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JulsSmile [24]
3 years ago
12

A smooth 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 o

f 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.
Physics
1 answer:
Marrrta [24]3 years ago
4 0

Answer:

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

KE_i=\dfrac{1}{2}mu^2

KE_i=\dfrac{1}{2}0.3\times 10^2

KEi= 15 J

Final kinetic energy

KE_f=\dfrac{1}{2}mv^2

KE_f=\dfrac{1}{2}0.3\times 5^2

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

 V=\sqrt{r.g}

Now by putting the values

V=\sqrt{0.8\times 10}

V= 2.82 m/s

So kinetic energy KE

KE=\dfrac{1}{2}mV^2

KE=\dfrac{1}{2}0.3\times 2.82^2

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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