Double pendulum struck at distance a from pivot

[JohanL]

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

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

When i have solved similar problems i have used that

$$<br /> (\frac {dT} {d\dot{q}})_f - (\frac {dT} {d\dot{q}})_i = F_x<br />$$

But i don't think this works now.

Any ideas on how to continue?

 

[arildno]

The new velocity of the second rod's C.M must satisfy:
$$\omega\vec{k}\times{l_{1}\vec{i}_{r}-\omega\vec{k}\times{\frac{l_{2}}{2}}\vec{i}_{r}=\vec{v}_{c.m.2}$$
where $$\vec{i}_{r}$$ is the unit vector down along the pendulum system.

Note that if there had been a net impulse couple acting in the joint, i.e, rod 1 imparting an impulse to rod 2, and rod 2 imparting an equal, but oppositely directed impulse on rod 1, then rod 1 would have experienced a change in its angular velocity.
Thus, no such impulse couple was present.
But, therefore, the impulse $$\vec{I}$$ striking at "a" is solely responsible for the perceived change in the momentum of rod 2, that is:
$$\vec{I}=m_{2}(\omega\vec{k}\times(l_{1}-\frac{l_{2}}{2})\vec{i}_{r}-\omega\vec{k}\times(l_{1}+\frac{l_{2}}{2})\vec{i}_{r})=-m_{2}\omega{l}_{2}\vec{k}\times\vec{i}_{r}$$

But, this must be consistent with the change in angular momentum rod 2 experience as a result of $$\vec{I}$$ striking at "a":
Measured, from the C.M of rod 2, we must have:
$$a(-\vec{i}_{r})\times\vec{I}=-\mathcal{I}_{C.M}2\omega\vec{k}$$
where the moment of inertia with respect to the C.M fulfills: $$\mathcal{I}_{C.M}=\frac{m_{2}l_{2}^{2}}{12}$$

Solving for "a", we get $$a=-\frac{l_{2}}{6}$$, i.e, it is below the center of mass (the distance from the joint is therefore $$\frac{2}{3}l_{2}$$

 

[thepatientmental]

To solve this double pendulum mechanics problem, we can use the Lagrangian method to find the equations of motion for the system. The Lagrangian, L, is defined as the difference between the kinetic energy (T) and the potential energy (V) of the system.

In this case, the kinetic energy of the system is given by:

T = 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}

And the potential energy is given by:

V = 1/2*(m_1 + 2m_2)gl_1cos\theta_1 + 1/2*m_2gl_2cos\theta_2

Next, we can use the Euler-Lagrange equations to find the equations of motion for the system:

(d/dt)(∂L/∂\dot{\theta_1}) - (∂L/∂\theta_1) = F_1

(d/dt)(∂L/∂\dot{\theta_2}) - (∂L/∂\theta_2) = F_2

Where F_1 and F_2 are the forces acting on the first and second pendulum, respectively.

To find the value of a that will result in the rods having angular velocity w and -w after the lower rod is struck, we can use the conservation of angular momentum. This means that the total angular momentum of the system before and after the impact should be equal.

Using this information, we can find the values of F_1 and F_2 that will result in the desired angular velocities. From there, we can solve for the value of a that satisfies the conservation of angular momentum equation.

Overall, solving this problem involves using the Lagrangian method and conservation of angular momentum to determine the equations of motion and the value of a that will result in the desired angular velocities for the double pendulum system.