A bicycle racer is going downhill at 12.9 m/s when, to his horror, one of his 2.

Question

A bicycle racer is going downhill at 12.9 m/s when, to his horror, one of his 2.39 kg wheels comes off when he is 71.0 m above the foot of the hill. We can model the wheel as a thin- walled cylinder 85.0 cm in diameter and neglect the small mass of the spokes. Part A How fast is the wheel moving when it reaches the foot of the hill if it rolled without slipping all the way down? n/s Submit My Answers Give Up Part B How much total kinetic energy does the wheel have when it reaches the bottom of the hill Submit My Answers Give Up Continue

Explanation / Answer

given, iniitla speed while going downhill, u = 12.9 m/s

wheel mass, m = 2.39 kg

distnace from the foot of the hill, h = 71 m

diameter of wheel, 2r = 0.85 m

a. if the wheel rolls down without slipping let its final speed be v

then

from conservation of energy

0.5mu^2 + 0.5Iu^2/r^2 + mgh = 0.5mv^2 + 0.5Iv^2/r^2

now herer I is moment of inertia of the wheel , I = 0.5Mr^2

hence

0.5mu^2 + 0.25mu^2 + mgh = 0.5mv^2 + 0.25mv^2

0.75u^2 + gh = 0.75v^2

0.75*12.9^2 + 9.81*71 = 0.75*v^2

v = 33.092 m/s

b. total kinetic energy of a rolling cyclider = 0.5mv^2 + 0.5Iv^2/r^2 = 0.5mv^2 + 0.5(0.5mr^2)v^2/r^2 = 0.75mv^2

KE total = 0.75*2.39*33.092^2 = 1962.93173172 J

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