# Equation substitution

1. Jul 18, 2011

### exidez

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:

• ###### InvertedPend.jpg
File size:
65.6 KB
Views:
64
Last edited by a moderator: May 5, 2017
2. Jul 18, 2011

### tiny-tim

Last edited by a moderator: Apr 26, 2017
3. Jul 18, 2011

### exidez

ahhh of course they are!

Why do i always forget the fundamentals!!

4. Jul 18, 2011

### exidez

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..

$\frac{d^2}{dt^2}(x+Lsin(\theta))$

First working out the first derivative

$\frac{d}{dt}(x+Lsin(\theta)) = \frac{d}{d\theta}\frac{d\theta}{dt}(x+Lsin(\theta))$

$\frac{d}{dt}(x+Lsin(\theta)) = \frac{d\theta}{dt}Lcos(\theta)$

So the second derivative is:

$\frac{d}{dt}(\frac{d\theta}{dt}Lcos(\theta))$

$\frac{d\theta}{dt}\frac{d}{d\theta}(\frac{d\theta}{dt}Lcos(\theta))$

Using the product rule:

$\frac{d\theta}{dt}(\frac{d}{d\theta}(\frac{d\theta}{dt})Lcos(\theta)-\frac{d\theta}{dt}Lsin(\theta))$

$\frac{d}{d\theta}(\dot{\theta})\dot{\theta}Lcos( \theta)-\dot{\theta}^{2}Lsin(\theta))$

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

$L\ddot{\theta}cos(\theta) - \dot{ \theta}^{2}Lsin(\theta))$

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??

5. Jul 18, 2011

### tiny-tim

hi exidez!

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

you can just do d/dt (θ') = θ''

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

6. Jul 19, 2011

### exidez

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 ??

7. Jul 19, 2011

### tiny-tim

hi exidez!

(try using the X2 icon just above the Reply box )
show us how you got that

8. Jul 19, 2011

### exidez

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:

• ###### P1012483.jpg
File size:
22.3 KB
Views:
55
9. Jul 19, 2011

### tiny-tim

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

10. Jul 19, 2011

### exidez

The two equations:

$F-H=M\ddot{x}+k\dot{x})$

$H=m\frac{d^{2}}{dt^{2}}(x-Lsin(\theta))$

Derivation:

$\ddot{x}=\frac{1}{M}(F-H-k\dot{x})$

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

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

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

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

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

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

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

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

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

The text says it should be:

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

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

11. Jul 19, 2011

### exidez

The Two equations:

$V-mg=m\frac{d^{2}}{dt^{2}}(Lcos(\theta))$

$I\ddot{\theta}+c\dot{\theta}=VLsin(\theta)-HLcos(\theta)$

Derevation:

$\ddot{\theta}=\frac{1}{I}(VLsin(\theta)-HLcos(\theta)-c\dot{\theta})$

$\ddot{\theta}=\frac{1}{I}(VLsin(\theta)-HLcos(\theta)-c\dot{\theta})$

$V=m\frac{d^{2}}{dt^{2}}(Lcos(\theta))+mg$

$V=m\frac{d}{dt}(\frac{d}{dt}(Lcos(\theta)))+mg$

$V=m\frac{d}{dt}(\frac{dx}{dt}\frac{d}{dx}(Lcos( \theta)))+mg$

$V=m\frac{d}{dt}(0)+mg$

$V=mg$

$H=m\frac{d}{dt}(\frac{d}{dt}(x+Lsin(\theta)))$

$H=m\frac{d}{dt}(\frac{dx}{dt}\frac{d}{dx}((x+Lsin(\theta))))$

$H=m\frac{d}{dt}(\frac{dx}{dt}(1+0))$

$H=m\frac{d}{dt}(\dot{x}(1+0))$

$H=m\frac{d}{dt}(\dot{x})$

$H=m\ddot{x}$

Now sub new H and V into theta double dot:

$\ddot{\theta}=\frac{1}{I}(mgLsin(\theta)-m\ddot{x}Lcos(\theta)-c\dot{\theta})$

$\ddot{\theta}=\frac{1}{I}(mL(gsin(\theta)-\ddot{x}cos(\theta))-c\dot{\theta})$

The text says it should be:

$\ddot{\theta}=\frac{1}{I+L^{2}m}(mL(gsin(\theta)-\ddot{x}cos(\theta))-c\dot{\theta})$

12. Jul 19, 2011

### tiny-tim

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'

13. Jul 19, 2011

### exidez

ahhhhh yes. Then i get a term involving $\ddot{x}$, take it over the other side, factor the common $\ddot{x}$ out and take what is in the bracket on the other side.

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