- #1

JohanL

- 158

- 0

a double pendulum made up of two rods...look at the image below.

The lower rod is struck at a distance a from the point connecting the rods (straight arrow in the image). Before that both rods are at the equilibrium postion and have angular velocity w. Determine a so that the rods have angular velocity w and -w after the lower rod have been struck.

solution:

The lagranian is

[tex]

L = 1/6*(m_1 + 3m_2)l_1^2\dot{\theta_1}^2 + 1/6*m_2l_2^2\dot{\theta_2}^2 + 1/2*m_2l_1l_2cos(\theta_1 -\theta_2})\dot{\theta_1}\dot{\theta_2} + 1/2*(m_1 + 2m_2)gl_1cos\theta_1 + 1/2*m_2gl_2cos\theta_2

[/tex]

When i have solved similar problems i have used that

[tex]

(\frac {dT} {d\dot{q}})_f - (\frac {dT} {d\dot{q}})_i = F_x

[/tex]

But i don't think this works now.

Any ideas on how to continue?

The lower rod is struck at a distance a from the point connecting the rods (straight arrow in the image). Before that both rods are at the equilibrium postion and have angular velocity w. Determine a so that the rods have angular velocity w and -w after the lower rod have been struck.

solution:

The lagranian is

[tex]

L = 1/6*(m_1 + 3m_2)l_1^2\dot{\theta_1}^2 + 1/6*m_2l_2^2\dot{\theta_2}^2 + 1/2*m_2l_1l_2cos(\theta_1 -\theta_2})\dot{\theta_1}\dot{\theta_2} + 1/2*(m_1 + 2m_2)gl_1cos\theta_1 + 1/2*m_2gl_2cos\theta_2

[/tex]

When i have solved similar problems i have used that

[tex]

(\frac {dT} {d\dot{q}})_f - (\frac {dT} {d\dot{q}})_i = F_x

[/tex]

But i don't think this works now.

Any ideas on how to continue?