How are equations used to derive the final solution in this image?

[exidez]

In the attached image, how are equations 1, 2, 3 and 4 used to come to the final equation of 5 and 6? I am suspecting it has something to do with the derivative with respect to t, but I don't 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 if further information is needed.

 

[tiny-tim]

hi exidez!

… I am suspecting it has something to do with the derivative with respect to t, but I don't know how they remove it to get the final solution.

they're using the https://www.physicsforums.com/library.php?do=view_item&itemid=353" d/dt (sinθ) = dθ/dt d/dθ (sinθ) = θ'cosθ etc

try it ​

 

[exidez]

ahhh of course they are!

Why do i always forget the fundamentals!

 

[exidez]

Ok so now I am lost again.. I tried...

If i just work out one, i can get the other but i still can't 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 isn't quite the answer I am 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??

Your help is much appreciated

 

[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 ​

 

[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 ??

 

[tiny-tim]

hi exidez!

(try using the X² icon just above the Reply box )

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..

show us how you got that

 

[exidez]

it would be faster if i just took a photo of my working

I changed the variable of M to M_c 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...

 

[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 ​

 

[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

 

[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})

 

[tiny-tim]

ah, this is where you're going wrong …

\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})

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' ​

 

[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 learned and revised a lot just going through this understanding. Thanks again