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Falling object simulation

  1. Mar 9, 2016 #1
    1. The problem statement, all variables and given/known data
    This is an academic project related to a falling object simulation.
    I have come to the point where I'm trying to calculate the per degree angle of the falling object seen in the attached images. The object has known height and length in pixels and is fixed on a pivot.

    2. Relevant equations
    PE = mgh
    KE = ½Iω2
    I = ⅓ml2

    3. The attempt at a solution
    Will calculate the difference in PE between two angles, eg: ΔPE = PE(90°) - PE(89°) = KE
    So, ΔPE = KE
    Δ(mgh) = ½Iω2
    Δ(mgh) = ½(⅓ml22 (cancel m)
    Δ(gh) = 1/6 l2 ω2
    6 Δ(gh) = l2 ω2
    ...
    ω = √6Δ(gh)/l2

    Also, h is the height of the COM calculated at every angle using h = sin(angle)*hyp
    hyp is the distance of centroid to the rotation point.

    ω is measured in rad/sec
    I'm confused on how to go back from ω to the actual animation of the falling object and also if what I calculated is correct.

    Any help is very much appreciated.
    Thanks
     

    Attached Files:

  2. jcsd
  3. Mar 9, 2016 #2

    BvU

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    Hello George,
    You are trying to calculate what ?
    And why would you want to do that?

    As comments: I don't see much wrong, but you should realize the pivot point is moving when you round off the bottom of your object. And the moment of inertia is no longer that for a rectangular block.
    I wouldn't use 'back' here ? Unless you have calculated ##\phi(t)## already ....
     
  4. Mar 9, 2016 #3
    Hello BvU

    Thanks for the comments.

    I'm trying to calculate the KE at every angle starting from 90° to 0° clockwise. I assume that KE will give me ω. This ω is the angular velocity at every angle. Is this assumption correct?
    KE must be equal to the difference in PEs, that's why I use this. May I ask whether you have a different method to solve this?

    Finally, I have ω, which is rad/sec, my simulation will work in 25fps, so in every frame I will have to consume 25 degrees - 1 degree per frame.
    I'm confused on how to assign ω per frame.

    Thanks
     
  5. Mar 9, 2016 #4

    BvU

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    your background isn't in physics ?

    What would you do to simulate an object that someone lets go at 1.25 m from the floor and that drops to the ground in 0.5 s ?
    Or a projectile trajectory ?
    It seems to me you then get a uniform rotation - 25 degrees/sec, so 3.67 seconds and it's flat on the floor.

    The thing to do is to solve the equation of motion. In the above example (dot means ##d\over dt## -- are you familiar with differentiation ? -- ): $$
    \ddot y = -g \quad \Rightarrow \quad \dot y = \dot y_0 - gt \quad \Rightarrow \quad y = y_0 + \dot y_0 t - {1\over 2} gt^2$$But now something similar for the orientation (##\theta##) for the falling rod.

    If you insist, your approach can be made equivalent to numerically solving the equation of motion: You get ##\Delta \omega \over \Delta \theta ## in steps of one degree and - with initial conditions - you can add them up to get ##\omega(\theta) ##. And an angle of 1 degree is traversed in ##{2\pi\over 360\;\omega}## giving you the time (number of frames) from one angle to the next (1 degree further) angle.
     
  6. Mar 9, 2016 #5
    Let me clarify something: There two phases here, one to calculate the velocity at every degree, that is, for 90° ω=0 ...for 59° ω=2.89 ... for 19° ω=3.78 etc.

    Then, having the above stored in memory, I'll have to draw the rod and simulate the falling action. The question is how I will go from ω to fps. Is so confusing!
     
  7. Mar 9, 2016 #6

    BvU

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    I thought you had fps fixed at 25 :wink: !

    Basically you want ##\theta(t)##, right ? You can draw the thing at an angle ##\theta## and time runs with 0.04 s between two frames.

