A bicycle racer is going downhill at 11.0 m/s when, to his Horror, one of his 2.

Question

A bicycle racer is going downhill at 11.0 m/s when, to his Horror, one of his 2.25-kg wheels comes off as he is 75.0 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. (a) 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 How far will it roll before it stops if the friction coefficient of the road is .35 and it does not hit anything on the way

Explanation / Answer

a) using energy conservation,

initial PE + KE = final KE + PE

(2.25x 11^2 /2 ) + ( 2.25 x r^2 x (11/r)^2 /2 ) + (2.25 x 9.81 x 75)

= (2.25 v^2 /2) + (2.25 x r^2 x (v/r)^2 / 2) + 0

1927.69 = 2.25v^2

v = 29.27 m/s

b) KE = (2.25 x 29.27^2 / 2) + (2.25 x 29.27^2 / 2) = 1927.69 J

c) using work energy theorem,

work done by friction = change in KE

- f . d = 0 - 1927.69

(0.35 x 2.25 x 9.81)d = 1927.69

d = 249.53 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.