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 0.75 kg and a radius of 0.62 m around her waist. The rotation axis is perpendicular to the plane of the hoop, but approximately 0.40 m from the center of the hoop. a. What is the rotational inertia of the hoop in this case? b. If the hula hoop is rotating with an angular speed of 13.7 rad/s, what is its rotational kinetic energy?

Explanation / Answer

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

IH = IR + Mh2 = MR2 + Mh2 = 0.75*(0.622 + 0.402) = 0.4083 Kg.m2

b) Rotational Kinetic Energy is given by:

KE = (1/2)I(omega)2 = 0.5*0.4083*13.7^2 = 38.32 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.