What Are the Dynamics of a Bicycle Wheel with External Forces?

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The discussion focuses on the dynamics of a bicycle wheel under various conditions, specifically addressing the calculations of wheel revolutions per minute (rpm), normal force on a mosquito on the wheel, the velocity of mud flying off the wheel, and angular deceleration during braking. The calculations reveal that the wheels rotate at approximately 127.4 rpm, while the normal force on the mosquito is influenced by both gravitational and centripetal forces. The velocity of the mud is determined to be twice the forward speed of the bicycle, and the angular deceleration requires knowledge of the braking force applied over a distance of 10 meters.

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A bicycle is traveling along a flat straight road with a steady
horizontal velocity of 4m/s.
Assume the front and rear wheels have a radius of 0.3m and they
do not slip on the road.

a. How many rpms (revolutions/minute) do the wheels make?[1]
-4m/s *60sec/1min = 240m/min
Circumference: 2pi*r= 0.6pi = 1.884 m/rev
240m/min divided by 1.884m/rev = 127.4rpm

im pretty sure that's right...


b. A mosquito of mass m lands on the top of the wheel as it is going down the
road. When it is on the top of the wheel, what is the normal force felt by the
mosquito. Neglect the force of the air.[1]
-N-mg=ma
-N=ma+mg
N= -(ma+mg) <--- NOT 100% SURE ITS RIGHT.

c. Suppose a piece of mud flies off the very top of the wheel. What is it velocity, Vx and Vy with respect to the ground? Note definition of x-y coordinates. [1]
I HAVE NO IDEA WHERE TO START!

d. How long will the mud take to reach the ground? [1]
THIS ONE EITHER...

e. If the bicycle stops using a constant breaking force over a distance of 10 m, what is the angular deceleration of the wheel?[1]
 
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(a) looks good.
(b) When the wheel is not turning the normal force holding the bug against the wheel is mg. When the bug is moving in circular motion on the wheel, it will feel less pressed against the wheel - as you do when riding in a car going fast over a hill. So it must be mg - ma, where a is the centripetal acceleration.
(c) As the wheel turns, its center is moving at 4 m/s with respect to the ground. The bottom of the wheel is not slipping on the ground, so its speed must be zero. The top of the wheel appears to be moving forward with respect to the rider so it must be going faster than 4 m/s. It turns out to be twice as fast.

(d) With the answer to (c) you can do this. Just projectile motion.
(e) Too bad they don't give the force; you will have an F in your answer.
Just convert the linear motion into circular motion. See http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html#drot for conversion formulas.
 

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