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Why does vertically falling rain make a slanted steaks on a window?

by PHYSMajor
Tags: falling, rain, slanted, steaks, vertically, window
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PHYSMajor
#19
Jan13-13, 09:30 AM
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Quote Quote by Doc Al View Post
Are you sure it said velocity and not speed?

If the car is moving at 60 mph with respect to the road, then the rain will have a horizontal velocity of 60 mph with respect to the car.

Can you please give the name of your textbook and the problem number.

Okay, so I have two texts here, with similar questions. I shall quote one of the texts, and try to show you which parts confuse me, and why. I'm almost certain I haven't understood reference frames properly.

Problem: "A car travels due east with a speed of 50.0km/h. Rain-drops are falling at a constant speed vertically with respect to the Earth. The traces of the rain on the side windows of the car make an angle of 60.0 with the vertical. Find the velocity of the rain with respect to (a) the car and (b) the Earth."

Solution: (a) - I drew a rt. angle triangle with a 60.0 angle between the vertical and the horizontal. I'm supposed to find the length of the hypotenuse, which would give me the velocity of the raindrops. This gives sin (60.0) = (50.0km/h)/hyp, so hyp. = (50.0km/h)/sin(60.0) = 57.7km/h.

(b) - The book says that the answer is the length of the vertical, which is simply (50.0k/h)/tan (60.0) = 28.9km/h.

Now, I don't understand why the velocity of the raindrops with respect to the earth is the vertical, and not the hypotenuse of my triangle. Likewise, why isn't the velocity of the drops with respect to the car the vertical instead of the hypotenuse? The answers seem reversed to me. Going back to the analogy of the dog and person running side by side at a velocity (a m/s) with respect to the earth, the velocity of the dog with respect to the person would be 0, as the origin is placed on the person and the person itself is moving. With this analogy, wouldn't the drop, which has a horizontal velocity because it is on the car, have a horizontal velocity with respect to the earth, meaning that with respect to the earth, the drop has both a horizontal and a vertical velocity, and thus the velocity in part (b) should be the length of the hypotenuse and not the vertical? I get the impression that once the drop is on the car we are to assume that it has the same horizontal velocity as the car, and because the horizontal velocity of the car was measured with respect to the earth, this should also be the case with the drop.

The way I've phrased the question may make it difficult to understand. It's the best I could do.
cepheid
#20
Jan13-13, 11:04 AM
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The velocity of the drop relative to the earth is the drop velocity seen by an observer who is stationary with respect to the earth. So it's the velocity seen by an observer standing on the side of the road,for example. This is vertical, since it's explicitly stated in the problem that the drops are falling straight down (no wind).

The velocity relative to the car is the velocity seen by an observer who is stationary relative to the car. Eg the driver. I'll use the notation v_a/b to mean "velocity of 'a' relative to 'b'". In this case, to get the drop velocity in the car frame, we just add velocities like so

v_drop/car = v_drop/earth + v_earth/car

In words: the velocity of the drop relative to the car is equal to the velocity of the drop relative to the earth plus the velocity of the earth relative to the car. Note that this is a vector sum.

In his case, since v_drop/earth is vertically
downward, and v_earth/car is westward, the resultant has both downward and westward components. The drop trajectory appears slanted from the car.
Chestermiller
#21
Jan13-13, 11:11 AM
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Quote Quote by Doc Al View Post
The key point is that it slides at an angle due to the relative velocity.

The problem ignores air resistance. Or did you miss that?

Don't complicate a simple problem.
No, I didn't miss it. I just think it's silly to ignore an effect that in real life would be so important. My main issue would be with the problem statement.
PHYSMajor
#22
Jan13-13, 11:19 AM
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Quote Quote by cepheid View Post
The velocity of the drop relative to the earth is the drop velocity seen by an observer who is stationary with respect to the earth. So it's the velocity seen by an observer standing on the side of the road,for example. This is vertical, since it's explicitly stated in the problem that the drops are falling straight down (no wind).
It's true that the v_drop/earth (using your notation here) is vertical before the drops reach the side window. However, when they are on the side window and have "acquired" the horizontal velocity of the car while the car is moving, doesn't v_drop/earth now have a horizontal component as well?

