# Rain and Centifugal Force Puzzler

1. Jun 21, 2007

### jambaugh

Here is a puzzler I once posed for a fellow physics student. I have the correct answer... but can you figure it out?

Picture a box shaped car with open side windows on a circular track. Suppose it is raining but there is no wind so the rain falls straight down. When the car is stationary the rain will just miss the bottom edge of the side windows so that no rain gets inside...but just barely.

Now for the question. When the car begins moving around the circular track will rain get into the car?

It would seem from a stationary observer's frame that since the rain is vertical and the windows are vertical no rain should ever get into the car.

It would seem from the perspective of someone inside the car that there is a centrifugal acceleration pulling all objects away from the center of the track and by Einstein's equivalence principle this is no different from a gravitational
force doing the same thing. So a few of the rain drops should "curve" in through the window.

Or should they?

So which is it?

2. Jun 21, 2007

### rcgldr

The tracks are exerting a centripetal force on the train car, accelerating it inwards, perpendicular to it's direction of travel. You'd also have turbulent air flow at the edges of the windows.

3. Jun 21, 2007

### jambaugh

Let us work in an idealized setting and neglect wind or turbulence. If you like make it a lunar car and dust falling straight down because I'm up there sprinkling it. Stick to the key issue.

And yes of course the tracks are exerting a centripetal force on the car. Hence it is in an accelerating frame. What happens to the rain/dust which isn't being accelerated by the tracks?

4. Jun 21, 2007

### D H

Staff Emeritus
Neither line of reasoning is "correct".

The first line of reasoning is obviously false. From the point of view of a stationary observer, rain will fall into the inner window. Suppose the windows run the length of the car and the line from the center of rotation to the center of the car is normal to the car's velocity vector. Now examine what happens to a raindrop is exactly even with the upper front corner of the inner window. The distance from the center of the track to this raindrop is greater than the distance from the center of the track to the car. The car will intercept this raindrop will enter the car as it falls.

The second line of reasoning is also incorrect. Einstein's equivalence principal applies only to the inside of a black box. This is not a black box. (It has windows!) Moreover, there is no reason to even think about relativity in a non-relavistic situation.

The velocity of the same raindrop discussed above has a rearward component when observed by someone in the car. However, this velocity vector is still parallel to the car window. Therefore, from an observer inside the car the car stays dry.

5. Jun 21, 2007

### rcgldr

Only if the window is infinitely short from front to back. If the window is at the center of the car, the front part is rotating inwards, while the back part is rotating outwards.

6. Jun 21, 2007

### D H

Staff Emeritus
You do know what means, don't you?

7. Jun 21, 2007

### StatusX

Here's a similar question. There is a ball at rest on the grass inside the track. Will it roll towards you due to centrifugal force?

The answer is obviously no. The ball would appear to orbit the center of rotation at a constant speed, being kept in place by the coriolis force, which acts radially, since:

$$\vec F_{cor} = 2m \vec \omega \times \vec v = 2m (\omega \hat z) \times (v \hat \theta) = -2m \omega v \hat r$$

The centrifugal force is $mv^2/r = m \omega v$, also in the r direction, so is more than balanced by the coriolis force, and the particle maintains its circular orbit.

Of course, its silly to go through these calculation: of course the ball will appear to move in a circle if it is at rest and the frame is rotating.

The vertical motion of the rain makes no difference, since the transformation from the rotating to non-rotating frame doesn't affect the z component.

And DH, jambaugh is perfectly correct to appeal to the equivalence principle here, although he doesn't compute the forces correctly. This applies generally, not just in boxes or relativistic situations. I'll try to explain this more later when I have more time.

Last edited: Jun 21, 2007
8. Jun 21, 2007

### jostpuur

This doesn't seem very relevant for the problem. What if the window was slightly curved, so that its surface would be a part of a big cylinder? So that the window's surface would appear stationary when car moves along the circular track.

I say the solution to the paradox is, that when you use non-inertial frames, you must not only use centrifugal force, but all pseudo forces. Coreolis force is relevant for the rain drops in this case.

EDIT: Three minutes late! :/

9. Jun 21, 2007

### LHarriger

I think the rain could get in but not for the reason listed. Take a look at the attached picture which is supposed to be a highly exagerated top view of the window (heavy line) at two different points along the circular path. The small dot inbetween the window lines is a rain drop falling parrallel to the window. If the time it takes the window to sweep through this path is less than the time it take for the rain drop to fall from the top of the window to the bottom of the window, then the drop will be swept into the car.

#### Attached Files:

• ###### Window.bmp
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10. Jun 21, 2007

### Cyrus

I agree with this logic.

11. Jun 21, 2007

### StatusX

Your analysis is right lharrigger. But the reason the rain gets in the car is not due to centrifugal force, as the OP seemed to imply, but to the particular geometry of this problem. For example, if the window was curved so as to always lie on the same circle, no rain would get it.

12. Jun 22, 2007

### rcgldr

To keep the train car dry inside, every cross section component of the train car needs to be oriented into the direction of travel relative to the rain. Acceleration is not a factor, just the orientation of the window versus the velocity vector. It's similar to having an angle of attack of zero degrees at all points along the window.

13. Jun 22, 2007

### jambaugh

And there you have it. I like this problem for the reason that it is insufficient to just pick one of the two offered explanations but rather point out why each is not quite correct.