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5 September, 00:55

A bug is on the rim of a 78 rev/min, 12 in. diameter record. The record moves from rest to its final angular speed in 2.91 s. Find the bug's centripetal acceleration 1.5 s after the bug starts from rest. (1 in = 2.54 cm). Answer in units of m/s 2.

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  1. 5 September, 01:00
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    Centripetal Acceleration = 2.701 m/s²

    Explanation:

    Given:

    Starting from rest so initial angular velocity ωi = 0 rad/s

    Final angular velocity ωf = 78 rev/min = 78 * 2π / 60 rad/s = 8.168 rad/s

    Radius of Rim = d/2 = 12/6 = 6 in = 6 * 2.54 cm = 15.24 cm = 0.1524 m

    angular acceleration α = (ωf - ωi) / t = (8.168 rad/s - 0 rad/s) / 2.91 s

    α = 2.807 rad/s²

    now to find angular velocity at 1.5 s we have

    ω = α * t = 2.807 rad/s² * 1.5 s = 4.210 rad/s

    so centripetal acceleration ac = ω² r = (4.210 rad/s) ² * 0.1524 m

    ac = 2.701 m/s²
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