Quote Quote by cepheid View Post
The velocity relative to the car is the velocity seen by an observer who is stationary relative to the car. Eg the driver. I'll use the notation v_a/b to mean "velocity of 'a' relative to 'b'". In this case, to get the drop velocity in the car frame, we just add velocities like so

v_drop/car = v_drop/earth + v_earth/car
Wouldn't the correct formula be v_drop/earth = v_car/earth + v_drop/car? This is the formula I have been using all along. The reason is because I am assuming the earth is stationary, and the car is moving relative to the earth. The drop is also moving relative to the earth, but after it is on the car's window, it would have the horizontal velocity of the car. This would mean (as it seems to me) that relative to the earth, the velocity of the drop is the vector sum of the velocity of the car relative to the earth, and the drop relative to the earth.

Please take note that I am not concerned with the velocity of the drop before it has reached the window. Before it is "on" the window, it only has a vertical velocity, I understand that part perfectly. However, after it is on the window, the velocity of the drop should have a horizontal component, in addition to its original vertical component, no?
Doc Al
#23
Jan13-13, 11:48 AM
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Quote Quote by PHYSMajor View Post
Please take note that I am not concerned with the velocity of the drop before it has reached the window.
But you should be.
Before it is "on" the window, it only has a vertical velocity, I understand that part perfectly.
Vertical with respect to the ground, not the car. You want to know with what velocity the drop hit the car with respect to the car.
However, after it is on the window, the velocity of the drop should have a horizontal component, in addition to its original vertical component, no?
You are confusing yourself by worrying about the fate of the drops once they are on the car. Of course, once the drops are on the car and moving along with it (which will be affected by wind and such), they will now have a horizontal velocity with respect to the road since the car has accelerated them. I suggest that you forget about that and first understand the velocity of the drops as seen from the car just before they hit.
cepheid
#24
Jan13-13, 12:39 PM
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Quote Quote by PHYSMajor View Post
It's true that the v_drop/earth (using your notation here) is vertical before the drops reach the side window. However, when they are on the side window and have "acquired" the horizontal velocity of the car while the car is moving, doesn't v_drop/earth now have a horizontal component as well?
For the second time: this doesn't happen instantaneously. The drops are going to streak across the window at an angle because they hit it at an angle, because that is their velocity relative to the car while falling. This is the point Doc Al has been trying to make for two pages now.


Quote Quote by PHYSMajor View Post
Wouldn't the correct formula be v_drop/earth = v_car/earth + v_drop/car? This is the formula I have been using all along.
Your equation is not different from mine, since v_car/earth = -v_earth/car.

Quote Quote by PHYSMajor View Post
Please take note that I am not concerned with the velocity of the drop before it has reached the window. Before it is "on" the window, it only has a vertical velocity, I understand that part perfectly. However, after it is on the window, the velocity of the drop should have a horizontal component, in addition to its original vertical component, no?
Eventually the drop velocity it will gain a horizontal component (in the Earth frame) but not right away. Hence, in the Earth frame, the car's surface slides horizontally past the drop for a while, before friction gets rid of their relative horizontal motion, and the drop starts being carried along with the car.

EDIT: In the car frame, the drop strikes at an angle and streaks along the car body at that angle, but it's continuously slowing while doing so. Eventually it comes to a stop (no more horizontal motion across the car body) and just continues falling straight down the car body (unless friction is enough to kill this vertical component too).
A.T.
#25
Jan13-13, 04:05 PM
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Quote Quote by Chestermiller View Post
If it weren't for the "no slip boundary condition" and surface tension, the drop would not leave a streak on the window.
And let's not forget the big bang. Without that, no window, no water so no streak.
jtbell
#26
Jan13-13, 04:21 PM
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Quote Quote by Doc Al View Post
The rain clouds are not traveling along inside the car, are they?
If the car is being driven by Joe Btfsplk, the rain clouds might well be traveling along above the car!

http://www.ojaipost.com/wp-content/u...oe-Btfsplk.jpg

http://en.wikipedia.org/wiki/Joe_Btfsplk
sophiecentaur
#27
Jan14-13, 04:47 AM
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It might be better to start with a simpler model than a raindrop on a car. Say a falling ball bearing and how that might appear to the passengers, looking sideways - work up from there. People have been throwing in more and more factors and it just gets too confusing if you don't deal with one thing at a time.


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