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Collision with an object after circular motion.

  1. Dec 10, 2011 #1
    1. The problem statement, all variables and given/known data

    "A bar line AB has a length of 1,5m and a mass of 0,5kg.
    It is stuck in the point B in a horizontal fixed hub, and it can oscillate freely with no friction in the verical plane.
    at time t=0, the bar line, which was still (not moving, inertia), is set free to rotate.
    When it reaches the vertical position it hits a body of mass m2 = 0,2kg.
    The bar line immediately stops, while the body m2 begins to move after the collision with a horizontal velocity.

    Find: The angular velocity of the bar line, the linear velocity of its free extreme point (so A, i guess), and the velocity of m2 after the collision.
    2. Relevant equations
    We have then:

    AB = 1,5m
    m1 = 0,5kg
    m2 = 0,2kg
    v0 = 0m/s



    3. The attempt at a solution
    Alright, I didn't do too much actually, I am just starting out with these kind of problems.
    I understand that as soon as the bar line is set free, it starts a circular motion, and the final velocity depends on the gravity acceleration ( I guess ). And that the impact will be at angle teta = pi/2.

    Now, I don't really know where to start from.
    I am thinking about finding the center of mass, is that the right path to begin with?
     

    Attached Files:

  2. jcsd
  3. Dec 10, 2011 #2

    gneill

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    Staff: Mentor

    Find the center of mass of the bar and determine how much its (gravitational) potential energy changes between the horizontal and vertical positions. This will tell you how much rotational KE it will gain. Use your knowledge of rotational mechanics to determine its angular velocity, etc.
     
  4. Dec 10, 2011 #3
    Since there is no talking about density I can assume the center of mass is just L/2 so:

    x = 0,75.
    Right? Now I understand that I have to do something with energy, so like:

    mgh = 1/2mv^2

    But I have to decompose the motion in x and y. I am not sure if this is correct but this is my basic idea.
    But I don't understand how may I find the angles in order to get the x and y components of the height.
     

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  5. Dec 10, 2011 #4

    gneill

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    Staff: Mentor

    That's good.

    No, no decomposition is required. The change in gravitational PE for the center of mass of the rod goes into the rotational KE of the rod. There's only one rotational velocity, so no components required (a nice feature of rotational motion!) What's the formula for rotational KE?
     
  6. Dec 10, 2011 #5
    It'd be 1/2Iw^2
    where
    I = momentum of inertia
    So it'd be:

    1/3ML^2 => 1/3 0,5kg * (0.75)

    (I am not sure, but this should be it?)

    Then I can do the following:

    mgh = 1/2Iw^2

    Where I have I, m and g.
    I can assume that h = 0,75 because it'd be the radius, and so I could find the angular velocity?
     
  7. Dec 10, 2011 #6

    gneill

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    Staff: Mentor

    You have the correct formula for the moment of inertia of a thin rod about an end. But for some reason you've plugged in half the length of the given rod. That is not correct. You must use the full length of the rod, L = 1.5 m. The center of mass position doesn't enter into this formula. Otherwise, you've got the right idea!
    Yes. h is the change in height of the center of mass of the rod (so you might consider calling it "Δh").
     
  8. Dec 10, 2011 #7
    Ok so:

    1/3ML^2 => 1/3 0,5kg * (1.50)

    I = 0,25.

    mgh = 1/2Iw^2
    w = square root of... (2mgh)/I

    So
    w = 5,42 rad/sec

    Velocity = radius * w => velocity = 1,50 * 5.42 = 8.13m/s

    And then the other stuff is just an elastic collision which I should know how to handle.
    Just one thing, in the velocity up here, I did right plugging in the radius "1,50" or it should've been 0.75?

    Thanks a lot!
     
  9. Dec 10, 2011 #8

    gneill

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    Staff: Mentor

    Oops. Looks like you forgot to square the length.
    The radius is correct; You want the distance from the center of rotation to the point of interest. If you sort out the small error in the moment of inertia calculation you should be good.
     
  10. Dec 10, 2011 #9
    woops, my bad :)... okay got it! Thanks a lot. That's appreciated.
     
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