Answer:
(a) The torque produced by the net thrust is given by the cross product of the thrust force and the radius vector from the center of the circle to the airplane:
τ = r x F
The magnitude of the torque is:
|τ| = r * |F| * sin(θ)
where θ is the angle between the radius vector and the thrust force. Since the thrust force is perpendicular to the radius vector, we have θ = 90°, and sin(θ) = 1. Therefore:
|τ| = r * |F| = (30.1 m) * (0.801 N) = 24.1 N⋅m
So the magnitude of the torque produced by the net thrust is 24.1 N⋅m.
(b) The angular acceleration of the airplane is given by:
α = τ / I
where I is the moment of inertia of the airplane about its center of mass. Since the airplane is flying in a horizontal circle, we can assume that the axis of rotation is perpendicular to the plane of motion and passes through the center of the circle. In this case, the moment of inertia is simply the mass times the square of the radius:
I = mr^2 = (0.746 kg) * (30.1 m)^2 = 674.5 kg⋅m^2
Therefore:
α = τ / I = (24.1 N⋅m) / (674.5 kg⋅m^2) = 0.0357 rad/s^2
So the magnitude of the angular acceleration of the airplane is 0.0357 rad/s^2.
(c) The translational acceleration of the airplane tangent to its flight path is given by:
a = r * α
where r is the radius of the circle. Therefore:
a = (30.1 m) * (0.0357 rad/s^2) = 1.08 m/s^2
So the magnitude of the translational acceleration of the airplane tangent to its flight path is 1.08 m/s^2.