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 0.73 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?

Answer 1

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:

$I_H = I_R + Mh^2$

$= MR^2 + Mh^2$

$= 0.73 \times (0.60^2 + 0.38^2)$

= 0.73 × (0.36 +  0.1444)

= 0.368 $kg\ mg^2$

b. Now the rotational kinetic energy is

$= Half \times Inertia \times omega^2$

$= 0.5 \times 0.368 \times 14.1^2$

= 36.58 J

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