(a) 24.6 Nm
The torque produced by the net thrust about the center of the circle is given by:
[tex]\tau = F r[/tex]
where
F is the magnitude of the thrust
r is the radius of the wire
Here we have
F = 0.795 N
r = 30.9 m
Therefore, the torque produced is
[tex]\tau = (0.795 N)(30.9 m)=24.6 N m[/tex]
(b) [tex]0.035 rad/s^2[/tex]
The equivalent of Newton's second law for a rotational motion is
[tex]\tau = I \alpha[/tex]
where
[tex]\tau[/tex] is the torque
I is the moment of inertia
[tex]\alpha[/tex] is the angular acceleration
If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is
[tex]I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2[/tex]
And so we can solve the previous equation to find the angular acceleration:
[tex]\alpha = \frac{\tau}{I}=\frac{24.6 Nm}{707.5 kg m^2}=0.035 rad/s^2[/tex]
(c) [tex]1.08 m/s^2[/tex]
The linear acceleration (tangential acceleration) in a rotational motion is given by
[tex]a=\alpha r[/tex]
where in this problem we have
[tex]\alpha = 0.035 rad/s^2[/tex] is the angular acceleration
r = 30.9 m is the radius
Substituting the values, we find
[tex]a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2[/tex]
