Physics Question. Plese explain Steps. Bicycle Forces of motion?

Question

Bicycle and rider weigh 250 Lbs
Force applied at the pedal is 25 Lbs
Wheel Diameter is 20"
Pedal length from center of drive sprocket is 10"
Drive pedal sprocket has 20 teeth
Idle driven sprocket has 35 Teeth
What is the force applied by the tire on the road?
What is the acceleration?
What is the velocity after 20 seconds?
What is the coefficent of friction supposed to be for this to happen?
Must explain the steps

Answer 1

I am assuming that the crank and the tires have no mass, and that their is no friction within the bike, and that the _rolling_ friction against the road is zero. Otherwise, there is not enough information to answer the questions.

> What is the force applied by the tire on the road?

One way to look at this is by conservation of energy. Consider the work you do when you turn the pedal crank by 1 revolution. Using work=force × distance, you get:

Work_in = (F_in)(D_in)
= (25lbs.)(2π)(10in.)

When turn the pedal by one revolution, the tire turns (35/20) =(7/4) revolution, because of the gear ratio. So the distance the bike travels in that time is:

D_out = (7/4)(π)(20in.)

The work done by the road on the bike (Work_out) should be the same as the work you put in with your feet (Work_in), assuming no friction within the bike.

Work_out = work_in
(F_out)(D_out) = (F_in)(D_in)
(F_out)(7/4)(π)(20in.) = (25lbs.)(2π)(10in.)

Solve for "F_out" and you have the answer to part (a).

>What is the acceleration?

Use "F_out = ma"

What is the velocity after 20 seconds?

Use "Δv = at"

> What is the coefficent of friction supposed to be for this to happen?

F_out is the same as the frictional force of the road against the tire. Its maximum possible value is (F_normal)(μ). In other words:

F_out ≤ (F_normal)(μ)

F_normal is just the weight of the bike + rider (which they give you); and you calculated F_out earlier. Solve for μ.