5. BICYCLE WHEELS Consider pedaling a typical bicycle. By pushing down on the pe

Question

5. BICYCLE WHEELS Consider pedaling a typical bicycle. By pushing down on the pedal the main sprocket is made to rotate. The chain transfers this rotation to the rear sprocket which causes the rear wheel to turn. A typical bicycle wheel has a radius of 35 cm and a mass of 1.8 kg. The pedals are situated approximately 12 cm from the center of main sprocket. The main sprocket has a radius of 5 cm and the rear sprocket has a radius of 2 cm. Imagine a 70 kg person is starting a bicycle from rest and uniformly accelerates to 11 m/s. (a) Determine an appropriate model for a rotating bicycle wheel and use that to estimate the moment of inertia for the wheel. (b) If the sprockets are essentially massless, what is the change in angular momentum when the bicycle accelerates to the given speed? Do you need to consider both wheels or is just one sufficient? (c) What is the maximum torque the person can exert? Where is this torque exerted? Using this torque, what is the minimum time required for the bicycle to reach the 11 m/s? (d) What linear acceleration does the result from part (c) imply? Comment on the reasonableness of this acceleration. Is it significantly different (either larger or smaller) than you expected? What does this mean for how reasonable this model for an accelerating bicycle is?

Explanation / Answer

a) The wheel of bicycle can be modeled as a circular ring (hoop) of mass M and radius R

(http://en.wikipedia.org/wiki/List_of_moments_of_inertia)

I = MR^2 =1.8*0.35^2 =0.2205 (kg*m^2)

b) the linear speed is bicyvle is the same as the linear speed of a point at the exterior of one wheel.

V =w*R

The change in angular momentum for one wheel (from zero speed to V) is

Delta(L) =I*w =I*V/R

For the entire bicycle (having 2 wheels) the change in angular momentum is

Delta(Ltot) =2L =2I*V/R = 2*0.2205*11/0.35 =13.86 (kg*m^2/s)

c)

The person is exerting toque on the main sprocket. the maximum torque exerted is (D is the distance to pedal from sprocket center)

T =F*D =G*D =m*g*D =70*9.81*0.12 =82.4 (N*m)

Assuming there are no losses in transmission of torque the minim time to accelerate until V=11 m/s is

T =Delta(L)/time

time t = Delta(L)/T = 13.86/82.4 =0.168 sec

d)

linear acceleration is a =V/ t= 11/0.168 =65.4 m/s^2

This value is significantly larger than the real acceleration of a bicycle. This happens because when accelerating, the person needs not only to change the total angular momentum of the wheels, but also to change its own linear momentum (in other words the person needs to overcome its own inertia and self accelerate its own mass). The above model with just 2 weels that need to be accelerated from rest dramatically fails.

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

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