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.
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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.
 
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\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.
 
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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.
 
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