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

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Homework Help Overview

The problem involves a bicycle traveling at a speed of 12 km/h on a wet road, focusing on how far ahead the front axle is from where the wheel jets water it picks up. The radius of the wheel is given, and the scenario includes considerations for the presence and absence of a mudguard affecting water separation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations for time and distance of water drops, with initial attempts focusing on the scenario with a mudguard. Questions arise regarding the conditions without the mudguard and the angle of water separation.

Discussion Status

Some participants have provided corrections and clarifications regarding the speed conversion and the initial calculations. There is ongoing exploration of how to approach the scenario without the mudguard, with one participant seeking hints for further understanding.

Contextual Notes

Participants are navigating the complexities of projectile motion and the effects of the mudguard on water separation, indicating a need for clarity on assumptions regarding the angle of separation and the initial conditions of the problem.

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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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