A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so that its pendul

Question

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so that its pendulum motion takes 3.00 s. How far from the center of the rod should the pivot be located?

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Sigridomena 5 years 2021-07-18T09:54:05+00:00 1 Answers 272 views 0

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  1. thienan
    0
    2021-07-18T09:55:38+00:00
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    Answer:

    0.087 m

    Explanation:

    Length of the rod, L = 1.5 m

    Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

    time period, T = 3  s

    the formula for the time period of the pendulum is given by

    T = 2\pi \sqrt{\frac{I}{mgd}}    …. (1)

    where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

    Moment of inertia of the rod about the centre of mass, Ic = mL²/12

    By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

    I = Ic + md²

    I = \frac{mL^{2}}{12}+ md^{2}

    Substituting the values in equation (1)

    3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

    9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

    12d² -26.84 d + 2.25 =  0

    d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

    d=\frac{26.84\pm 24.75}{24}

    d = 2.15 m , 0.087 m

    d cannot be more than L/2, so the value of d is 0.087 m.

    Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

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