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26 April, 01:39

A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the) bottom of a hill, starting from rest. If the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill? A) 3.57mB) 4.28 mC) 3.14 mD) 2.68 m

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  1. 26 April, 03:13
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    (A) = 3.57 m

    Explanation:

    from the question we are given the following:

    diameter (d) = 3.2 m

    mass (m) = = 42 kg

    angular speed (ω) = 4.27 rad/s

    from the conservation of energy

    mgh = 0.5 mv^{2} + 0.5Iω^{2} ... equation 1

    where

    Inertia (I) = 0.5mr^{2}

    ω = / frac{v}{r}

    equation 1 now becomes

    mgh = 0.5 mv^{2} + 0.5 (0.5mr^{2}) (/frac{v}{r}) ^{2}

    gh = 0.5 v^{2} + 0.5 (0.5) (v) ^{2}

    4gh = 2v^{2} + v^{2}

    h = 3v^{2} : 4 g ... equation 2

    from ω = / frac{v}{r}

    v = ωr = 4.27 x (3.2 : 2)

    v = 6.8 m/s

    now substituting the value of v into equation 2

    h = 3v^{2} : 4 g

    h = 3 x (6.8) ^{2} : (4 x 9.8)

    h = 3.57 m
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