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Hello, I am having trouble applying Lagrangian to this problem:

A uniform thin circular rubber band of mass ##M ## and spring constant k has an original radius ##R##. Now it is tossed into the air. Assume it remains circular when stabilized in air and rotates at angular speed ##\omega## about its center uniformly. Which of the following gives the new radius of the rubber band?

The answer for the equilibrium solution is $$(4\pi^2kR)/(4\pi^2k − M\omega^2 ) $$

This can be found relatively easily through force analysis and Newton's 2nd law.

When trying to find the same solution through Lagrangian, I found the solution comes out to be ##(4\pi^2kR)/(4\pi^2k + M\omega^2 ) ##

Here is how I am using Lagrangian, suppose the radius of the ring is ##r## as it's spinning and it's radial velocity is ##\dot r##

$$L = T-V = 1/2 M v^2 - 1/2 k x^2 = 1/2 M ((r \omega)^2 + \dot r^2) - 1/2 k ( 2 \pi r - 2\pi R)^2$$

Applying Eular-Lagrange equation:

$$\frac{d}{d t} \frac{\partial L}{\partial \dot r} = \frac {\partial L}{\partial r}$$

Because ##\omega## is a function of ##r## and the angular momentum of the system cannot change, this allows the Lagrangian to be written in terms of angular momentum ##\cal L##.

$$L = T-V = 1/2 M v^2 - 1/2 k x^2 = 1/2 {\cal L}^2/r^3 + 1/2 M \dot r^2 - 1/2 k ( 2 \pi r - 2\pi R)^2$$

The algebra quickly leads to

$$M \ddot r = - {\cal L}^2/M r^3 - 4 \pi^2 k (r - R) = - M r \omega^2 - 4 \pi^2 k (r - R)$$

This leads to the equilibrium solution when ##\ddot r = 0##

$$r = (4\pi^2kR)/(4\pi^2k + M\omega^2 ) $$

What am I doing wrong?

Thanks,

A uniform thin circular rubber band of mass ##M ## and spring constant k has an original radius ##R##. Now it is tossed into the air. Assume it remains circular when stabilized in air and rotates at angular speed ##\omega## about its center uniformly. Which of the following gives the new radius of the rubber band?

The answer for the equilibrium solution is $$(4\pi^2kR)/(4\pi^2k − M\omega^2 ) $$

This can be found relatively easily through force analysis and Newton's 2nd law.

When trying to find the same solution through Lagrangian, I found the solution comes out to be ##(4\pi^2kR)/(4\pi^2k + M\omega^2 ) ##

Here is how I am using Lagrangian, suppose the radius of the ring is ##r## as it's spinning and it's radial velocity is ##\dot r##

$$L = T-V = 1/2 M v^2 - 1/2 k x^2 = 1/2 M ((r \omega)^2 + \dot r^2) - 1/2 k ( 2 \pi r - 2\pi R)^2$$

Applying Eular-Lagrange equation:

$$\frac{d}{d t} \frac{\partial L}{\partial \dot r} = \frac {\partial L}{\partial r}$$

Because ##\omega## is a function of ##r## and the angular momentum of the system cannot change, this allows the Lagrangian to be written in terms of angular momentum ##\cal L##.

$$L = T-V = 1/2 M v^2 - 1/2 k x^2 = 1/2 {\cal L}^2/r^3 + 1/2 M \dot r^2 - 1/2 k ( 2 \pi r - 2\pi R)^2$$

The algebra quickly leads to

$$M \ddot r = - {\cal L}^2/M r^3 - 4 \pi^2 k (r - R) = - M r \omega^2 - 4 \pi^2 k (r - R)$$

This leads to the equilibrium solution when ##\ddot r = 0##

$$r = (4\pi^2kR)/(4\pi^2k + M\omega^2 ) $$

What am I doing wrong?

Thanks,

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