Question

Linearize the equation

Answer 1

\[\large F = l \ddot\theta(\frac{-(M+m)}{cos(\theta)} + mcos\theta) - ml\dot \theta ^2sin(\theta) + (M+m)g\frac{sin(\theta)}{cos(\theta)} \]

Or just
\[\large F = l \ddot\theta(\frac{-(M+m)+ mcos^2\theta}{cos(\theta)}) - ml\dot \theta ^2sin(\theta) + (M+m)gtan(\theta) \]

I am to linearize this equation, so assume \(\large \theta\) is small and \(\large \dot \theta\) is also small

\[\large sin\theta \approx \theta\]
\[\large cos\theta \approx 1 - \frac{\theta^2}{2}\]

From that, I arrive at

\[\large F = l \ddot\theta(\frac{-(M+m)+ m(1 - \frac{\theta^2}{2})^2}{1 - \frac{\theta^2}{2}}) + (M+m)g\theta \]

Which leads to

\[\large F = l \ddot\theta(\frac{-M-m+ m(1 - \theta^2 + \frac{\theta^4}{4})}{1 - \frac{\theta^2}{2}}) + (M+m)g\theta \]

NOT sure where to go from here, i was thinking maybe the top was going to be some type of series and simplify but I'm not seeing it

Point out my mistakes! :D

Answer 2

It's worth noting that it SHOULD simplify down to

\[\large F = -Ml\ddot \theta + (M+m)g\theta\]

Answer 3

THOUGHT!

What if instead of writing this in terms of \(\large cos\) what if I just do \(\large cos(\theta) = \frac{sin\theta}{tan\theta}\) ??

Since we know when theta is small...both \(\large sin\) and \(\large tan\) \(\large \approx \theta\)

Answer 4

If I do that...

\[\large F = l \ddot\theta(\frac{-(M+m)}{cos(\theta)} + mcos\theta) - ml\dot \theta ^2sin(\theta) + (M+m)g\frac{sin(\theta)}{cos(\theta)} \]

Becomes

\[\large F = l \ddot\theta(-(M+m) \frac{tan\theta}{sin\theta} + m\frac{sin\theta}{tan\theta}) + (M+m)g\tan\theta\]

Or just, letting sin and tan go to theta

\[\large F = l \ddot\theta(-(M+m) + m) + (M+m)g\theta\]
\[\large F = l \ddot\theta(-M) + (M+m)g\theta\]

WOO!

Answer 5

Sometimes it's better just to write it out for the 5th time haha

Answer 6

what grade are you in

Answer 7

I'm thinking if you linearize, you would use
\[ \cos \theta \approx 1 \]

Answer 8

That's what I originally thought @phi however upon doing a quick search for small thetas, I saw that both sin and tan went to theta...however cos went to that formula I had above

If what you're saying is true, this would have been a much quicker process haha

Answer 9

I mean, conceptually it DOES make sense, which is why I was surprised when I saw it approached something other than theta

Answer 10

yes, cos = 1 - theta^2 is more accurate, but that is 2nd order
I assume we toss out the theta^2 term
and also we ignore
\[ m l \dot{\theta}^2 \sin \theta\]

Answer 11

Exactly, and yeah, luckily here it lead to the same result using just a little manipulation

I appreciate the insight @phi !

Answer 12

looking at it, if we used \( 1 - \theta^2 \) we also multiply that by \( \ddot{\theta} \)
and we get a 3rd order term, which we can definitely ignore.