Given Information:
Velocity = v = 150 m/s
angle = θ = 12°
Required Information:
Radius of curvature = R = ?
Answer:
Radius of curvature = [tex]R = 1.079 \times 10^{4} \: m[/tex]
Explanation:
Please refer to the attached diagram,
[tex]Fcos(\theta) = m \cdot g\\Fcos(12^{\circ}) = m \cdot g \:\:\:\:\:\:eq. 1[/tex]
Where m is the mass of the plane and g is the gravitational acceleration.
[tex]Fsin(\theta) = \frac{mv^{2}}{R}\\Fsin(12^{\circ}) = \frac{m \cdot v^{2}}{R}\:\:\:\:\:\:eq. 2[/tex]
Where v is the velocity of the plane and R is the radius of curvature of the curved path of the airplane.
Dividing eq. 2 by eq. 1 yields,
[tex]tan(12^{\circ}) = \frac{v^{2}}{R\cdot g }[/tex]
[tex]since \: \frac{sin(\theta)}{cos(\theta)} = tan(\theta)[/tex]
[tex]tan(12^{\circ}) = \frac{v^{2}}{R\cdot g }\\\\R = \frac{v^{2}}{g\cdot tan(12^{\circ}) }\\\\R = \frac{150^{2}}{9.81\cdot 0.212 }\\\\R = 10793\\R = 1.079 \times 10^{4} \: m[/tex]
Therefore, the radius of curvature of the curved path of the airplane is 1.079×10⁴ m
