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19 September, 08:23

A physical pendulum consists of a uniform solid disk (of radius R 2.35 cm) supported in a vertical plane by a pivot located a distance d 1.75 cm from the center of the disk. The disk is dis - placed by a small angle and released. What is the period of the resulting simple harmonic motion?

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  1. 19 September, 09:22
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    T = 0.3658

    Explanation:

    The expression to use to calculate the period is the following:

    T = 2π √I/Mgd (1)

    Where:

    I: moment of Innertia of pendulum

    g: gravity acceleration (9.81 m/s²)

    d: distance of the pivot

    M: mass of the disk.

    Before we do anything, we will find first the moment of Innertia of the pendulum.

    This can be calculated with the following expression:

    I = 1/2 MR² + Md² (2)

    At the moment we don't have the mass of the disk, but we don't need it, we will express I in function of M, and then, it will be canceled with the M of expression (1). Calculating M we have (Remember that the units of radius and distance should be in meter):

    I = 1/2 M (0.0235) ² + M (0.0175) ²

    I = (2.76x10^-4) M + (3.06x10^-4) M

    I = (5.82x10^-4) M (3)

    Now, we will replace this value in equation (1):

    T = 2π √ (5.82x10^-4) M / (9.81) * (0.0175) M - --> Here M cancels out

    T = 2π √ (5.82x10^-4) / (9.81) * (0.0175)

    T = 2π * 0.0582

    T = 0.3658 s

    This is the period of the pendulum
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