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Energy conservation law problem

  1. Jun 9, 2007 #1
    1. The problem statement, all variables and given/known data

    Here is the drawing which shows the situation:

    http://img507.imageshack.us/my.php?image=drawingae3.jpg

    The mass of each ball is equal to m. A rod which connects the balls is weightless. A vertical wall affects a ball which is below with force F. I need to find this force when the angle between the rod and ground is alfa (the rod is vertical at the beginning).


    3. The attempt at a solution

    Stress in the rod T.

    Potential energy of the right ball is
    PE = mgR sin(α)

    Kinetic energy of the right ball is
    KE = 1/2 m v²(α)

    Conservation of energy:
    KE + PE = Eo
    1/2 m v²(α) + mgR sin(α) = mgR
    v²(α) = 2 gR (1 - sin(α))

    Centripetal accleration:
    a(α) = v²(α)/R = 2g(1 - sin(α))

    Balance of force along the rod:
    ma(α) = mg sin(α) - T(α)
    2mg(1 - sin(α)) = mg sin(α) - T(α)
    mg(2 - 3sin(α)) = -T(α)
    T(α) = mg(3sin(α) - 2)


    Answer:
    F(α) = T(α) cos (α) = mg(3sin(α) - 2)cos(α)

    But the last equation means that if α<arcsin(2/3) = 41.8 deg, the left ball will no longer touch the wall. Is it possible? Maybe there are some mistakes in the solution?

    Thanks in advance.
     
  2. jcsd
  3. Jun 9, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    What about the tangential acceleration of the ball?
     
  4. Jun 9, 2007 #3
    Well does this mean that the right ball gets some rotational kinetic energy because it rotates around the left ball (in my solution this energy is not counted) and the left ball doesn't get any rotational kinetic energy because it is located at the rotation centre? Maybe then I should use moment of inertia, but I have no idea how to implement this.
     
  5. Jun 9, 2007 #4
    I think you have made a mistake in this step:

    Balance of force along the rod:
    ma(α) = mg sin(α) - T(α)

    it should be:

    mg sin (alpha) - T/cos(alpha) = 2mg(1-sin(alpha))

    think about the forces on the rod (or applied through the rod): there is weight, and "T/cos(alpha)" .( The actual force is T but there is also a vertical force applied. Because they change accordingly with the angle, the force applied through the rod is T/cos(alpha)

    then rearranging that I got,

    3mg*sin(alpha)*cos(alpha) - 2mg*cos(alpha)=T
     
  6. Jun 9, 2007 #5

    Doc Al

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    Staff: Mentor

    No need to consider the rotational energies of the balls--assume they are small. I assume that the problem states that there is no friction between the left ball and the wall or floor, and that you are to find the horizontal force exerted by the vertical wall. My suggestion was to consider the complete acceleration of the right ball, not just the centripetal acceleration. Find the horizontal component of the ball's acceleration and use it to find the horizontal force on the ball.
     
  7. Jun 11, 2007 #6
    Thanks for explanation. But could anyone give a suggestion about counting horizontal component of the ball's acceleration. And I have no idea how this tangential acceleration interrelates with centripetal acceleration.
     
  8. Jun 11, 2007 #7

    Doc Al

    User Avatar

    Staff: Mentor

    The total acceleration is the combination of centripetal and tangential components. You can find the horizontal component of the total acceleration by finding the horizontal components of the centripetal and tangential pieces. (Figure out the tangential acceleration of the ball using Newton's 2nd law for rotation.)
     
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