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1 April, 19:36

A 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 20 N over 1.2 seconds 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

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  1. 1 April, 19:55
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    Tangential velocity = 10.9 m/S

    Explanation:

    As per the data given in the question,

    Force = 20 N

    Time = 1.2 S

    Length = 16.5 cm

    Radius = 33.0 cm

    Moment of inertia = 1200 kg. cm^2 = 1200 * 10^ (-4) kg. m^2

    = 1200 * 10^ (-2) m^2

    Revolution of the pedal : revolution of wheel = 1

    Torque on the pedal = Force * Length

    = 20 * 16.5 10^ (-2)

    = 3.30 N m

    So, Angular acceleration = Torque : Moment of inertia

    = 3.30 : 12 * 10^ (-2)

    = 27.50 rad : S^2

    Since wheel started rotating from rest, so initial angular velocity = 0 rad/S

    Now, Angular velocity = Initial angular velocity + Angular Acceleration * Time

    = 0 + 27.50 * 1.2

    = 33 rad/S

    Hence, Tangential velocity = Angular velocity * Radius

    = 33 * 33 * 10^ (-2)

    = 10.9 m/S
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