Question

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?

Answer 1

Answer:

 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