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.

Question

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.

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Thu Thảo 4 years 2021-07-31T14:49:18+00:00 1 Answers 155 views 0

Answers ( )

  1. MichaelMet
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    2021-07-31T14:50:56+00:00
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    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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