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Simple Tangential Acceleration Problem

  1. Feb 20, 2006 #1
    Problem:
    A uniform rod of length L and mass M is pivoted at one end (O) and is released from rest in a horizontal position. The moment of inertia of the rod about one end is I = (mL^2)/3 . The initial tangential acceleration fo the far end A of the rod would be?

    This is part of my review for my AP Physics C test. I thought first maybe I could use torque in some way

    T= rF = LMg
    And [tex] T[/tex] [tex]= I \alpha [/tex]

    So, I equated the torque statements thus having

    [tex] LMg [/tex][tex] = (1/3)(ML^2)(\alpha) [/tex]

    This lead me to the result of 3g/L = [tex] \alpha [/tex]
    and by the identity [tex] a = r \alpha [/tex] I got 3g = acceleration. This is not the right answer. The correct answer is 3g/2, as opposed to my 3g. Where did the 1/2 come from? Thanks!

    Wait...I think I figured it out...is it hte fact that in T = rF, the r is referring to 1/2 the Length, not the full length?
     
    Last edited: Feb 20, 2006
  2. jcsd
  3. Feb 20, 2006 #2
    Yes, the gravitational force always acts on the center of mass of an object. In this case, since the rod is of uniform mass density, the center of mass is also the center of the rod.
     
  4. Feb 20, 2006 #3
    Excellent. Thank you very much!

    However, I have one more pesky question that I actually am not too sure how to work. Any advice would be helpful.

    A 800 Newton student has an apparent weight of 600 Newtons at the top of a vertical circular loop of a roller coaster. The car has a uniform speed v and is upside dwn inside the circular path of the roller coaster. The apparent weight of the student when the car is at the lowest point would be...

    I attempted to draw myself a diagram. I said the centripetal force would be pointing towards the center, the weight would be pointing downwards, but wasn't too sure how the "weight the student feels" would be doing. I assumed downwards in direction also, but I don't know. (Note: this is me analyzing when the kid is at the top of the roller coaster, just so I can familiarize myself with the situation.)

    Any advice or help would be kindly appreciated!
     
  5. Feb 20, 2006 #4
    By "apparent weight" they mean what a scale that the student was sitting on would read. So gravity is pulling the student down with a force of 800N but the net force on him (due to gravity and the acceleration of the roller coaster) is 600N into his seat (i.e. straight up).
     
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