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Newton's Second Law - Polar Coordinates

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data

    problem.png

    Given that:

    theta_dot = 6 rad/sec
    m_A = 0.8kg
    u_k = 0.40

    The problem also mentions that movement is at a constant angular rate so I think that means:

    r_doubleDot = 0
    theta_doubleDot = 0

    Lastly, at an instant:

    r_dot = 800mm/sec = 0.8m/sec

    2. Relevant equations


    equation.png
    (wrote these out in word)

    3. The attempt at a solution

    So I think I have to use polar coordinates for this question...

    - In polar coordinates I think the only force acting in F_r is the friction force (towards the bar)
    - In the F_theta direction i think the only force acting is the force moving the rod (I called it F_push)

    from equation 2 and 3 I have:

    F_r.png (4)

    and from equations 1 and 3 I have:

    F_theta.png (5)

    This is as far as I have gotten with the problem. The issue is that I'm not sure if I have my force equation set up correctly or if i'm missing some piece of information because I don't know how I would find r (polar coordinates confuse me a lot)

    Just as a side note, I know that the final answer is supposed to be 4.39N.

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Feb 18, 2015 #2

    ehild

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    The collar has got weight and it also contributes to the "pushing force".
     
  4. Feb 18, 2015 #3
    I'm not too sure how I would incorporate the weight in the force equation though. All I can really think of doing is:

    Fpush_New.png

    However this seems a bit strange to me because I can't see how the weight of the object would contribute to the forces in the direction of theta

    Also, I haven't used the fact that

    u_k = 0.40
     
  5. Feb 18, 2015 #4

    ehild

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    You need the force of friction and it is the force of interaction between the rod and the collar, multiplied by u_k.
    The collar experiences the horizontal force maθ from the rod and also the vertically upward force mg. The horizontal force accelerates it horizontally. It is needed to increase its tangential velocity when it moves outward on the rod, so as it maintain the angular velocity when the radius increases. The vertical force is the reaction force of its weight.
    Both forces contribute to the force of interaction between the rod and collar, and to the force of friction, as well. These forces are perpendicular to each other, so they add as vectors. You simply added them, which is wrong.
     
  6. Feb 20, 2015 #5
    Alright so I think I finally have an answer! So my new force equations are:

    force_Equations.png

    These forces act perpendicular to each other, so to find the net force of interaction you can use the Pythagorean Theorem.

    And like you said, the force of friction

    This gives me the equation

    Ffriction.png

    Now all that was left was to plug in my numbers:

    Ffriction_answer.png

    I think my biggest mistake was that I didn't account for the vertical upwards force due to the weight of the collar.
     
  7. Feb 20, 2015 #6

    ehild

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    Well done! :)
     
  8. Feb 20, 2015 #7
    Thank you so much for the help! I truly appreciate it :)
     
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