Pendulum in Freefall from CN Tower: Help!

[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!

 

[hemmul]

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!

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 ;)

 

[Andrew Mason]

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!

My first question would be: why would you care about this during the last 10 seconds of your life?

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

 

[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.

 

[Andrew Mason]

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.

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

 

[pinkie]

My first question would be: why would you care about this during the last 10 seconds of your life?

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

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

 

[Andrew Mason]

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

Yes. As in "stopped".

AM

 

[ZapperZ]

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

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

Zz.

 

[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.

 

[hemmul]

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.

huh! i suggest you change password

 

[SpaceTiger]

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.

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

 

[Astronuc]

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

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, let's 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.

 

[SpaceTiger]

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.

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).

 

[SpaceTiger]

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.

 

[Andrew Mason]

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.

You may be right. Can you show this mathematically?

AM

 

[SpaceTiger]

You may be right. Can you show this mathematically?

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.

 

[Andrew Mason]

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

 

[SpaceTiger]

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).