A man stands at the center of a platform that rotates . without friction with an

Question

A man stands at the center of a platform that rotates . without friction with an angular speed of 3.62 rev/s. Hid arms are outstretched, and he holds a heavy weight in each hand. The moment of inertia of the man, the extended weights, and the platform is 8.4 kg*m^2. When the man pulls the weights inward toward his body, the moment of inertia decreases to 1.5 kg*m^2. What is the resulting angular speed of the platform? What is the kinetic energy of the system before the man pulls the weights to his body? What is the kinetic energy of the system after the man pulls the weights to his body?? What is the change in kinetic energy of the system?

Explanation / Answer

a) Angular momentum of the system is conserved.

I11 = I22

=> 8.4 * 3.65 = 1.52

=> 2 = 20.44 rev/s

b) KEinitial = I12/2 = 8.4 * (3.65 * 2)2 / 2 = 2209.0 J

c) KEfinal = I22/2 = 1.5 * (20.44 * 2)2 / 2 = 12370.4 J

d) KE = 12370.4 - 2209.0 = 10161.4 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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