120 points] A man stands on a platform that is rotating (without friction) with

Question

120 points] A man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 kg.m2 If by moving the bricks the man decreases the rotational inertia of the system to 2.0 kg.m2 What are 4. a. The resulting angular speed of the platform b. The ratio of the new kinetic energy of the system to the original kinetic energy? c. What source provided the added kinetic energy?

Explanation / Answer

(A) Applying angular momentum conservation,

Li = Lf

Ii wi = If wf

(6)(1.2) = (2) wf

wf = 3.6 rev/s

(B) KE = I w^2 /2

ratio = KEf / Kei = (2)(3.6^2)/(6)(1.2^2)

= 3  

(C) This comes from man. Man is doing work on system as he bringe bricks closer.  

this work done = gain in KE

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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