    And you are somewhere already: starting with ##\theta = 0## isn't very interesting: nothing happens...

    But if you have a list with omegas life is easy ! Say you start at 89 degrees and draw Frame 1 for t = 0. Look up omega and do ##\theta(0.04) = \theta(0) + 0.04 \omega## (be sure to convert back to degrees !)
    Look up (or interpolate) ##\omega(\theta(0.04))## and repeat! Happy falling !
     
  8. Mar 9, 2016 #7
    The list of ω doesn't look great though, I would expect to see some higher acceleration at the end - when the rod reaches the ground - rather than at the start of the fall. See the diagram of velocity-frame no.
     

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  9. Mar 9, 2016 #8

    BvU

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    In the equation of motion there is a ##\sin\theta## And your series 4 shows something looking like that ? But I don't really know what 1, 2,3 and 4 stand for...
     
  10. Mar 9, 2016 #9
    Series 1 shows the ω for every degree (0-90)
    sin(θ) is used to calculate the height of COM
     

    Attached Files:

  11. Mar 10, 2016 #10

    BvU

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    Oh, well, series 1 also looks like a sine on that scale....
     
  12. Mar 10, 2016 #11
    It doesn't look right. I would expect a different graph.
     
  13. Mar 10, 2016 #12

    BvU

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    Namely ?
     
  14. Mar 10, 2016 #13
    I would expect to see some higher acceleration at the end - when the rod reaches the ground - rather than at the start of the fall
     
  15. Mar 10, 2016 #14

    BvU

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    Yes you said that. But you still haven't said which is which or what you do to get these results.

    In post #1, if you write ΔPE = KE, do you understand that that in fact should be ΔPE = ΔKE ?
     
  16. Mar 10, 2016 #15
    I was advised to use ΔPE = KE by another fellow in this forum. Is not my idea, I'm afraid.
    To get these results, I do what I say in Post#1, nothing more...
     
  17. Mar 10, 2016 #16

    BvU

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    Well, how do you think you describe the change from potential energy to kinetic energy for a falling object ?
    With ΔPE = KE, or with ΔPE = ΔKE ?

    I know lots of other fellows (and a few ladies) in PF and none of those would advise the former. Could it be you misinterpreted ?
     
  18. Mar 10, 2016 #17
    Perhaps I was misinterpreted...
    If GPE loss = KE gain, then loss in PE is calculated by the difference in PEpre - PEpost
    Since PE = mghCOM, is mainly h the only changing parameter here. But there is something missing and that's why I get the graph series 1 as the sin graph.
     
  19. Mar 10, 2016 #18

    BvU

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    Can you show your steps ? I still don't know which is which or what you do to get these results.
    Or why you think it's wrong, for that matter. It may well be correct.

    As long as you do ΔPE = ΔKE with respect to the same reference point. If that is the starting position in both cases you are relatively OK. Note that you run into trouble with the initial conditions very easily this way! ##\ \ \theta(0) = 0 \ \ \& \ \ \omega(0) = 0\quad ## is a (non-stable) equilibrium point: theoretically nothing happens.
    If (your post #1) $$ {\rm PE} =mg{l\over 2} \sin \theta \quad
    $$then $${\Delta \rm PE} = mg{l\over 2}(1-\sin \theta) \quad
    $$and you easily get $$\omega^2 = {3g\over l} (1-\sin\theta)\quad $$which does increase nicely for ##\theta \downarrow 0##. So I don't really see a problem -- even for this very cumbersome method :smile: .

    --
     
    Last edited: Mar 11, 2016
  20. Mar 10, 2016 #19

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  21. Mar 10, 2016 #20

    BvU

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    I see that, but I don't see why you think your thingy is wrong becasue you still haven't said what it shows. I only see a meaningless series 1, 2, 3, 4 and no idea what is on the abscissa. Should I read from left to right or from right to left ?
    Can you show your steps ?
     
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