# Homework Help: Rolling downhill

1. Nov 11, 2007

### bigtymer8700

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?

2. Nov 11, 2007

### mdk31

No, that looks correct.

3. Nov 11, 2007

### bigtymer8700

here was my attempt
Ki=136.125J
Ui=1653.75J
Kf=1/2mv^2+ 1/2Iw^2
Uf=0

for im 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

4. Nov 11, 2007

### bigtymer8700

am i messing up on calculating the ang velocity (w)?