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Gyroscopic Effect on Two Wheeler
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$sin \ \theta = \frac{a}{h}$

$a \ = \ hsin \ \theta$

$cos \ \theta \ = \ \frac{b}{h}$

$b \ = \ h \ cos \ \theta$

$\sum M_o \ = \ 0 \curvearrowright$ +

$M_g \ (hsin \ \theta) \ – \ Fc . \ (h cos \ \theta) \ – \ c \ = \ 0$

$M_g h \ sin \ \theta \ - \ \frac{Mv^2}{R} \ h cos \ \theta \ – {\ { \frac{v}{r} . \ \frac{v}{R} \ cos \theta \ [ 2 I_w \pm \ I_E. \ G] }} \ = \ 0$

$Mgh \ sin \ \theta \ = \ \frac{Mv^2}{R} .h \ cos \ \theta \ + \ \frac{v^2}{R.r} \ cos \ \theta \ (2 I_w \ \pm \ I_E \ G)$

$\frac{sin \ \theta}{cos \ \theta} = tan \ \theta \ = \ \frac{v^2}{R} \ [ mh \ + \ \frac{(2 \ I_w \ + \ I_E \ G)}{r} \times \frac{1}{Mgh}$

SUMMARY:

$C = W_w . \ W_P \ cos \ \theta \ [ 2 \ I_w \ I_E . \ G]$

where, $W_w = \frac{v}{r}$ & $W_p = \frac{v}{R}$

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