Question 13.2: An aircraft having a weight of 250 kN and a tricycle underca...

The horizontal and vertical inertia forces m a_{x} and m a_{y} act at the CG, as shown in Fig. 13.5; m is the mass of the aircraft; and a_{x} and a_{y} its accelerations in the horizontal and vertical directions, respectively. Then, resolving forces horizontally,

from which

Now, resolving forces vertically,

which gives

Then,

a_{y}=\frac{950}{m}=\frac{950}{250 / g}=3.8 g  (i)

Now, taking moments about the CG,

I_{ CG } \alpha-1,200 \times 1.0-400 \times 2.5=0  (ii)

from which

Note that the units of I_{ CG } \alpha are mkNrad but radians are non-dimensional, so are not included but implicit. Hence,

\alpha=\frac{I_{ CG } \alpha}{I_{ CG }}=\frac{2,200 \times 10^{6}}{5.65 \times 10^{8}}=3.9 rad / s ^{2}  (iii)

From Eq. (i), the aircraft has a vertical deceleration of 3.8 g from an initial vertical velocity of 3.7 m/s. Therefore, rom elementary dynamics, the time, t, taken for the vertical velocity to become zero is given by

v=v_{0}+a_{y} t  (iv)

in which v=0 and v_{0}=3.7 m / s. Hence,

from which

In a similar manner to Eq. (iv), the angular velocity of the aircraft after 0.099 s is given by

in which \omega_{0}=0 and \alpha=3.9 rad / s ^{2}. Hence,

that is,