Question

student decides to give his bicycle a tune up. He flips it upside down (so there’s no friction with the ground) and applies a force of 25 N over 1.0 second to the pedal, which has a length of 16.5 cm. If the back wheel has a radius of 33.0 cm and moment of inertia of 1200 kg·cm^2, what is the tangential velocity of the rim of the back wheel in m/s? Assume he rides a fixed gear bicycle so that one revolution of the pedal is equal to one revolution of the tire. (Round your answer to 1 decimal place for entry into eCampus. Do not enter units. Example: 12.3)

Answer 1

Answer:

11.34 m/s

Explanation:

Force F = 25 N

Lenght of application l = 16.5 cm = 0.165 m

Time of application t = 1 sec

Radius of wheel r = 33.0 cm = 0.33 m

Moment of inertia I = 1200 kg·cm^2 = 0.12 kg-m^2

Let us assume he rides a fixed gear bicycle so that one revolution of the pedal is equal to one revolution of the tire

Solution:

Torque T on pedal = F x l = 25 x 0.165 = 4.125 N-m

Rotational impulse = T x t = 4.125 x 1 = 4.125 N-m-s

Initial momentum of wheel = 0 (since it start from rest)

Final momentum of wheel = I x w

Where w = angular speed

I x w = 0.12w

Change of momentum = 0.12w - 0 = 0.12w

Rotational impulse = momentum change

4.125 = 0.12w

w = 4.125/0.12 = 34.375 rad/s

Tangential velocity of wheel = angular speed x radius of wheel

V = w x r = 34.375 x 0.33 = 11.34 m/s