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 h, where h is less than R, the radius of the hoop. Suppose Maria spins a hula hoop with a mass of 0.74 kg and a radius of 0.68 m around her waist. The rotation axis is perpendicular to the plane of the hoop, but approximately 0.46 m from the center of the hoop.

(a) What is the rotational inertia of the hoop in this case?
______kg · m2

(b) If the hula hoop is rotating with an angular speed of 12.5 rad/s, what is its rotational kinetic energy?

_______J

Explanation / Answer

a) Using parallel axis theorem we can write the following equation

IH = IR + Mh2 = MR2 + Mh2 = 0.74*(0.682 + 0.462) = 0.499 Kg.m2

b) Rotational Kinetic Energy is given by;

KE = (1/2)I(omega)2 = 0.5*0.499*12.5^2 = 38.98 J

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