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 0.375-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 7.00 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. rev

Explanation / Answer

Given,

Radius, r = 0.6 m

Mass, m = 0.375 Kg

Initial speed, u = 7 m/s

Final speed, v = 5.50 m/s

A)

Mechanical energy tronsformed,

W = 0.5x m[u^2 - v^2] = 0.5 x 0.375 x (7^2 - 5.5^2) = 3.516 J

B)

System started with 9.19 J of energy and loses 5.67 J

So, 9.19/5.67 = 1 revolution

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