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Angular Motion In A Plane

  1. Nov 24, 2005 #1
    Hey, I have this question here, the hardest one on this sheet, and I've been stuck on it all day now. I'd really appreciate maybe just a hint, as I want to do it myself but I just need a kickstart.

    I drew three free body diagrams for each position of the pendulum, then marked off tension and the force of gravity. I ended up with three equations but it never went anywhere and I truly am stuck on this. Any hints would be appreciated. Thanks.
  2. jcsd
  3. Nov 24, 2005 #2


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    You won't be needing a free-body diagram for this one. It's easy enough to do in your head :P (the FBD part that is)

    2 Concepts

    qn1: what is the change in potential energy and gain in kinetic energy from 40 degrees to the 20 degrees?

    qn2: what is the force needed to keep such a mass with such kinetic energy in circular motion?
  4. Nov 24, 2005 #3
    Yea you are right about the FBD's in my head, I was just stuck and decided to draw a picture :biggrin:

    For question 1, I set the length to one just to play around with the question. By doing that, I found that the heigh at point at (40degrees) was 0.234m, and at point B it was 0.06m. By solving the equation
    PEa = PEb + KEb, I got v = 1.85 m/s.

    Then I got the idea I could use Fnet = mv^2/r - maybe I will try solving for the mass now or something. Anyways I will keep trying with this.

    I have no idea if that is what you were suggesting? I'm just playing around with it trying to see if something can be done.

    I also converted the 60 degree trip in total to rads, and got 1.05 rads. Am I getting warmer or colder? lol

    For question 2, I'm not done with the first question, but I willget back to you.

    Thanks for your help so far :smile:
  5. Nov 24, 2005 #4


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    As long as your doing along the lines of [tex]PE = mgLcos\theta[/tex] you're on the right track.
  6. Nov 24, 2005 #5
    Okay, I really think I'm on to something here.

    PE = mgLcos(theta)

    Code (Text):

    PEa = PEb + KEb
    mgH = mgH + mv^2/2
    cancel the m's
    gL0.234 = gL0.06 + v^2/2
    2.29L = 0.588L + v^2/2
    3.404L = v^2
    Then I subbed that into Fnet = mv^2/R

    Fnet = m3.404L/R

    and then I cancelled the L and the R since I assumed they were the same thing, and I'm left with Fnet = 3.404m

    Last edited: Nov 24, 2005
  7. Nov 24, 2005 #6


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    Yes, very legitimate. I'm just a bit curious with the 0.234 and 0.06 you put behind the L's. If those are checked, I think you just hit jackpot. :D

    Oh, Fnet isn't the complete end of the story yet though. Some basic FBD stuff to do yet to find the tension as it's also fighting against gravity.
  8. Nov 24, 2005 #7
    My thinking when I did the 0.234 and 0.06 was this: (I could be wrong)

    When we've done these questions in class before (it's a grade 12 physics class), when the pendulum is on y axis going straight down, it is assumed it's height is zero. So if there is a 40 degree line that is going off of the vertical, then you do the cosine thing and get the number 0.766.This is the height from the end of the 40 degree line to the x axis above. So for me, I subtracted this number from 1 and got 0.234.

    Same procedure for the other one too. Does this make sense to you or have I made an error?
  9. Nov 24, 2005 #8
    1/2 mv squared = t - mg
    is that the final formula that would work
    with v squared being whatever we get from the energy equations?
    Last edited: Nov 24, 2005
  10. Nov 24, 2005 #9
    I drew my new FBD, and came up with this.

    Fnet = 0.348mg <----I went back and redid it without adding in the gravity, wanted to leave it as a letter just like in the answer.

    So my new equation according to my FBD was:

    0.348mg = T - mg
    T = 1.348mg

    I'm off slightly. Can anyone see where I made a mistake?
  11. Nov 24, 2005 #10


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    Alright, so you've assumed that the bottom of the pendulum is h=0. That's perfectly fine.

    Now that you've justified the velocity at this point. You need to be a bit more careful with the FBD at 20 degrees. What direction is Fnet? What direction is T, the tension? What direction is gravity?
  12. Nov 24, 2005 #11
    i see what you mean and it worked
    but only if i apply the cos20 to mg
    not if i apply it to T
    why's that?
    thanks for your time :)
  13. Nov 24, 2005 #12
    Ahhh okay I did it now!

    Because this FBD will be tilted, the gravity will be -mgcostheta which makes it -0.940mg.

    So the equation would be
    0.348mg = T - 0.940mg
    1.29mg = T

    Thanks so much for your help man! Appreciate it!
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