A smooth circular hoop with a radius of 0.700 m is placed flat on the floor. A 0

Question

A smooth circular hoop with a radius of 0.700 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 10.00 m/s. After one revolution, its speed has dropped to 4.00 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.
J

(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

a) Mechanical energy converted into internal energy =change in energy = initial kinetic energy - final kinetic energy

= (1/2) m(Vf2 -Vi2)

m= 0.450 kg

Vf = 4 m/s

Vi = 10 m/s

=> change in energy = 18.9 J

b) if frctonal force is constant

Vf2 -Vi2 = 2a x

in first case

42 -102 = 2a (1)

in second case

0 -102 = 2a(n)

dividing we get

84/100 = 1/n

n= 100/84 =1.19 revolutions

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