A person of mass 73kg stands at the center of a rotating merry-go-round platform

Question

A person of mass 73kg stands at the center of a rotating merry-go-round platform of radius 3.5m and moment of inertia 950kg kg m^2 . The platform rotates without friction with angular velocity 0.90rad/s . The person walks radially to the edge of the platform. A)Calculate the angular velocity when the person reaches the edge. B)Calculate the rotational kinetic energy of the system of platform plus person before the person's walk. C)Calculate the rotational kinetic energy of the system of platform plus person after the person's walk.

Explanation / Answer

I final = 950 + mr^2 = 950 + 73*3.5^2 = 1844.25 kg m^2 (a) I1 w1 = I2 w2 950*0.9 = 1844.25*w2 w2 = 0.46 rad/s (b) KE initial = 1/2 I w^2 = 1/2 * 950 * 0.9^2 = 384.75 J (c) KE final = 1/2 I w^2 = 1/2 * 1844.25 * 0.46^2 = 195.12 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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