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written 5.4 years ago by |
$sin \ \theta = \frac{a}{h}$
$a \ = \ hsin \ \theta$
$cos \ \theta \ = \ \frac{b}{h}$
$b \ = \ h \ cos \ \theta$
$\sum M_o \ = \ 0 \curvearrowright$ +
$M_g \ (hsin \ \theta) \ – \ Fc . \ (h cos \ \theta) \ – \ c \ = \ 0$
$M_g h \ sin \ \theta \ - \ \frac{Mv^2}{R} \ h cos \ \theta \ – {\ { \frac{v}{r} . \ \frac{v}{R} \ cos \theta \ [ 2 I_w \pm \ I_E. \ G] }} \ = \ 0$
$Mgh \ sin \ \theta \ = \ \frac{Mv^2}{R} .h \ cos \ \theta \ + \ \frac{v^2}{R.r} \ cos \ \theta \ (2 I_w \ \pm \ I_E \ G)$
$\frac{sin \ \theta}{cos \ \theta} = tan \ \theta \ = \ \frac{v^2}{R} \ [ mh \ + \ \frac{(2 \ I_w \ + \ I_E \ G)}{r} \times \frac{1}{Mgh}$
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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