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 34

Question

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 34 N over 0.6 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

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Orla Orla 6 months 2021-08-12T14:38:24+00:00 1 Answers 1 views 0

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  1. Verity
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    2021-08-12T14:39:47+00:00
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    Answer:

    V=9.2565m/s

    Explanation:

    From the question we are told that:

    Force F = 34 N  

    Time t = 0.6 s

    Length of pedal l_p=16.5cm \approx0.165m

    Radius of wheel r = 33 cm = 0.33 m

    Moment of inertia, I = 1200 kgcm2 = 0.12 kg.m2

    Generally the equation for Torque on pedal \mu is mathematically given by

    \mu=F*L\\\mu=34*0.165

    \mu=5.61N.m

    Generally the equation for  angular acceleration \alpha is mathematically given by

     \alpha=\frac{\mu}{l}

     \alpha=\frac{5.61}{0.12}

     \alpha=46.75

    Therefore Angular speed is \omega

    \omega=\alpha*t

    \omega=(46.75)*(0.6)

    \omega=28.05rad/s

    Generally the equation for  Tangential velocity V is mathematically given by

    V=r\omega

    V=(0.33)(28.05)

    V=9.2565m/s

     

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