1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     

    Attached Files:

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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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...
     

    Attached Files:

  10. Jul 19, 2011 #9

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equation substitution
  1. Tricky substitution (Replies: 3)

  2. Reversing substitution (Replies: 5)

Loading...