How much ahead the front axle does the wheel jet the water that it pick

AI Thread Summary
A bicycle traveling at 12 km/h on a wet road raises the question of how far ahead the front axle the wheel jets water it picks up. Initial calculations determined that with a wheel radius of 35 cm, the time for water droplets to fall to the ground is approximately 0.38 seconds, resulting in a distance of 1.27 meters when corrected for speed. The discussion highlights the complexity of the scenario without a mudguard, suggesting that water droplets may separate at an angle before reaching the highest point of the wheel. The approach involves treating the problem as a two-dimensional projectile motion, requiring the calculation of time of flight and distance based on the angle of separation. This problem emphasizes the importance of understanding projectile motion in real-world applications.
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Homework Statement



A bicycle with a speed 12 km/h is driving along a leveled wet road. How much ahead the front axle (see picture) does the wheel jet the water that it picks up from the floor?The radius of the wheel is r = 35 cm. Let’s suppose that the drops of water that are separating from the highest point of the wheel are flying the farthest; and we suppose that the drops are not separating from the lower part of the wheel because of the mudguard of the wheel is preventing them. What is the solution for the bicycle without the mudguard under the same condition?

Homework Equations



∆y= y- y_0= v_0t- ½ gt²
∆x= x- x_0= v_0t

The Attempt at a Solution



First I calculated time:
∆y= y- y_0= v_0y*t- ½ gt²
y_0= 0
v_0y= 0
and I set for y= (-0.7m), because the drop is falling to the ground

y= -½ gt²
t= 0.38 s

Now I calculated the distance:
∆x= x- x_0= v_0x*t
x_0= 0
v_0x= 12 m/s
t= 0.38 s

x= v_0x*t
x= 4.53 m
 
Last edited:
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It's 12 km/h, not 12 m/s. You didn't get the part without the mudguard. I think the furthest drops will escape before the highest point.
 
Yes, I made a typo, it's 12 km/h= 3.33 m/s!

And x= 1.27 m

Thank you very much!
 
willem2 said:
You didn't get the part without the mudguard. I think the furthest drops will escape before the highest point.

Yeah, I have no idea how to approach this part! Can you give me a hint?

Thanks!
 
If the water separates at an angle \phi before the highest point, it will be at an altitude of 2R - R cos(\phi), it will have a speed of ... (split in x and y components) and start out at a distance ... behind the front axle.

Once you know that, it's just a 2d projectile problem: find the time of flight, and from that find the distance. The answer will depend on \phi of course, and you need to find the maximum by differentiaton.
 
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