阿吉 發問時間: 科學其他:科學 · 7 年前

求解兩題普通物理題目

1.A pendulum bob of mass m of length L has its motion interrupted by a peg vertically beneath the support at a distance 3L/4 from the support. (a) if the bob released from a horizontal position. What is the tension in the rope at the top of its motion after it has hit the peg? (b) Show that the angle to the vertical at which the pendulum should be released so that the tension at the top of the circuit just vanishes is given by cos Θ = 3/8.

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2.Find the CM of an uniform semicircular plate of radius R and area density x kg/m2 as shown in the Fig.

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1 個解答

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  • 7 年前
    最佳解答

    1. Firstly we calculate the velocity of the bob when it reaches the top: By the formula "potential energy lost"="kinetic energy", we have mg(L-1/4L*2)=1/2mv^2. After calculation, we got v=√(gL) at the top of its movement.

    Secondly, we use Newton's second law which says F=ma. Here, when the bob reaches the top, there are two forces which applies to it, they are: the weight of the bob and the tension of the rope, both of which are pointing downwards. And by the formula for circular motion, we have a=v^2/R where R is the radius of the movement, equals to 1/4L here(after it has hit the peg).

    Therefore, we can write N+mg=m(√(gL))^2/(1/4L)

    After simplification, we obtain N=3mg.

    For the second question, we have nothing to change but the "potential energy lost" in the first formula. Here, it is equal to mg(1/2L-Lcosθ). Making a graph will let you understand better.

    The rest is exactly the same, and finally we want to find cosθ while N=0.

    After simplification, we obtain cosθ=3/8.

    2.I didn't understand your question, sorry.

    2013-12-07 20:05:57 補充:

    All right I've finally understood it... CM means center of mass...

    Firstly we can deduce directly that it is situated on the Y-axis since it's an axis of symmetry.

    I denote λ the density in order that we don't confuse it with the x, which signifies the abscissa.

    The ordinate of the center of mass

    2013-12-07 20:10:12 補充:

    is determined by the formula [S(R,-R)S(√(R^2-x^2),0) λy dy dx] / (λ*π*R^2/2), where S signifies an integral.

    the result is 4R/3π.

    So its coordinates are (0,4R/3π).

    參考資料: Myself
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