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 7.50 m/s. After one revolution, its speed has dropped to 5.00 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. 3 X Your response differs from the correct answer by more than 100%. J (b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion. 3 X Your response differs from the correct answer by more than 10%. Double check your calculations, rev

Explanation / Answer

a) workdone by friction = -change in kinetic energy of the particle

= 0.5*m*(v2^2 - v1^2)

= 0.5*0.375*(5^2 - 7.5^2)

= -5.86 J

b) distance travelled in one revoltuion, d = 2*pi*r

= 2*pi*0.6

= 3.77 m

acceleration of the particle, a = (v2^2 - v1^2)/(2*d)

= (5^2 - 7.5^2)/(2*3.77)

= -4.14 m/s^2

remaining distance travelled before comimg to rest = (vf^2 - v2^2)/(2*a)

= (0^2 - 5^2)/(2*(-4.14))

= 3.02 m

no of revolutions made by the particle = 3.02/(2*pi*0.6)

= 0.801

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