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A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0.350-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 5.50 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.

Respuesta :

MrRoyal

Answer:

a. The energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules

b. The total number of revolutions the particle makes before stopping is 1.43 revolutions

Explanation:

a.

Given

m = mass of particle = 0.350-kg

u = initial speed of 10.00 m/s

v = final speed = 5.50 m/s

r = radius = 0.600 m

We assume that the floor is horizontal;

This means that F = mg.

We also assume the rotational kinetic energy to be negligable.

Having listed the assumptions, we proceed as follows;

Let ∆E represent The energy transformed from mechanical to internal in the particle hoop.

This is given by

∆E = KE1 - KE2

Where KE1 = ½mu²

KE2 = ½mv²

So, ∆E = KE1 - KE2 becomes

∆E = ½mu² - ½mv²

∆E = ½m(u² - v²)

∆E = ½ * 0.350 * (10² - 5.5²)

∆E = 12.20625

∆E = 12.21J (Approximated)

Hence, the energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules

b.

Let n = number of revolutions

The relationship between n and the energy is

1/n = (KE1 - KE2)/KE1

Make n the subject of formula

n = KE1 / (KE1 - KE2)

n = ½mu² / (½mu² - ½mv²) --- Simplify

n = ½mu² / (½m(u² - v²)) ---- Divide through by ½m

n = u² / (u² - v²)

n = 10² / (10² - 5.5²)

n = 1.433691756272401

n = 1.43 rev

Hence, the total number of revolutions the particle makes before stopping is 1.43 revolutions

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