A bicycle wheel has a radius R = 32.0 cm and a mass M = 1.82 kg which you may assume to be concentrated on the outside radius. A resistive f

Question

A bicycle wheel has a radius R = 32.0 cm and a mass M = 1.82 kg which you may assume to be concentrated on the outside radius. A resistive force f = [07] N (due to the ground) is applied to the rim of the tire. A force F is applied to the sprocket at radius r such that the wheel has an angular acceleration of α = 4.50 rad/s2 . The tire does not slip.(a) If the sprocket radius is 4.53 cm, what is the force F (N)?(b) If the sprocket radius is 2.88 cm, what is the force F (N)?(c) What is the combined mass of the bicycle and rider (kg)?

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Bình An 3 months 2021-07-19T16:27:54+00:00 1 Answers 4 views 0

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    2021-07-19T16:29:07+00:00

    Answer:

    a) The force is 1056.93 N

    b) If the sprocket radius is 2.88 cm, the force is 1662.47 N

    c) The combined mass is 102.08 kg

    Explanation:

    a) The net torque is:

    \tau =I\alpha =mr^{2} \alpha

    Where

    m = 1.82 kg

    r = 32 cm = 0.32 m

    α = 4.5 rad/s²

    Replacing:

    \tau =1.82*0.32^{2} *4.5=0.839Nm

    The force is:

    \tau =rF-(147*0.32)\\F=\frac{\tau +(147*0.32)}{r}

    Where

    r = 4.53 cm = 0.0453 m

    Replacing:

    F=\frac{0.839+47.04}{0.0453} =1056.93N

    b) If r = 2.88 cm = 0.0288 m, replacing in the previous equation:

    F=\frac{0.839+47.04}{0.0288} =1662.47N

    c) The combined mass is:

    m=\frac{F}{a} =\frac{147}{0.32*4.5} =102.08kg

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