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Equation substitution

  1. Jul 18, 2011 #1
    In the attached image, how are equations 1, 2, 3 and 4 used to come to the final equation of 5 and 6? Im suspecting it has something to do with the derivative with respect to t, but I dont know how they remove it to get the final solution. Am I missing something incredibly simple that is not worth mentioning in the text?

    The image was taken from http://www.r8sac.org/files/SPC/bugeja.pdf [Broken] if further information is needed.
     

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    Last edited by a moderator: May 5, 2017
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  3. Jul 18, 2011 #2

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  4. Jul 18, 2011 #3
    ahhh of course they are!

    Why do i always forget the fundamentals!!
     
  5. Jul 18, 2011 #4
    Ok so now im lost again.. I tried...

    If i just work out one, i can get the other but i still cant get the one..

    [itex]\frac{d^2}{dt^2}(x+Lsin(\theta))[/itex]

    First working out the first derivative

    [itex]\frac{d}{dt}(x+Lsin(\theta)) = \frac{d}{d\theta}\frac{d\theta}{dt}(x+Lsin(\theta))[/itex]

    [itex]\frac{d}{dt}(x+Lsin(\theta)) = \frac{d\theta}{dt}Lcos(\theta)[/itex]

    So the second derivative is:


    [itex]\frac{d}{dt}(\frac{d\theta}{dt}Lcos(\theta))[/itex]

    [itex]\frac{d\theta}{dt}\frac{d}{d\theta}(\frac{d\theta}{dt}Lcos(\theta))[/itex]

    Using the product rule:

    [itex]\frac{d\theta}{dt}(\frac{d}{d\theta}(\frac{d\theta}{dt})Lcos(\theta)-\frac{d\theta}{dt}Lsin(\theta))[/itex]


    [itex]\frac{d}{d\theta}(\dot{\theta})\dot{\theta}Lcos( \theta)-\dot{\theta}^{2}Lsin(\theta))[/itex]

    This isnt quite the answer im supposed to get..
    its supposed to be:

    [itex]L\ddot{\theta}cos(\theta) - \dot{ \theta}^{2}Lsin(\theta))[/itex]

    close but no cigar.... any clues on where i went wrong or maybe there is something further with the derivative of theta dot with respect to theta??

    Your help is much appreciated
     
  6. Jul 18, 2011 #5

    tiny-tim

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    hi exidez! :smile:

    i think you're making things unnecessarily difficult by trying to do d/dθ (θ') …

    you can just do d/dt (θ') = θ'' :wink:

    d/dt (sinθ) = θ' d/dθ (sinθ) = θ'cosθ

    d/dt (θ'cosθ) = θ''cosθ + θ' d/dt(cosθ) = θ''cosθ + θ' θ' d/dθ(cosθ)

    = θ''cosθ - θ'2 sinθ :smile:
     
  7. Jul 19, 2011 #6
    Thanks tiny-tim,

    I understand it correctly now and i have the matching answer but with one tiny problems..

    If you look at equation 5 and 6they have a multiplyer at the front being 1/(I+L^2m) and 1/(M+m) respectively

    however, when in my answers they are simply 1/I and 1/M respectively as that is what the rearranged equations 3 and 4 give.. I did not get any division from chain/product rule that can give this answer...

    Do you know how they derived the I+L^2m or M+m on the numerator ??
     
  8. Jul 19, 2011 #7

    tiny-tim

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    hi exidez! :smile:

    (try using the X2 icon just above the Reply box :wink:)
    show us how you got that :smile:
     
  9. Jul 19, 2011 #8
    it would be faster if i just took a photo of my working

    I changed the variable of M to Mc to ensure i wasnt getting mixed up with m. They are two different masses. A little messy but you can see how i got 1/I and 1/M...
     

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  10. Jul 19, 2011 #9

    tiny-tim

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    in your left-hand column (the x'' equation), you've lost the extra x'' terms between lines 4 and 5

    in your right-hand column, a lot seems to have gone missing …

    it would be easier to check if you would type it out :redface:
     
  11. Jul 19, 2011 #10
    The two equations:

    [itex]F-H=M\ddot{x}+k\dot{x})[/itex]

    [itex]H=m\frac{d^{2}}{dt^{2}}(x-Lsin(\theta))[/itex]

    Derivation:

    [itex]\ddot{x}=\frac{1}{M}(F-H-k\dot{x})[/itex]

    [itex]\ddot{x}=\frac{1}{M}(F-m\frac{d^{2}}{dt^{2}}(x-Lsin(\theta))-k\dot{x})[/itex]

    [itex]\ddot{x}=\frac{1}{M}(F-m\frac{d}{dt}(\frac{d}{dt}(x+Lsin(\theta)))-k\dot{x})[/itex]

