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2 November, 14:53

A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 53.2 m. At the instant it makes an angle of 34.1° with the vertical as it falls, what are (a) the radial acceleration of the top, and (b) the tangential acceleration of the top. (Hint: Use energy considerations, not a torque.) (c) At what angle θ is the tangential acceleration equal to g? Assume free-fall acceleration to be equal to 9.81 m/s2

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  1. 2 November, 15:05
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    From an energy balance, we can use this formula to solve for the angular speed of the chimney

    ω^2 = 3g / h sin θ

    Substituting the given values:

    ω^2 = 3 (9.81) / 53.2 sin 34.1

    ω^2 = 0.987 / s

    The formula for radial acceleration is:

    a = rω^2

    So,

    a = 53.2 (0.987) = 52.494 / s^2

    The linear velocity is:

    v^2 = ar

    v^2 = 52.949 (53.2) = 2816.887

    The tangential acceleration is:

    a = r v^2

    a = 53.2 (2816.887)

    a = 149858.378 m/s^2

    If the tangential acceleration is equal to g:

    g = r^2 3g / sin θ

    Solving for θ

    θ = 67°
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