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Gyroscopic Effect on Shaft Rotating Inside Bearing:
1 Answer
written 5.6 years ago by |
$C = I_w. W_p$
$I = m k^2$
$I = \frac{mr^2}{2} = \frac{4 \times 0.075^2}{2} = 0.011 \ kgm^2$
$w = \frac{2 \pi \times 800}{60} = 83.7755 \ rad/sec$
$w_p = \frac{2 \pi Np}{60} = \frac{2 \pi \times 50}{60} = 5.23$ rad/sec
$C = 0.011 \times 83.7755 \times 5.23$
$C = 4.81 \ Nm$
$\sum M_A = 0 \curvearrowright +$
$- R_B \times 0.080 + 39.24 \times 0.040 \ – \ 4.81 = 0$
$R_B = -40.50N = 40.50N \ (\downarrow)$
$EM_B = 0 \curvearrowright +$
$R_A \ (0.080) – 39.24 \ (0.040) \ – \ 4.81 \ = \ 0$
$R_A \ = \ 79.74 \ N (\uparrow)$