A 3-kg rock swings in a circle of radius 5 m. making one complete trip every 2 seconds. What is the centripetal acceleration of the rock?

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  • oubaas
    Lv 7
    5 months ago

    ω = 2PI/T = 2PI/2 = PI rad/sec

    ac = ω^2*r = 3.1416^2*5 = 49.35 m/sec^2

  • 5 months ago

    Centripetal acceleration = v^2 ÷ r

    During the two seconds, the rock moves a distance of the circumference of a circle that has a radius.5 meters..

    d = 2 * π * 5 = 10 * π meters

    v = d ÷ t = 10 * π ÷ 2 = 5 * π

    The speed of the rock is approximately 15.7 m/s.

    Centripetal acceleration = 25 * π^2 ÷ 5 = 5 * π^2

    This is approximately 49.3 m/s^2. I hope this is helpful for you.

  • 5 months ago

    Centripetal acceleration is v²/r where v is velocity given by time taken to traverse the circumference in one rotation which is 2π*5 / 2 m/s = 5π² m/s.

    So we substitute (5π)² /5 to get Centripetal acceleration is 5π² m/s²

    Given that the mass of the rock is 3 kg, we may determine the force due to centripetal acceleration using the formula Force = mass x acceleration

    F = ma, = 3*5π² = 148.0 m/s²

  • 5 months ago

    The radius is 5 m, so the circumference is:

    C = 2πr

    C = 10π

    The rock completes one revolution in 2 seconds, so its velocity is:

    v = C / t

    v = 10π / 2

    v = 5π

    We can now find the centripetal acceleration:

    a = v² / r

    a = (5π)² / 5

    a = 5π²

    a ≈ 49.3 m/s²

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