A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0

Question

A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0.350-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 9.50 m/s. After one revolution, its speed has dropped to 5.50 m/s because of friction with the floor.
(a) Find the energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution.
(b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.

Explanation / Answer

Radius of the hoop r = 0.6 m Mass of the particle m = 0.35 Kg Initial speed u = 9.50 m/s Final speed v = 5.50 m/s (a) Mechanical energy tronsformed W = (1/2)m[u2 - v2]   (in one revolution)                                                      = 0.5(0.35 kg)[(9.50 m/s)2-(5.50 m/s)2]                                                      = 10.5 J (b) The system has a total energy = (1/2)mu2                                                = 15.8 J The energy lost per revolution is 10.5 J Therefore number of revolutions is n = 15.8 J/ J                                                            = 1.5 revolutions                                                = 15.8 J The energy lost per revolution is 10.5 J Therefore number of revolutions is n = 15.8 J/ J                                                            = 1.5 revolutions

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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