A smooth circular hoop with a radius of 0.500 m is placed flat on the floor. A 0

Question

A smooth circular hoop with a radius of 0.500 m is placed flat on the floor. A 0.400-kg particle slides around the inside edge of the hoop. The particle is given the initial speed of 8.00 m/s. After one revolution, its speed has dropped to 6.00 m/s because of the 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

It is W=m(v2^2-v1^2)/2=5.6(J). Friction force. F=5.6/0.5*2pi=1.8(N) minimum speed of the particle so that it still can go up the top of the hoop. v^2/r=g. so v=sqrt(r*g)=2.21 (m/s). delta W=0.2*(6^2-2.21^2)=6.22 5.6

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