A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0
ID: 2007377 • Letter: A
Question
A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0.475-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 8.50 m/s. After one revolution, its speed has dropped to 4.50 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 = 0.475 Kg Initial speed u = 8.50 m/s Final speed v = 4.50 m/s (a) Mechanical energy tronsformed W = (1/2)m[u^2 - v^2] (in one revolution) = 0.5 * 0.475 * [ 72.25 - 20.25] = 0.5 * 0.475 * 52 = 12.35 J (b) Minimum velocity to go up the top of the hoop v = sqrt [rg] = sqrt[0.6*9.8] = sqrt[5.88] = 2.424 m/s W = (1/2)m[(8.50)^2 - (2.424)^2] = 0.2375[72.25 - 5.875] = 15.76 J The minimum enegy required to complete two revolutions = 2 * 12.35 = 24.7 J (which is greater than 15.76 J) So its clear that the particle can coplete one revolution only. = 15.76 J The minimum enegy required to complete two revolutions = 2 * 12.35 = 24.7 J (which is greater than 15.76 J) So its clear that the particle can coplete one revolution only.Related Questions
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