# Mechanics - pendulum

1. May 6, 2005

### 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

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

When i have solved similiar problems i have used that

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

But i dont think this works now.

Any ideas on how to continue?

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2. May 6, 2005

### 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 $$\frac{2}{3}l_{2}$$

Last edited: May 6, 2005
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