(.18) = 1.875w2

    w = [tex]\sqrt{3.04}[/tex]

    How do i relate this to velocity and time. I can use v = wr, but how can i solve for t when v = 2 m/s

    2 = .3w, therefore i need to find t when w = 2/.3
     

    Attached Files:

  2. jcsd
  3. May 14, 2009 #2
    I'm having trouble understanding the image. Can you get a better one or explain the image further?
    Anyway, it seems you are making things much more complicated than they have to be.
    Draw a FBD, and apply Newton's second law in linear and angular forms to each mass. The rest should just be kinematics.
     
  4. May 15, 2009 #3
    The spool has two radiuses. one is .18m and one is .3m. they start from rest. when the blocks are released, block A weighs more so it will fall, thus lifting block B. When will block a reach a speed of 2 m/s.
     
  5. May 15, 2009 #4
    But ya i dont think i needed most of what i posted first. Heres what im at

    I = mk2 = 30(.25)2 = 1.875

    25(.3) - 10(.18) = 1.875w2

    w = [tex]\sqrt{3.04}[/tex]

    w = v/r

    [tex]\sqrt{3.04}[/tex](.3) = v

    how do i get a t in there
     
  6. May 15, 2009 #5
    I think that I would take a different approach. You know that

    [tex]\sum M =I\alpha[/tex]

    and there is a relationship between alpha and the acceleration of the blocks.

    So you can write the sum of the moments in terms of a

    Since [itex]a=\frac{dv}{dt}[/itex] you should be able to find t using the initial conditions.

    Edit: Was there a reason you thought that it was an Impulse & Momentum problem? I am only wondering in case I overlooked something.
     
  7. May 15, 2009 #6
    okok i think i got it

    I = 1.875

    Iw2 + + [tex]\sum\int[/tex]Mdt = Iw2

    0 + [tex]\int[/tex](25(.3) - 10(.18)dt = 1.875w2

    5.7t = 1.875w2 therefore w = [tex]\sqrt{3.04t}[/tex]

    w = v/r

    [tex]\sqrt{3.04t}[/tex](.3) = v = 2

    3.04t = 4

    t = 1.32 seconds

    Is this right
     
  8. May 15, 2009 #7
    I thought it was impulse and momentum because the chapter is called

    Planar Kinetics of a Rigid Body: Impulse and Momentum
    Section 19.2 Principle of Impulse and Momentum

    Im assuming there is more than one way to solve it, im just suppose to use impulse and momentum
     
  9. May 15, 2009 #8
    Edit: I can multiply :rofl: I think you're good!

    You can always do it the other way as a check.

    You have all the numbers necessary.
     
    Last edited: May 15, 2009
  10. May 15, 2009 #9
    but u do agree with my answer based on my method
     
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