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What is the horizontal force exerted by the pivot on the rod

  1. Oct 11, 2016 #1
    1. The problem statement, all variables and given/known data
    A rod swings freely about a pivot. Is has mass M and length L. The average force and duration of hit are Fo and t respectively. A) What is the speed of the com of the rod just after It is hit? B) what is the horizontal component of the force exerted by the pivot on the rod? At what value of x is the center of percussion?

    2. Relevant equations
    V = (L^2) (omega) / (12x)
    I of rod = (1/12)ML^2
    X is distance down rod of applied force


    3. The attempt at a solution
    A has been found. I think C) is one third - and if it is that means I don't need help with b. To find B I just used the equation from A and multiplied it by mass and divided by time and reduced to m(L^2)*alpha/(12x). I set this equal to I*alpha for the pivot - the I of which I THINK is just mr^2. The moment of inertia of the pivot needs to be accounted for because it exerts a force on the rod in this case. This gave me 1/3 for part c. Is this correct? Because I used r for the pivot as L/2 (same L as rod) which can't possibly make sense...
     
  2. jcsd
  3. Oct 11, 2016 #2

    haruspex

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    The pivot is at one end? It is hanging vertically?
    What hit? Is it struck perpendicularly at the free end?
     
  4. Oct 11, 2016 #3
    It hangs freely. It's hit with an average force f which causes it to rotate about the pivot. I now know that part B and C are wrong. It is number 2
     

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  5. Oct 11, 2016 #4

    haruspex

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    Thanks for posting the original text, that's much clearer.
    Please show your working for part b. The outline you gave is not detailed enough.
     
  6. Oct 12, 2016 #5
    Okay. Uh, part A is wrong...the moment of inertia is actually ML^2/3. To go about part A this is what I did... please tell me if this is correct because I have absolutely no confidence in my ability to do physics at this point. I don't know what to do for B - all I know is that the force at the bottom is equal to the force at the top which is determined by the torque of the pivot point at the top and the force that is solved for by rearranging part A for the bottom. I know that x needs to be 2L/3, I just need to show this. I can't come up with a correct equation to equate the torque and force respectively. And if part A is wrong I can say goodbye to this entire problem, anyway. So that is my first priority right now.
     

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  7. Oct 12, 2016 #6
    There is no part C, lol.
     
  8. Oct 12, 2016 #7

    haruspex

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    What makes you say that? I also get ##\frac {3xF_0\Delta t}{2ML}##. The only thing I see wrong with your answer in the attachment is that you have renamed ##\Delta t## to just t.
    If that were true there would be no acceleration of the rod's mass centre.
     
  9. Oct 12, 2016 #8
    Wow. That is honestly incredibly reassuring. Okay then... now onto B. Valid??
     

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  10. Oct 12, 2016 #9

    haruspex

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    Your image is a bit faint for my old eyes, but the final answer is right.
     
  11. Oct 12, 2016 #10
    Awesome. Thank you so much.
     
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