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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aquialaska aquialaska · Answer 1 · 2020-05-02T08:21:49+02:00

Answer:

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