A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 9.0 m/s. (

Question

A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 9.0 m/s. (a) Determine the translational kinetic energy of its center of mass. J (b) Determine the rotational kinetic energy about its center of mass. J (c) Determine its total energy.

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Trúc Chi 4 years 2021-08-12T16:28:08+00:00 2 Answers 17 views 0

Answers ( )

  1. thienhuong
    0
    2021-08-12T16:29:39+00:00
    Reply

    Answer:

    a) 567J

    b) 283.5J

    c) 850.5J

    Explanation:

    given

    Mass of the cylinder, m = 14kg

    Speed of mass, v = 9m/s

    To determine the Translational Kinetic Energy, we use KE = 1/2mv²

    KE(trans) = 1/2 * 14 * 9²

    KE(trans) = 567J

    To determine the Rotational Kinetic Energy, we use = 1/2Iw²

    KE(rot) = 1/2Iw² = 1/2 * 1/2mr² * (v/r) ²

    KE(rot) = 1/4 * mv²

    KE(rot) = 1/4 * 14 * 9²

    KE(rot) = 283.5J

    To determine the Total Energy, we sum up both the transnational and rotational energies = KE(trans) + KE(rot)

    Total energy = 567J + 283.5J

    Total energy = 850.5J

  2. binhan
    0
    2021-08-12T16:29:45+00:00
    Reply

    Answer:

    a) 567J

    b) 283.5J

    c)850.5J

    Explanation:

    The expression for the translational kinetic energy is,

    E_r = \frac{1}{2} mv^2

    Substitute,

    14kg for m

    9m/s for v

    E_r = \frac{1}{2} (14) (9)^2\\= 567J

    The translational kinetic energy of the center of mass is 567J

    (B)

    The expression for the rotational kinetic energy is,

    E_R = \frac{1}{2} Iw^2

    The expression for the moment of inertia of the cylinder is,

    I = \frac{1}{2} mr^2

    The expression for angular velocity is,

    w = \frac{v}{r}

    substitute

    1/2mr² for I

    and vr for w

    in equation for rotational kinetic energy as follows:

    E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2

    = \frac{mv^2}{4}

    E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J

    The rotational kinetic energy of the center of mass is 283.5J

    (c)

    The expression for the total energy is,

    E = E_r + E_R\\\\

    substitute 567J for E(r) and 283.5J for E(R)

    E = 567J + 283.5\\= 850.5J

    The total energy of the cylinder is 850.5J

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