A jet pilot takes his aircraft in a vertical loop.f the jet is moving at a speed of 1500 km/h at the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.0 g's.
not 5
v^2=gr
r=(v^2)/g
r=(416.67)^2/(6*9.8)
r=2952.56...m
(1500km/h=416.666666 m/s)
To determine the minimum radius of the circle, we need to find the centripetal acceleration at the lowest point of the loop and set it equal to 6.0 g's.
Centripetal acceleration is given by the formula:
a = v^2 / r
where:
a = centripetal acceleration
v = velocity
r = radius
To convert the speed from km/h to m/s, we divide it by 3.6. So, the velocity in m/s would be:
1500 km/h ÷ 3.6 = 416.67 m/s (approximately)
Now, we can substitute the values into the formula:
6.0g = (416.67 m/s)^2 / r
Since 1 g is equal to the acceleration due to gravity (9.8 m/s^2), we can convert 6.0 g's into m/s^2:
6.0 g's × 9.8 m/s^2/g ≈ 58.8 m/s^2
Now, rearranging the formula to solve for r, we have:
r = (v^2) / a
Substituting the values:
r = (416.67 m/s)^2 / 58.8 m/s^2
Calculating this:
r ≈ 2941.5 m
Therefore, the minimum radius of the circle is approximately 2941.5 meters.