Question

A bicycle racer is going downhill at 14.0 m/s when, to his horror, one of his 2.50-kg wheels comes off when he is 67.0 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? Express your answer in meters per second.

Answer 1

Using law of conservation of energy,

mgh + 0.5 mv^2 + 0.5 (mR^2)(v^2/R^2) = 0.5 mv1^2 + (0.5 mR^2)(v1^2/R^2)

mgh + mv^2 = mv1^2

v1^2 = v^2 + gh

V1^I = 14^2 + (9.8 x 67)

Velocity, v = 29.20 m/s

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