Question

A bicycle racer is going downhill at 15.0 m/s when, to his horror, one of his 2.42-kg wheels comes off when he is 73.0 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. 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. How much total kinetic energy does the wheel have when it reaches the bottom of the hill? Express your answer in joules.

Answer 1

here,

initial speed of wheel , u = 15 m/s

mass of wheel ,m = 2.42 kg

height , h = 73 m

diameter , d = 85 cm

radius , r = d/2 = 42.5 cm = 0.425 m

let the final speed of wheel be v

using conservation of energy

(0.5 * m * v^2 + 0.5 * I * w^2 ) = (0.5 * m* u^2 + 0.5 * I * w0^2) + m * g * h

(0.5 * m * v^2 + 0.5 * m * r^2 * (v/r)^2 ) = (0.5 * m* u^2 + 0.5 * m * r^2 * (u/r)^2) + m * g * h

v^2 = u^2 + g * h

v^2 = 15^2 + 9.81 * 73

v = 30.7 m/s

the speed of wheel at the bottom of hill is 30.7 m/s

the total kinetic energy does the wheel have when it reaches the bottom of the hill , KE = (0.5 * m * v^2 + 0.5 * I * w^2 )

KE = (0.5 * m * v^2 + 0.5 * m * r^2 * (v/r)^2 )

KE = 2.42 * 30.7^2 J = 2.28 * 10^3 J