The motion of spinning a hula hoop around one\'s hips can be modeled as a hoop r

Question

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/?, where h is less than R, the radius of the hoop. Suppose Maria spins a hula hoop with a mass of 0.77 kg and a radius of 0.66 m around her waist. The rotation axis is perpendicular to the plane of the hoop, but approximately 0.44 m from the center of the hoop. What is the rotational inertia of the hoop in this case? kg middot m^2 If the hula hoop is rotating with an angular speed of 13.9 rad/s, what is its rotational kinetic energy? J

Explanation / Answer

from the parallel axis theorem the rotational inertia of the given case is

I = IR+ Mh2= MR2+ Mh2= 0.77*(0.662+ 0.462) = 0.4983Kg.m2

(b)

formula for rotational kinetic energy is

K = 1/2 * I w^2

= 1/2 * 0.4983 * ( 13.9)^2 = 48.14 J

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