Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

13.1 Aircraft Inertia Loads 385

whichgives

may=950kN

Then

ay=

950

m

=

950

250 /g

=3.8g (i)

NowtakingmomentsabouttheCG

ICGα− 1200 ×1.0− 400 ×2.5=0(ii)

fromwhich

ICGα=2200mkN

Hence,

α=

ICGα

ICG

=

2200 × 106

5.65× 108

=3.9rad/s^2 (iii)

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

v=v 0 +ayt (iv)

inwhichv=0andv 0 =3.7m/s.Hence,

0 =3.7−3.8×9.81t

fromwhich

t=0.099s

InasimilarmannertoEq.(iv),theangularvelocityoftheaircraftafter0.099sisgivenby

ω=ω 0 +αt

inwhichω 0 =0andα=3.9rad/s^2 .Hence,

ω=3.9×0.099

thatis,

ω=0.39rad/s