How Does the Galilean Transformation Affect Raindrop Perception in a Moving Car?

[jdefrancesco]

Homework Statement

In a Summer's day, there's no wind, and start to rain. So the drops fall vertically for an observer on the ground. A car has a velocity of 10 Km/h and the driver see that the drops are coming perpendicularly to the windshield. If 60° is the angle between the windshield and the horizontal, determine:
1) The velocity of the drops seen from the earth.
2) The velocity of the drops when hits the windshield.

Homework Equations

$$v_D^C$$ = velocity of the drop for the driver

$$v_C^G$$ = velocity of the car for an observer on the ground

$$v_D^G$$ = velocity of the drop for an observer on the ground

$$v_D^G = v_D^C + v_C^G$$

The Attempt at a Solution

$$v_D^C = Xcos(330°) \hat{i} + Xsen(330°) \hat{j}$$

$$v_C^G = -10 \hat{i}$$

And because the drops are falling vertically:

$$Xcos(330°) \hat{i} - 10 = 0$$

$$X = 11,55 km/h$$

Then, (1):

$$|v_D^G |= Xsin(330) = -5,77 km/h$$

And finally (2):

$$|v_D^C| = 11,55 km/h$$

I don't know if it is correct :P

 

[anathema]

Thank you for your post. I would like to provide some feedback on your solution attempt.

Firstly, your attempt at solving the problem seems to be on the right track. However, there are a few things that need to be addressed.

1) The first equation, v_D^G = v_D^C + v_C^G, is not entirely correct. This equation is used for vector addition, where the final velocity of the drop (v_D^G) is equal to the sum of the velocity of the drop for the driver (v_D^C) and the velocity of the car for an observer on the ground (v_C^G). However, in this problem, we are only interested in the velocity of the drop for an observer on the ground, which can be calculated using the velocity of the drop for the driver and the velocity of the car for an observer on the ground. So the correct equation to use here is v_D^G = v_D^C - v_C^G.

2) In your solution, you have used the angle 330° for the direction of the drop's velocity. However, this angle should be 60°, as mentioned in the problem statement. Also, the angle should be measured from the horizontal, not the vertical.

3) The equation Xcos(330°) \hat{i} - 10 = 0 is not correct. This equation is assuming that the drop's velocity is in the same direction as the car's velocity, which is not the case. The correct equation to use here is Xcos(60°) \hat{i} - 10 = 0.

4) The final answer for the velocity of the drop for an observer on the ground (|v_D^G|) should be a positive value, as the drop is falling towards the ground. So the correct answer would be 5.77 km/h.

5) The final answer for the velocity of the drop when it hits the windshield (|v_D^C|) should be the same as the initial velocity of the drop, which is 11.55 km/h. So the correct answer for (2) would be 11.55 km/h.

I hope this helps to clarify any confusion and provides a more accurate solution. Keep up the good work and keep practicing your problem-solving skills!

 

[nlightner]

The Galilean transformation is a mathematical tool used in classical mechanics to transform coordinates and velocities between different frames of reference. In this problem, the Galilean transformation is used to determine the velocity of raindrops falling from a stationary observer's perspective and from the perspective of a moving car.

Your attempt at a solution is correct. By using the Galilean transformation, you were able to calculate the velocity of the raindrops as seen by the observer on the ground, as well as the velocity of the raindrops when they hit the windshield of the moving car. This shows how the perspective of an observer can affect their perception of velocity and how the Galilean transformation can be used to account for this.

It is important to note that the Galilean transformation is only valid for low velocities and does not account for relativistic effects. In situations where velocities approach the speed of light, the Galilean transformation is replaced by the Lorentz transformation. Overall, the Galilean transformation is a useful tool in classical mechanics, but it has its limitations and cannot fully explain all physical phenomena.