Answer :
Answer:
[tex]_T}=24.57Nm[/tex]
ω = 0.0347 rad/s²
a ≅ 1.07 m/s²
Explanation:
Given that:
mass of the model airplane = 0.741 kg
radius of the wire = 30.9 m
Force = 0.795 N
The torque produced by the net thrust about the center of the circle can be calculated as:
[tex]_T } = Fr[/tex]
where;
F represent the magnitude of the thrust
r represent the radius of the wire
Since we have our parameters in set, the next thing to do is to replace it into the above formula;
So;
[tex]_T}=(0.795)*(30.9)[/tex]
[tex]_T}=24.57Nm[/tex]
(b)
Find the angular acceleration of the airplane when it is in level flight rad/s²
[tex]_T}=I \omega[/tex]
where;
I = moment of inertia
ω = angular acceleration
The moment of inertia (I) can also be illustrated as:
[tex]I = mr^2[/tex]
I = ( 0.741) × (30.9)²
I = 0.741 × 954.81
I = 707.51 Kg.m²
[tex]_T}=I \omega[/tex]
Making angular acceleration the subject of the formula; we have;
[tex]\omega = \frac{_T}{I}[/tex]
ω = [tex]\frac{24.57}{707.51}[/tex]
ω = 0.0347 rad/s²
(c)
Find the linear acceleration of the airplane tangent to its flight path.m/s²
the linear acceleration (a) can be given as:
a = ωr
a = 0.0347 × 30.9
a = 1.07223 m/s²
a ≅ 1.07 m/s²