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Work Energy Method for Rotational Motion

  1. Jul 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi guys, I would like to ask for this particular question, The GPE is 0 for all position is because I set datum as the original position, GPE = 0 and when the string got detached from the axle, there is no GPE for datum 2. Am I right?



    Thanks
     

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  3. Jul 15, 2012 #2

    Doc Al

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    Doesn't the height of mass B change?
     
  4. Jul 15, 2012 #3
    You mean there is GPE? o_0

    Because i see the working for E2 and E3 equation the GPE is 0 and they got the answer
     
  5. Jul 15, 2012 #4

    Doc Al

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    Again, what matters is change in GPE. And since mass B changes height, there is a change in GPE regardless of what you choose as your reference point.
     
  6. Jul 15, 2012 #5

    Doc Al

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    Not sure what you mean.
     
  7. Jul 15, 2012 #6
    Since you say height changes = there is GPE, how come the GPE in the working for E1 and E2 are 0 and able to get the answer.
     

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  8. Jul 15, 2012 #7

    Doc Al

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    For that part of the problem (the slowing down of A after the mass B has detached) only the energy of A is relevant. And the GPE of A doesn't change. (The height of B changes as it falls, not the height of A.)

    And since you are given the angular acceleration of A as the mass is falling, there's no need to bother with the changing GPE of B even for the first part of the problem.
     
  9. Jul 15, 2012 #8
    GPE of A doesn't change because it's a wheel.

    So the "E" equation is all about wheel A?
     
  10. Jul 15, 2012 #9

    Doc Al

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    The GPE doesn't change because its center of mass doesn't change height.
    Right. They are equating the work done on A to its change in energy.
     
  11. Jul 15, 2012 #10
    Then for other questions how am I going to know whether to lnclude the GPE or not??

    This is so hard :cry:
     
  12. Jul 15, 2012 #11

    Doc Al

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    It just depends on the system you choose to analyze. Which is often up to you. (Many problems can be solved in multiple ways.)

    In this particular problem it seems to me that they gave you too much information. They didn't need to give you the acceleration. But perhaps they wanted you to use a certain method of solving it.
     
  13. Jul 15, 2012 #12
    Can I say that if the question give me info on parcel B and nothing on A, there will be GPE?

    And also my lecturer says that everything is same for work energy for linear and rotational. But the only difference is that kinetics for rotational has an additional "1/2Iω2

    I am confused on when to include linear K and rotational K. Can you tell me how do i spot?


    Thanks alot..
     
  14. Jul 15, 2012 #13

    Doc Al

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    If you choose to analyze mass B there will certainly be a change in GPE since it moves vertically.
    For rotational K, something with mass has to be rotating. For example: A spinning wheel or a rolling cylinder.

    For linear K, something needs to translate--its center of mass has to move.
     
  15. Jul 15, 2012 #14
    We need angular acceleration to find angular velocity at the end of 8rev. I'm pretty sure they have assumed α is constant, so we shall use ω202=2*α*Δθ. But the info on mass B is totally irrelevant.
     
  16. Jul 15, 2012 #15
    For the above mentioned question, how come linear K is not included? The parcel is attached to the axle and the parcel moves, there should be linear K right?
     
  17. Jul 15, 2012 #16
    Find final omega at the end of 8 rev using:

    ω202=2αΔθ. (ω0=0).Make sure to convert revolution to radians.

    Find Io=k2mA, k: radius of gyration

    T=0.5mAvA2+0.5Ioω2

    Since A doesn't have any translational motion, vA=0. Now you can find T(kinetic energy) easily.

    For part B, ƩMo=Ioα, Find α from here(which is negative if we take clockwise positive).

    Use ω202=2αΔθ again to find Δθ and finally convert it to revolutions. Note also that your final ω now is zero and your initial ω is what you found earlier(at the end of 8rev).
     
    Last edited: Jul 15, 2012
  18. Jul 15, 2012 #17

    Doc Al

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    You are given the angular acceleration, but you don't need it. You should be able to figure it out from the other information given.
     
  19. Jul 15, 2012 #18
    You mentioned that "It just depends on the system you choose to analyze. Which is often up to you"

    For the 1st diagram, at first i tried to include A & B in my E equation but you said that I can choose a system (1) which is A to solve since they gave me alot of info on A.


    For the diagram below, I solved this question with my E equation having A and B (2 System)

    My initial understanding is that you need to include whatever thing you see in the picture and solve since it's connected together.

    But you said that i can choose the system and solve it, so can this question be solve by just using 1 system eg. A only?
     

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  20. Jul 15, 2012 #19

    Doc Al

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    Since you are given the angular acceleration, you don't have to worry about the kinetic energy of the falling mass. All you need to analyze is mass A, which has no linear kinetic energy. (If the angular acceleration wasn't given, and you wanted to use energy methods, then you'd have to include the kinetic energy of the falling mass in your analysis.)
     
  21. Jul 15, 2012 #20

    Doc Al

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    Good.
    Not always. Sometimes they give you extra information that makes it easier. (Like they did in your original problem.)
    No. (I said that you can often choose the system. But sometimes you have no choice.)

    In your original problem they gave you all the information needed so you just had to analyze mass A. But that's not the case here.
     
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