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14 June, 22:43

Consider two sizes of disk, both of mass M. One size of disk has radius R; the other has radius 2R. System A consists of two of the larger disks rigidly connected to each other with a common axis of rotation. System B consists of one of the larger disks and a number of the smaller disks rigidly connected with a common axis of rotation. If the moment of inertia for system A equals the moment of inertia for system B, how many of the smaller disks are in system B? 1

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  1. 15 June, 01:44
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    number of smaller disks in system B = 4

    Explanation:

    Moment of Inertia of smaller disk is:

    I_small = (1/2) MR²

    Moment of inertia of larger disk is;

    I_large = (1/2) M (2R) ² = 2MR²

    Now total moment of Inertia of system A will be 2 of the larger disk I_large

    Thus, I_A = 2 x (2MR²) = 4MR²

    While moment of inertia of system B will consist of 1 number of larger disk and n number of smaller disk.

    Thus;

    I_B = 2MR² + n ((1/2) MR²)

    Thus, I_B = MR² (2 + (n/2))

    Since the two systems have equal moment of inertia, we will ewuate I_A to I_B. Thus,

    I_A = I_B gives;

    4MR² = MR² (2 + (n/2))

    MR² will cancel out to give;

    4 = 2 + (n/2)

    n/2 = 4 - 2

    n/2 = 2

    n = 4
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