    [itex]\ddot{x}=\frac{1}{M}(F-m\frac{d}{dt}(\frac{d \theta}{dt}\frac{d}{d \theta}(x+Lsin(\theta)))-k\dot{x})[/itex]

    [itex]\ddot{x}=\frac{1}{M}(F-m\frac{d}{dt}(\frac{d \theta}{dt}(Lcos(\theta)))-k\dot{x})[/itex]


    [itex]\ddot{x}=\frac{1}{M}(F-m\frac{d}{dt}(\dot{\theta}Lcos(\theta))-k\dot{x})[/itex]

    [itex]\ddot{x}=\frac{1}{M}(F-m\ddot{\theta}Lcos(\theta)+m\dot{\theta}\frac{d}{dt}(Lcos(\theta))-k\dot{x})[/itex]

    [itex]\ddot{x}=\frac{1}{M}(F-m\ddot{\theta}Lcos(\theta)+m\dot{\theta}^{2}\frac{d }{d\theta}(Lcos(\theta))-k\dot{x})[/itex]


    [itex]\ddot{x}=\frac{1}{M}(F-m\ddot{\theta}Lcos(\theta)-m\dot{\theta}^{2}Lsin(\theta)-k\dot{x})[/itex]


    [itex]\ddot{x}=\frac{1}{M}(F-mL(\ddot{\theta}cos(\theta)-\dot{\theta}^{2}sin(\theta))-k\dot{x})[/itex]

    The text says it should be:

    [itex]\ddot{x}=\frac{1}{M+m}(F-mL(\ddot{\theta}cos(\theta)-\dot{\theta}^{2}sin(\theta))-k\dot{x})[/itex]

    the same issue is with the theta double dot. So what ever i am doing wrong here is the same issue with theta double dot. I will type the theta dot out in the next post
     
  12. Jul 19, 2011 #11
    The Two equations:

    [itex]V-mg=m\frac{d^{2}}{dt^{2}}(Lcos(\theta))[/itex]

    [itex]I\ddot{\theta}+c\dot{\theta}=VLsin(\theta)-HLcos(\theta)[/itex]

    Derevation:

    [itex]\ddot{\theta}=\frac{1}{I}(VLsin(\theta)-HLcos(\theta)-c\dot{\theta})[/itex]

    [itex]\ddot{\theta}=\frac{1}{I}(VLsin(\theta)-HLcos(\theta)-c\dot{\theta})[/itex]

    [itex]V=m\frac{d^{2}}{dt^{2}}(Lcos(\theta))+mg[/itex]

    [itex]V=m\frac{d}{dt}(\frac{d}{dt}(Lcos(\theta)))+mg[/itex]

    [itex]V=m\frac{d}{dt}(\frac{dx}{dt}\frac{d}{dx}(Lcos( \theta)))+mg[/itex]

    [itex]V=m\frac{d}{dt}(0)+mg[/itex]

    [itex]V=mg[/itex]

    [itex]H=m\frac{d}{dt}(\frac{d}{dt}(x+Lsin(\theta)))[/itex]

    [itex]H=m\frac{d}{dt}(\frac{dx}{dt}\frac{d}{dx}((x+Lsin(\theta))))[/itex]

    [itex]H=m\frac{d}{dt}(\frac{dx}{dt}(1+0))[/itex]

    [itex]H=m\frac{d}{dt}(\dot{x}(1+0))[/itex]

    [itex]H=m\frac{d}{dt}(\dot{x})[/itex]

    [itex]H=m\ddot{x}[/itex]

    Now sub new H and V into theta double dot:

    [itex]\ddot{\theta}=\frac{1}{I}(mgLsin(\theta)-m\ddot{x}Lcos(\theta)-c\dot{\theta})[/itex]


    [itex]\ddot{\theta}=\frac{1}{I}(mL(gsin(\theta)-\ddot{x}cos(\theta))-c\dot{\theta})[/itex]

    The text says it should be:


    [itex]\ddot{\theta}=\frac{1}{I+L^{2}m}(mL(gsin(\theta)-\ddot{x}cos(\theta))-c\dot{\theta})[/itex]
     
  13. Jul 19, 2011 #12

    tiny-tim

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    ah, this is where you're going wrong …
    you're replacing d/dt (x) by θ' d/dθ (x), which comes out as zero …

    you can't do that!!

    i] d/dt (x) is x', no need to involve θ at all

    ii] the chain rule only works if (in this case) x is a function of θ, and it isn't

    try again, just putting d/dt (x) = x' :smile:
     
  14. Jul 19, 2011 #13
    ahhhhh yes. Then i get a term involving [itex]\ddot{x}[/itex], take it over the other side, factor the common [itex]\ddot{x}[/itex] out and take what is in the bracket on the other side.

    I got the correct answer now! Also For the [itex]\ddot{\theta}[/itex].
    Thank a lot for you help. It is greatly appreciated! I had learnt and revised a lot just going through this understanding. Thanks again
     
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