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?

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