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9 June, 00:55

In a laboratory test of tolerance for high acceleration, a pilot is swung in a circle 13.0 m in diameter. It is found that the pilot blacks out when he is spun at 30.6 rpm (rev/min).

a) At what acceleration (in SI units) does the pilot black out?

b) If you want to decrease the acceleration by 12.0% without changing the diameter of the circle, by what percent must you change the time for the pilot to make one circle?

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  1. 9 June, 01:07
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    a) a = 10,270 m/s² and b) 34%

    Explanation:

    Let's start by reducing the units to the SI system

    w = 13 rpm (2pirad / 1rev) 1 min / 60s) = 1,257 rad / s

    R = D / 2 = 13.0 / 2 = 6.50 m

    a) The formula for centripetal acceleration is

    a = w² R

    a = 1,257 2 6.50

    a = 10,270 m / s²

    b) Let's calculate how much 12% of the acceleration is and subtract them

    12% a = 10.270 12/100 = 1.2324

    The new acceleration is

    a2 = a - 12% a

    a2 = 10.270 - 1.2324

    a2 = 9.0376 m / s²

    Let's calculate the time of a revolution, which we will call period for the two accelerations.

    a = v² / r

    a = (2 π R / T) ² / R = 4 π² R / T²

    T = √4 π² R/a

    T = 2π √ (R/a)

    First acceleration

    T1 = 2 π √6.50/10.270

    T1 = 4.999s

    Acceleration reduced 12%

    T2 = 2 π √6.5/9.0376

    T2 = 6.721 s

    Period change

    ΔT = T2-T1 = 4.999 - 6.721 = - 1.72 s

    Let's calculate the change from the initial period

    % = Δt / T1 100

    % = 1.72 / 4.999 100

    % = 34%

    The period must be reduce by this amount
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