Ask Question
11 April, 23:12

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)

+4
Answers (1)
  1. 12 April, 03:11
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers