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The wheel has a weight of 5.50 lb, a radius of r=13.0 in, and is rolling in such a way that the center hub, O, is moving to the right at a constant speed of v=17.0 ft/s. Assume all the mass is evenly distributed at the outer radius r of the wheel/tire assembly. What is the total kinetic energy of the bicycle wheel?

Respuesta :

jorgezarate

Answer:

[tex]E_{k}=1589.5ftlb[/tex]  

Explanation:

[tex]E_{k}=E_{movement}+E_{rotational}\\[/tex]    

[tex]E_{k}=\frac{1}{2}mv^{2}+\frac{1}{2}Iw^{2}[/tex]     (1)

For this wheel:

[tex]w=\frac{v}{r}[/tex]

[tex]I=mr^{2}[/tex]:    inertia of a ring

We replace (2) and (3) in (1):

[tex]E_{k}=\frac{1}{2}mv^{2}+\frac{1}{2}(mr^{2})(\frac{v}{r})^{2}=mv^{2}=5.5*17^{2}=1589.5ftlb[/tex]  

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