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    2019 Fall - Exam2-Ans

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    antoinochang168899

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    2019 Fall - Exam2-Ans

    Uploaded by

    antoinochang168899
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    2019 Fall_Exam2-Ans

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    12 views2 pages

    2019 Fall - Exam2-Ans

    Uploaded by

    antoinochang168899

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
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    Answer sheet

    Part I, Note: 有效位數錯誤者，扣 0.5 分。


    Each of【?a】and【?b】is 2.5 points.
    A【1】 𝑚𝑔(1 − 𝜇√3)⁄(2𝑚 + 𝑀) B【1】 (1.8 or 2)103
    𝐼1 𝐼
    A【2】 (𝜔1 )2 ( 1 − 1) B【2】 (a) (3.9 or 4)102, (b) 0.39 or 0.4
    2 𝐼2

    A【3】 0.35 B【3】 (1.0 or 1)10


    𝐼1 𝐼
    A【4】 (a) (3.9 or 4)102, (b) 0.39 or 0.4 B【4】 (𝜔1 )2 ( 1 − 1)
    2 𝐼2

    A【5】 5.9 or 6 B【5】 (a) = (b) = 1.6 or 2


    A【6】 (a) = (b) = 1.6 or 2 B【6】 5.9 or 6
    A【7】 (1.0 or 1)10 B【7】 0.35
    2𝐿
    A【8】 (a)2π√𝐿𝐶, (𝑏) ln 2 B【8】 (a) 2, (b) 0
    𝑅
    A【9】 (1.8 or 2)103 B【9】 𝑚𝑔(1 − 𝜇√3)⁄(2𝑚 + 𝑀)
    2𝐿
    A【10】 (a) 2, (b) 0 B【10】 (a)2π√𝐿𝐶, (𝑏) ln 2
    𝑅
    A【11】 L/6 B【11】 1.3102
    A【12】 3.71 B【12】 7.0102
    A【13】 1.3102 B【13】 L/6
    A【14】 7.0102 B【14】 3.71
    𝑣 2 𝑥 2 (𝑥 2 +1) 𝑣 2 𝑥 2 (𝑥 2 +1)
    A【15】 1 − 𝑔𝐿(𝑥 2 +1)2 (2𝑥+3) B【15】 1 − 𝑔𝐿(𝑥 2 +1)2 (2𝑥+3)

    A【15】= B【15】
    𝑀𝑥𝐿𝑣 𝑀 𝑀 2 𝑣𝑥
    Angular momentum conservation: = [ 3 (𝑥𝐿)2 + 𝐿 ]𝜔 ⇒ 𝑣𝑥𝐿 = [(𝑥𝐿)2 + 𝐿2 ]𝜔, i.e., 𝜔 = 𝐿(𝑥 2 +1).
    3 3

    To find the maximum angle , use mechanical energy conservation:


    1 𝑀 𝑀 2 𝑀 𝐿
    [ (𝑥𝐿)2 + 𝐿] 𝜔2 = ( 3 𝑥𝐿 + 𝑀 2)𝑔(1 − 𝑐𝑜𝑠𝜃).
    2 3 3

    1 3
    This gives (𝑥 2 + 1)𝐿𝜔2 = 𝑔(𝑥 + 2)(1 − 𝑐𝑜𝑠𝜃).
    2

    Therefore,
    1 (𝑣𝑥)2 3 (𝑣𝑥)2 (𝑥 2 +1)
    (𝑥 2 + 1) = 𝑔 (𝑥 + 2) (1 − 𝑐𝑜𝑠𝜃), 𝑎𝑛𝑑 𝑐𝑜𝑠𝜃 = 1 − 𝑔𝐿(𝑥 2 +1)2 (2𝑥+3) .
    2 𝐿(𝑥 2 +1)2

    1/2
    Part II, Note: 有效位數錯誤者，扣 0.5 分。

    A1 = B3 .
    (a) It is given at the top of exam paper that the moment of inertia for a solid sphere about diameter equals
    2MR2 ⁄5 or 2[𝜌(4⁄3)𝜋𝑅 3 ]𝑅 2⁄5.
    A hollow sphere can be thought of as a solid sphere with its interior dug out.
    So the moment of inertia of this hollow sphere is
    2𝑚 𝑏 5 − 𝑎5
    𝐼 = 2[𝜌(4⁄3)𝜋𝑏 3 ]𝑏 2 ⁄5 − 2[𝜌(4⁄3)𝜋𝑎3 ]𝑎2 ⁄5 =
    5 𝑏 3 − 𝑎3

    𝑚
    where 𝜌 = (4⁄3)𝜋(𝑏3 −𝑎3 ) has been used.

    𝑔sin𝜃
    (b) The acceleration along the slope equals 𝑥̈ = 1+(𝐼⁄𝑚𝑏2 ).

    ℎ 1
    Then according to the formula for a constant acceleration, = 2 𝑥̈ 𝑡 2 , the travel time can be readily obtained
    sin𝜃

    2ℎ 1 2ℎ 𝐼
    as √𝑥̈ 𝑠𝑖𝑛𝜃 = 𝑠𝑖𝑛𝜃 √ 𝑔 [1 + 𝑚𝑏2 ].

    A2 = B1 .
    (a) 0 = √𝑘/𝑚 = 1.2 or 1 Hz.
    (b) Time to reduce to ½ : t = 2m*ln(2)/b = 21x101 s

    Period = 2√𝑚/𝑘 = 7.7 s


    Oscillation number = 27

    A3 = B2 .
    𝑃 1
    (a) 𝐼𝐴 = 𝐴 = 4𝜋∙2.02 = 2.0 × 10−2 𝑊 ∙ 𝑚−2
    𝑃 1
    (b) 𝐼𝐵 = 𝐴 = 4𝜋∙5.02 = 1.6 × 10−2 𝑊 ∙ 𝑚−2
    (c)  = 340/170 = 2 m so destructive interference
    Assume amplitudes of each waves at C are AA and AB
    AA2/AB2 = 1.99/1.59 = 1.25, AA/AB = 1.12
    Resulting amplitude = AAAB = 0.12AB
    I = (0.12AB/AB)2IB = (0.19)21.5910-2 = 2.210-4 W/m2

    2/2

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