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The motion of spinning a hula hoop around one's hips can be modeled as a hoop rotating around an axis not through the center, but offset from the center by an amount h, where h is less than R, the radius of the hoop. Suppose Maria spins a hula hoop with a mass of 0.73 kg and a radius of 0.60 m around her waist. The rotation axis is perpendicular to the plane of the hoop, but approximately 0.38 m from the center of the hoop.(a) What is the rotational inertia of the hoop in this case? ________ kg m^2 (b) If the hula hoop is rotating with an angular speed of 14.1 rad/s, what is its rotational kinetic energy?

Respuesta :

Answer and Explanation:

Based on the given information, the formula and the computation is given below:

a. The rotational inertia of the hoop is shown below:

[tex]I_H = I_R + Mh^2[/tex]

[tex]= MR^2 + Mh^2[/tex]

[tex]= 0.73 \times (0.60^2 + 0.38^2)[/tex]

= 0.73 × (0.36 +  0.1444)

= 0.368 [tex]kg\ mg^2[/tex]

b. Now the rotational kinetic energy is

[tex]= Half \times Inertia \times omega^2[/tex]

[tex]= 0.5 \times 0.368 \times 14.1^2[/tex]

= 36.58 J

We simply applied the above formula for rotational inertia and rotational kinetic energy in order to reach with the correct answer

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