Question

A bicycle racer is going downhill at 13.0 m/s when, to his horror, one of his 2.14-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.

How fast is the wheel moving when it reaches the foot of the hill if it rolled without slipping all the way down?

How much total kinetic energy does the wheel have when it reaches the bottom of the hill?

Answer 1

here,

the initial speed of wheel , u = 13 m/s

the mass of wheel , m = 2.14 kg

height , h0 = 67 m

diameter , d = 85 cm

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

let the speed of wheel at the bottom be v

using conservation of energy

0.5 * m * v^2 + 0.5 * I * w^2 = (0.5 * I * w0^2 + 0.5 * m * u^2) + m * g * h0

0.5 * m * v^2 + 0.5 * (m * r^2) * (v/r)^2 = (0.5 * (m * r^2) * (u/r)^2 + 0.5 * m * u^2) + m * g * h0

v^2 = u^2 + g * h0

v^2 = 13^2 + 9.81 * 67

solving for v

v = 28.7 m/s

the speed of wheel at the bottom is 28.7 m/s

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

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

KEt = m * v^2 = 2.14 * 28.7^2 J

KEt = 1768.2 J