
#19
Jan1313, 09:30 AM

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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. Raindrops 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. 



#20
Jan1313, 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. 



#21
Jan1313, 11:11 AM

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#22
Jan1313, 11:19 AM

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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? 



#23
Jan1313, 11:48 AM

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#24
Jan1313, 12:39 PM

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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). 



#25
Jan1313, 04:05 PM

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#26
Jan1313, 04:21 PM

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http://www.ojaipost.com/wpcontent/u...oeBtfsplk.jpg http://en.wikipedia.org/wiki/Joe_Btfsplk 



#27
Jan1413, 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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