# Pendulum question~

1. Mar 28, 2005

### kiwikiwi79

You are in an elevator at the top of the CN tower. You have a pendulum and allow it to oscillate. The elevator falls to the ground after someone cuts its support cables. What does the pendulum do? Ignore the air resistance acting on the pendulum and the elevator. Help please!!!

2. Mar 29, 2005

### hemmul

Well, while the elevator cable is ok the pendulum will oscillate. The forces that are applied to it are roughly speaking gravitation and tension. But after you cut the cable, if no air resistance etc, the only force that is left is tension. The further motion depends on the phase - i.e. the velocity of the pendulum at cut-moment. It will start/continue circular motion with that velocity.
Well, actually, if the rope is winded over the point of suspension - the pendulum will move along the Archimed Spiral ;)

3. Mar 29, 2005

### Andrew Mason

But the answer is that once you start falling, all parts of the pendulum accelerate at the same rate. If the pendulum was at maximum amplitude and not moving when the elevator started falling, it would simply maintain that position relative to the elevator and to you. But if the pendulum was moving when you started falling, it would continue moving horizontally until the pendulum string was taut and then it would stop.

AM

4. Mar 29, 2005

### Crosson

In a freely falling frame the sum of the outside torque is zero and so the angular momentum is conserved.

Therefore the pendulum will swing around in a circle at constant velocity for duration of the fall.

5. Mar 29, 2005

### Andrew Mason

If it is stopped (maximum amplitude) it has 0 angular momentum, so if fall began at that point, why would it swing in a circle?

AM

6. Mar 30, 2005

### pinkie

I am very new to this stuff, what do you mean by maximum amplitude? Is this the same as 0 velocity?

7. Mar 30, 2005

### Andrew Mason

Yes. As in "stopped".

AM

8. Mar 30, 2005

### ZapperZ

Staff Emeritus
Is there a reason why there's a similar, almost identical question like this in the HW section posted by you?

Zz.

9. Mar 30, 2005

### pinkie

I am guessing that there is someone else from my class on this board who posted here. I didn't notice this was here until today. I posted my question on the homework board yesterday.

10. Mar 30, 2005

### hemmul

huh! i suggest you change password :rofl:

11. Mar 31, 2005

### SpaceTiger

Staff Emeritus
Why would it stop if there was motion perpendicular to the tension?

12. Mar 31, 2005

### Staff: Mentor

Think about the equation(s) of motion. When the elevator is falling (accelerating), the effective force of gravity on the pendulum goes to zero. As a pendulum swings in a plane to maximum height, it stops (magnitude of velocity (speed) = 0), and then reverses direction. If the effective force goes to zero, the velocity of the pendulum will remain the same, which would be zero at maximum height in the planar motion.

Also, lets not forget the effect of angular momentum. There will be tension in the pendulum arm (string) as long as there is motion of the mass at the end of the arm and an effective gravitational force.

Look at the equation that describes tension in the pendulum arm.

Last edited: Mar 31, 2005
13. Mar 31, 2005

### SpaceTiger

Staff Emeritus
His post said if the pendulum was moving when you started falling, it would eventually stop. In the case you're describing, it stopped moving right before falling (maximum height).

14. Mar 31, 2005

### SpaceTiger

Staff Emeritus
It seems that, once you start falling, you're dealing simply with a central force problem. That is, you have a pendulum bob with some initial angular momentum and a tension which pulls the bob in a perpendicular direction by some function of the distance from the pivot point. There should be no other forces at work. Assuming there's no dissipation, the pendulum should enter an "orbit" of some kind about the pivot point. The details of this orbit would of course depend upon the length dependence of the tension.

15. Mar 31, 2005

### Andrew Mason

You may be right. Can you show this mathematically?

AM

16. Mar 31, 2005

### SpaceTiger

Staff Emeritus
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.

17. Apr 1, 2005

### Andrew Mason

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

Last edited: Apr 1, 2005
18. Apr 1, 2005

### SpaceTiger

Staff Emeritus
Yeah, I suppose that shows the validity of my treatment of the interior of the elevator as an inertial frame. Also, that deals with the special case of a perfectly tight rope. The treatment I described would have allowed for an arbitrary spring constant in the rope as well (but only after the fall starts).