Accelerated mathemathical pendulum

  • Thread starter skrat
  • Start date
  • #1
746
8

Homework Statement


Mathemathical pendulum with mass ##m## on a line ##L## is moving with constant acceleration in xz plane, ##\vec{a}=(a_x,a_z)##. Let's consider only the oscillations in xz plane. Find stationary value of ##\theta ## and find the frequency of small movements around the stationary angle if ##a_x=0##.

wfferer.JPG

Homework Equations





The Attempt at a Solution



Ok, let's put one coordinate system somewhere :D and another one on the top of line ##L##. Let's say that these two coordinate systems are connected with vector R.

Let's express everything in accelerated system:

##\vec{r}=(Lsin\theta , Lcos\theta )## and ##\vec{R}=(-\frac{1}{2}a_xt^2,-\frac{1}{2}a_yt^2)##

Both together give me ##v=(\dot{\theta }Lcos\theta -a_xt,-\dot{\theta }Lsin\theta -a_yt)## and ##v^2=\dot{\theta }^2L^2+t^2(a_x^2+a_y^2)+2\dot{\theta }Lt(sin\theta a_y-cos\theta a_x##

So ##L=T-V=\frac{1}{2}m(\dot{\theta }^2L^2+t^2(a_x^2+a_y^2)+2\dot{\theta }Lt(sin\theta a_y-cos\theta a_x)+mgLcos\theta ##.

I won't even write the equations from here one because this brings me to wrong simple solution:

##\ddot{\theta }+\frac{g}{l}\theta =0## which can not be ok. It looks like I am missing something when calculating the velocity? But what?
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,857
1,655
I'd have tried:
In the accelerated frame, rotate the axis to reflect the apparent direction of gravity - you should end up with the same relations as you are used to right?

To recover the problem-statement x and y directions, transform back.

Note: I originally interpreted the diagram as indicating that the pivot was subject to an applied force.
 
  • #3
746
8

So ##L=T-V=\frac{1}{2}m(\dot{\theta }^2L^2+t^2(a_x^2+a_y^2)+2\dot{\theta }Lt(sin\theta a_y-cos\theta a_x)+mgLcos\theta ##.

I won't even write the equations from here one because this brings me to wrong simple solution


That's the problem. If I would write everything one of you would for sure notice that I was not careful.

Anyway, calculating everything ok brings me to

##\ddot{\theta }+\frac{1}{L}(a_xcos\theta +a_ysin\theta +g sin\theta )=0##

And therefore stationary angle ##\theta _0=arctg(\frac{a_x}{a_y+g})##

And for ##a_x=0## also

##\ddot{\theta }+\frac{1}{L}(a_y+g)\theta =0##
 

Related Threads on Accelerated mathemathical pendulum

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
5
Views
1K
Replies
31
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
0
Views
2K
Replies
0
Views
5K
  • Last Post
Replies
3
Views
7K
Replies
0
Views
2K
Top