Answer:
a) 24.0 N.m b) 3.6*10⁻² rad/s² c) 1.07 m/s²
Explanation:
a) If the force that produces the torque is perpendicular to the tethering wire, we can determine its magnitude just as follows:
τ = F*r = 0.800 N * 30.0 m = 24.0 N*m (1)
b) We can express the torque we found above, using the rotational form of Newton´s 2nd Law, as follows:
τ = I* α (2)
where I is the rotational inertia regarding an axis passing through the center of the circle and α is the angular acceleration of the airplane.
If we consider the airplane as a point mass, the rotational inertia I can be calculated as follows:
I = m*r² = 0.750 Kg * (30.0)² m² = 675 Kg*m²
From (1) and (2), we can solve for α, as follows:
[tex]\alpha = \frac{T}{I} = \frac{24.0 N*m}{675.0 kg*m2} = (3.6e-2) rad/s2[/tex]
⇒α = 3.6*10⁻² rad/s²
c) Applying the definition of the angular velocity, and the definition of an angle, we can find the following realtionship between the linear and angular velocity:
v = ω*r
Dividing both sides by Δt, we can extend this relationship to the linear and angular acceleration, as follows:
a = α*r
a = 3.6*10⁻² rad/s²* 30.0 m = 1.07 m/s²
a. The torque that the net thrust produces about the center of the circle is 24 Newton.
b. The angular acceleration of the airplane is equal to [tex]0.036 \;rad/s^2[/tex]
c. The linear acceleration of the airplane tangent to its flight path is[tex]1.08 \;m/s^2[/tex]
Given the following data:
a. To find the torque that the net thrust produces about the center of the circle:
Mathematically, the torque produced by a perpendicular force is given by:
[tex]T = Fr[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]T = 0.8 \times 30[/tex]
Torque, T = 24 Newton
b. To find the angular acceleration of the airplane, we would use Newton's Second Law of rotational motion:
[tex]T = I\alpha[/tex]
But, [tex]I = mr^2[/tex]
[tex]I = 0.750 \times 30^2\\\\I = 0.750 \times 900[/tex]
Moment of inertia, I = [tex]675\;Kgm^2[/tex]
[tex]\alpha = \frac{T}{I} \\\\\alpha = \frac{24}{675}\\\\\alpha = 0.036 \;rad/s^2[/tex]
c. To find the linear acceleration of the airplane tangent to its flight path:
[tex]a = r\alpha \\\\a = 30 \times 0.036[/tex]
Linear acceleration, a = [tex]1.08 \;m/s^2[/tex]
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