Physics, 18.10.2019 17:20 mcsaggers2842

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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Answer from: griffinadrianne946

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 from: wsdafvbhjkl

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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