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

[hello_world30]

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.

 

[anuttarasammyak]

$$\omega_0^2 < 2g/l$$ is transformed to
$$l^2 \omega_0^2 < 2gl$$
$$\frac{1}{2}m l^2 \omega_0^2 < mgl$$
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.

 

[hello_world30]

$$\omega_0^2 < 2g/l$$ is transformed to
$$l^2 \omega_0^2 < 2gl$$
$$\frac{1}{2}m l^2 \omega_0^2 < mgl$$
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 ?

 

[anuttarasammyak]

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.