A smooth circular hoop with a radius of 1.000 m is placed flat on the floor. A 0

Question

A smooth circular hoop with a radius of 1.000 m is placed flat on the floor. A 0.450-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 7.50 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 rev Need Help? Read It

Explanation / Answer

a] Since the particle returns to the same point with a smaller velocity, the only non-conservative force here (friction) does the work in decreasing the velocity

so, energy transformed from mechanical to internal in the system will be:

E = - (1/2)m(v2 - u2) = (1/2)(0.45)(7.52 - 4.52) = 8.1 J

b]

Work Done by friction in bringing the particle to stop will be:

E = - [(1/2)mv2 - (1/2)mu2] = (1/2)(0.45)(7.5)2 = 12.656 J

but 8.1 J is the work done by friction in one loop.

therefore, number of revolutions required to bring the particle to rest will be: n = 12.656/8.1 = 1.5625 rev.

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