Solution:

$I_w = 2.4 \ kgm^2$

$d = 660 \ mm$

$I_E = 1.2 \ kg.m^2$

$G = 3:1$

$M = 2200 \ kg$

$h = 550 \ mm$

$a = 1.5 \ m$

$R = 80 \ m$

From figure (1) (beside)

$f_c = \frac{Mv^2}{R} = \frac{2200 v^2}{80} = 27.5 v^2$

$W_w = \frac{v}{r} = \frac{v}{0.33} = 3.030 \ V$

$w_p = \frac{v}{R} = \frac{V}{80} = 0.0125 \ V$

$\sum \ M_A = 0, \curvearrowright$ +

$- 2 \ Ri \ (a) \ + \ mg \ (\frac{a}{2}) \ – \ Fc(h) \ –\ c \ = \ 0$

$- 2 \ Ri \ (1.5) \ + \ 2200 \ \times \ 9.81 \ (\frac{1.5}{2}) \ – \ 27.5 \ v^2(0.55) \ – \ 0.499 v^2 \ = \ 0$

$Ri = \frac{16186.4 \ – \ 15.125 \ v^2 \ – \ 0.499 \ v^2}{3}$

For stability:

$R_i \geq 0$

Equate : $R_i = 0$

On solving:

$V = 32.18 \ m/s$

$V = 115.84 \ km/hr$ . . . .(ANS)