Asmooth circular hoop with a radius of 0.800 m is placed flat on the floor. a 0.300-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 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.
(b) what is the total number of revolutions the particle makes before stopping? assume the friction force remains constant during the entire motion.

a)11.25 J

b)Number of revolution = 1

Explanation:

Given that

Radius ,r= 0.8 m

m= 0.3 kg

Initial speed ,u= 10 m/s

final speed ,v= 5 m/s

a)

Initial energy

KEi= 15 J

Final kinetic energy

KEf=3.75 J

The  energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J

b)

The minimum value to complete the circular arc

Now by putting the values

V= 2.82 m/s

So kinetic energy KE

KE=1.19 J

ΔKE= KEi - KE

ΔKE= 15- 1.19 J

ΔKE=13.80 J

The minimum energy required to complete 2 revolutions = 2 x 11.25 J

                                                                                                    = 22.5 J

Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.

Number of revolution = 1

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