SpaceTiger said:
Well, I suppose I could for a specific length dependence of the tension (e.g. for 1/r2, the solutions are identical to those for planetary orbits), but the above is mostly just a motivation for the general treatment of the problem, so there's not a lot of math involved. If you agree with the following statements, I think the treatment would automatically follow:
1) The interior of the elevator is an inertial frame.
2) The only real force involved is the tension.
3) The tension is always along the radius vector from the pivot point.
The rest can be found in most classical mechanics textbooks.
I agree with you. The bob prescribes a circle about the pivot so long as it was in motion when the elevator started to fall.
Let the elevator start falling at the moment the bob amplitude = 0 (speed v is horizontal, tangential to radius). At this point, the tension in the string provides a centripetal deflecting force Fc= mv^2/r where r is the length of the string, m is the mass of the bob and v = its horizontal speed. The string provides a radial force and v is tangential, so there is no torque on the system. Since torque is 0, angular momentum mr^2\omega = mvr is constant. So v is constant.
Analysing it from an inertial frame: From the origin of an inertial frame at rest relative to the initial elevator frame (before the fall), draw radial vectors from the origin to the pendulum pivot and to the bob called \vec{r_1}, \vec{r_2} respectively. The direction of the bob from the pivot is \hat r and \vec r = r \hat r = \vec{r_2} - \vec{r_1}
Before the fall:
(1) \ddot{\vec{r_2}} = \vec g - gcos\theta\hat{r} - \frac{v^2}{r}\hat{r}
(2) \ddot{\vec{r_1}} = 0
Subtract 2 from 1:
\ddot{\vec{r}} = \vec g - gcos\theta\hat{r} - \frac{v^2}{r}\hat{r}
where \ddot{\vec{r}} is the second time derivative of the radial displacement vector from the pivot to the bob.
At the moment the elevator cables are cut:
(3) \ddot{\vec{r_2}} = \vec g - \frac{v^2}{r}\hat{r}
(4) \ddot{\vec{r_1}} = \vec g
Subtracting (4) from (3):
\ddot{\vec{r}} = - \frac{v^2}{r}\hat{r}
which is the equation for circular motion about the pivot.
AM
