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F=MA 2009 Exam # 18 (Pendulum, a small peg, and mass m)

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data
    https://aapt.org/physicsteam/2010/upload/2009_F-maSolutions.pdf
    # 18

    2. Relevant equations
    T = 2∏√(L/g)


    3. The attempt at a solution
    Honestly, I'm not sure how at all to do this question. All I know that the distance from the small peg to mass m is L/3 - so that should be accounted for. What throws me off is how the pendulum starts of rotating about the small peg and should end up rotating about the fixed pivot point. I don't know how to represent that in equation.
     
  2. jcsd
  3. Jan 29, 2013 #2
    You can take this as a combination of two motions, one about the pivot point and the other about the small peg. When it is released, it performs motion about the pivot point. Can you find the time it takes to reach the mean position?
     
  4. Jan 29, 2013 #3
    How would I do that?
     
  5. Jan 29, 2013 #4
    You have the time period of the motion. (See that the angular displacement is of 5 degrees only, it will perform SHM!). The time period of SHM is the complete time it takes to perform one complete oscillation. What time it takes to reach the mean position from the extreme position? T/4. Do you see why?
     
  6. Jan 29, 2013 #5
    Ya I see why. It's because it takes T/4 to get to the mean. T/4 to get to the right extreme, and then the reverse to get back. So it makes sense that its T/4.
    What's next?
     
  7. Jan 29, 2013 #6
    You should be able to do it now. When it reaches the mean position, it performs SHM about the small peg. It goes to the extreme and again back to mean. You can calculate this time by using the time period of SHM about the small peg. When it reaches the mean position, it head backs to its initial position.
     
  8. Jan 29, 2013 #7

    Simon Bridge

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    18. A simple pendulum of length L is constructed from a point object of mass m suspended by a massless string attached to a fixed pivot point. A small peg is placed a distance 2L/3 directly below the fixed pivot point so that the pendulum would swing as shown in the figure below[*]. The mass is displaced 5 degrees from the vertical and released. How long does it take to return to its starting position?

    [*] the diagram did not copy - it shows the setup as described. The peg is dircetly below the pivot. The pendulum is displaced to the right and it shows that the string wraps the peg when it tries to swing to the left.

    So there are two motions - one is a pendulum about the peg and the other about the pivot - they have to join up smoothly. I think, here, the lynchpin is to think about the motion in stages - the time taken to swing from +5deg to 0deg, the time to swing from 0 to the far left, then back again, and the time to swing back to +5deg.

    [edit - oh I see I was beaten to it :) ]
     
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