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17 July, 04:27

The moment of inertia of a solid cylinder about its axis is given by 0.5MR2. If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is

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Answers (2)
  1. 17 July, 06:31
    0
    Given that moment of inertia is

    I=0.5MR²

    K. E=?

    We know that

    w=v/R

    Also

    Translational K. Et is 1/2Mv²

    And rotational K. Er is Iw²/2

    Since w=v/R

    Then K. Er=Iv²/2R²

    Also I=0.5MR²

    K. Er=0.5MR²v²/2R²

    K. Er=Mv²/4

    Then the ration of rotational Kinetic energy to transitional Kinetic energy is given as

    K. Er/K. Et

    Mv²/4 : Mv²/2

    Mv²/4 * 2/Mv²

    Then the ratio is 1/2

    The ratio of the rotational kinetic energy to the translational kinetic energy is 1/2 or 0.5
  2. 17 July, 06:58
    0
    Ratio of rotational kinetic energy to translational kinetic energy is 1/2 or 0.5

    Explanation:

    Rotational kinetic energy Er = (1/2) (Iω^ (2))

    Where;

    ω is the angular velocity and I is the moment of inertia around the axis of rotation

    For a solid cylinder moment of Inertia (I) = 0.5 (mr^ (2)) or (mr^ (2)) / 2

    Also, angular velocity (ω) = v/r

    Where v is the linear velocity and r is the radius of curvature.

    Thus plugging these into the equation for Er, we get;

    Er = (1/2) [ (mr^ (2)) / 2][ (v/r) ^ (2) ]

    So, Er = (1/4) [mr^ (2) ][ (v/r) ^ (2) ] = (1/4) mv^ (2)

    Now, formula for translational kinetic energy is Et = (1/2) mv^ (2)

    Thus, ratio or Er to Et is;

    [ (1/4) mv^ (2) ]/[ (1/2) mv^ (2) ] = 1/2 or 0.5
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