Calculating Velocity & Kinetic Energy of Bicycle Racer Rolling Downhill

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The discussion focuses on calculating the final velocity and kinetic energy of a bicycle racer after losing a wheel while rolling downhill. The initial conditions include a speed of 11.0 m/s, a wheel mass of 2.25 kg, and a height of 75.0 m. The kinetic energy formula used is Kf = 1/2Mv^2 + 1/2(I)w^2, with the moment of inertia calculated as I = MR^2. The user attempts to apply energy conservation principles, calculating initial kinetic energy (Ki) and potential energy (Ui), and is trying to determine the final velocity (Vf) at the bottom of the hill. The calculations lead to a final velocity of 38.8 m/s, but there are uncertainties regarding the angular velocity (w) calculation.
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A bicycle racer is going downhill at 11.0m/s when, to his horror, one of his 2.25kg wheels comes off when he is 75.0m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0cm in diameter and neglect the small mass of the spokes.

I need to get the velocity at the bottom of the hill and KEtotal at the bottom

I know Kf=1/2Mv^2 + 1/2(I)w^2

Vo=11m/s
M=2.25kg
r=.425m

I know I=MR^2
I=.4064kg*m^2

would you use Ki+Ui=Kf+Uf to get w? or is there another way?
 
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No, that looks correct.
 
here was my attempt
Ki=136.125J
Ui=1653.75J
Kf=1/2mv^2+ 1/2Iw^2
Uf=0

for I am stuck on Kf
Its a thin cylinder so I=MR^2 for w did i calculate it correct? v=wr so 11/.425= 25.8rad/s
Ki +Ui=Kf + Uf
so 136.125J +1653.75J= 1/2mv^2 + .4064*(25.8)^2 +0

from here i solved for the v i plugged in 2.25kg into m and i got 38.8m/s as Vf
 
am i messing up on calculating the ang velocity (w)?
 
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