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

Answer 2

We have that for the Question "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?"

It can be said that

• The acceleration the pilot blacks out = $56.419m/s^2$

• Percent change for the pilot to make one circle = $13.2\%$

 

From the question we are told

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).

Converting the angular velocity to rad/sec

$w = \frac{30.6*2\pi}{60}\\\\= 3.2028rad/sec$

Acceleration,

$a = w^2R\\\\= w^2*\frac{D}{2}\\\\=w^2*\frac{11}{2}\\\\=3.2028^2*5.5\\\\=56.419m/s^2$

Time period, T

$T = \frac{2\pi}{w}\\\\= 2\pi*\sqrt\frac{R}{a}$

acceleration decreased by 22%

$T^1 = 2\pi*\sqrt\frac{R}{0.78a}\\\\T^1 = 2\pi*\sqrt\frac{R}{a} * 1.132\\\\T^1 = 1.132T$

Therefore, change in time

$\delta T = \frac{T_0-T_1}{T_1}*100\\\\= \frac{(1.132)T-T}{T}*100\\\\= 13.2\%$

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