Calculating Time for a Simple Pendulum

Click For Summary
The discussion revolves around calculating the time it takes for a simple pendulum to reach its lowest point when released from a horizontal position. Participants note that while finding velocity as a function of angle is straightforward, integrating to factor in time is more complex. The motion of the pendulum is described by a second-order differential equation, leading to the conclusion that the solution involves elliptic integrals, which can only be computed numerically. There is also mention of existing approximations for the period of a pendulum that extend beyond small-angle assumptions. The conversation highlights the ongoing exploration and refinement of methods for analyzing pendulum motion.
ursubaloo
Messages
5
Reaction score
0
Consider the following problem: (this isn't homework, I thought this problem up myself and I'm wondering how to do it)

-------------------------------------
You have a simple pendulum of mass M and a radius R, which is released from the horizontal. How much time does it take to reach the lowest point, as a function of R?
-------------------------------------

It's easy to find the velocity as a function of the angle, but I couldn't figure out a way to factor time into it. There is also an approximation of the period of an osscilating pendulum which is equal to 2pi*root(L/g), but that holds only for small angle values.
So how do you do it?
 
Physics news on Phys.org
It's easy to find the velocity as a function of the angle, but I couldn't figure out a way to factor time into it.

v=dx/dt-> just separate and integrate.
 
Write the total mechanical energy balance and then use the definition of velocity.

Daniel.

P.S.It's an elliptic integral of the first kind at the end.
 
The equation of motion for a simple pendulum is
\frac{d^2\theta}{dt^2}= -(g/l) sin(\theta).
Since t does not appear explicitely, if we let \omega be the angular speed, we can convert this to
\omega\frac{d\omega}{d\theta}= -(g/l) sin(\theta)
\omega d\omega= -(g/l) sin(\theta)d\theta
which can be integrated to give
\omega^2= (2m/l) cos(\theta)+ C
Taking \theta= \frac{\pi}{2} when \omega= 0
(releasing the pendulum from rest at the horizontal), we get
\omega^2= (2m/)(cos(\theta)- 1)
I presume that is what ursubaloo meant saying "It's easy to find the velocity as a function of the angle".

brentd49 is correct saying "just separate and integrate", except for the word "just". As dextercioby said, that's an elliptic integral and can only be done numerically.
 
Last edited by a moderator:
It's unbelieveble that there are people nowadays which don't use latex when writing an article & post it on "arxiv"... :

Daniel.
 
It's also sometimes unbelievable the feeble stuff some people publish. Elliptic integrals have been investigated for over a 100 years, and many approximations superior to the one given in this reference have been worked out. All they had to do was look.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 3 ·
Replies
3
Views
7K