A string is wrapped several times around the rim of a small hoop with radius 7cm and mass 2kg. The free end of the string is held in place a

Question

A string is wrapped several times around the rim of a small hoop with radius 7cm and mass 2kg. The free end of the string is held in place and the hoop is released from rest.
After the hoop has descended 80cm, calculate (a) the angular speed of the hoop and (b) the speed of its center.

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Bình An 6 months 2021-07-20T11:59:55+00:00 1 Answers 10 views 0

Answers ( )

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    2021-07-20T12:00:59+00:00

    Answer:

    Explanation:

    Given that,

    The radius of loop r = 7cm = 0.07m

    Mass of hoop M = 2kg

    The hoop is released from rest, the initial velocity is 0m/s

    A. Angular speed of the hoop after a descended of h= 80cm = 0.8m

    Applying the conservation of energy

    Ei + W = Ef

    Where,

    Ei = initial energy of the system, I.e the initial potential and kinetic energy

    Ef = final energy of the system I.e the final potential and kinetic energy

    W = 0, since no external force is acting on the systems

    Ui + K.Ei + 0 = Uf + K.Ef

    System was initially at rest, K.Ei = 0

    Uf = 0 at the zero level

    Then,

    Mgh = ½ •Icm•w² + ½M•Vcm²

    Icm = Mr², since it is circular loop

    Then,

    Mgh = ½ •Mr²•w² + ½M•Vcm²

    M cancels out

    gh = ½ •r²•w² + ½ Vcm²

    Since Vcm = wr

    Then, gh = ½ •r²•w² + ½ w²r²

    gh = r²w²

    w = √(gh/r²)

    w = √(9.81 × 0.8/0.07²)

    w = 40 rad/s

    b. Speed at the center

    Since Vcm = wr

    Vcm = 40×0.07

    Vcm = 2.8 m/s

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