I Proving that ##\omega_0^2 < 2g/l ## for a simple pendulum.

AI Thread Summary
The discussion focuses on proving that for a simple pendulum, if the initial angular velocity squared (ω0^2) is less than 2g/l, the string remains in tension and the pendulum's motion stays below the horizontal plane. The user attempts to derive this using the Lagrangian mechanics approach, starting with the kinetic and potential energy equations. They express the condition ω0^2 < 2g/l in terms of energy comparisons, indicating that the kinetic energy at the lowest point is less than the potential energy at the maximum height. The user seeks guidance on applying Lagrange multipliers or alternative methods to address the tension issue. The conversation highlights the intersection of energy conservation and dynamics in analyzing pendulum motion.
hello_world30
Messages
4
Reaction score
0
TL;DR Summary
Using Lagrange multipliers to prove tension as a force of constraint and that # \omega_0^2 < 2gl ##
Here is the problem :
A pendulum is composed of a mass m attached to a string of length l, which is suspended
from a fixed point. When hanging at equilibrium, the pendulum is hit with a horizontal
impulse that results in an initial angular velocity ω0. Show that if ω20 < 2g/l, the string
will always be in tension and that the motion will be confined below a horizontal plane
passing through the suspension point of the string.My attempt at the solution:

L = ## 1/2 ml^2\theta'^2 + mglcos(\theta) ##

Then solving for the Euler Lagrange equations I get ##T = m\theta^" l^2 + mglsin\theta ##

However, I do not know how to proceed from here.
 
Physics news on Phys.org
\omega_0^2 &lt; 2g/l is transformed to
l^2 \omega_0^2 &lt; 2gl
\frac{1}{2}m l^2 \omega_0^2 &lt; mgl
where m is mass of pendulum plomb.
LHS is kinetic energy at the bottom. RHS is potential energy from the bottom corresponding to pendulum angle of 90 degree.
 
  • Like
Likes hello_world30
anuttarasammyak said:
\omega_0^2 &lt; 2g/l is transformed to
l^2 \omega_0^2 &lt; 2gl
\frac{1}{2}m l^2 \omega_0^2 &lt; mgl
where m is mass of pendulum plomb.
LHS is kinetic energy at the bottom. RHS is potential energy from the bottom corresponding to pendulum angle of 90 degree.
is there a way to solve it using lagrange multipliers ?
 
I am not familiar with applying Lagrange method for such a code tension issue.
Balance of virtual centrifugal force, gravity and code tension would be an another method.
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...
Back
Top