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A moment Mosin wt is applied to the end of the end of the bat of figure.
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Determine the maximum value of Mo such that the steady state amplitude of angular oscillation does not exceed 10˚ if w = 500 rpm, k = 7000 $Nm^-1$ c = 650 $Nsm^-1$ , L = 1.2 m and the mass of the bar is 15kg.

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1] $w \ = \ \frac{0 \pi \times 500}{60} \ = \ 52.35 \ rad/sec$

$KE \ = \ \frac{1}{2} I_o \ \dot{θ}^2$

$= \frac{1}{2} \ [ \frac{mL^2}{12} \ + \ m \ (\frac{L}{4})^2 ] \dot{θ}^2 $

$Ieq \ = \ \frac{mL^2}{12} \ + \ \frac{ml^2}{16}$

$Ieq \ = \ 0.145 \ mL^2$

= 3.15 $m^4$

2] $PE \ = \ (PE)_LS \ + \ (PE)_RS$

$= \frac{1}{2} \ K \ (\frac{1}{4} \theta)^2 \ + \ \frac{1}{2} \times 2k \times (\frac{3L}{4} \theta)^2$

$Keq \ = \ [ \frac{Kl^2}{16} \ + \ \frac{18kL^2}{16}]$

$= \frac{kL^2}{16} \ + \ \frac{9kL^2}{8}$

Keq = 11970 N/m

3] $W_n \ = \ \sqrt{ \frac{K_{eq}}{I_{eq}}} \ = \ 61.64$ rad/sec

4] $W.D \ = \ - \int \ t \ c. \dot{y} \ dy$

$= - \int \ c. \ (\frac{L}{4} \theta) \ d \ (\frac{L}{4} \theta)$

$C_{eq} \ = \ \frac{cL^2}{16}$

$C_{eq} \ = \ 58.5 \frac{Ns}{m}$

$\xi \ = \ \frac{C_{eq}}{Cceq} = \frac{cL^2/16}{2 I_{eq} \ W_n} = 0.1506$

$r = \frac{w}{w_n} = \frac{52.35}{65.38} = 0.849$

5] $\theta_{st} = \frac{Mo}{K_{eq}}$

$\theta_{st} = \ 11970 \ = \ M_o$

6] $\frac{\theta}{\theta_{st}} = \frac{1}{\sqrt{ (1- r^2)^2 \ + \ (2 \xi r)^2}}$

$\frac{0.174}{\theta_{st}} \ = \ \frac{1}{\sqrt{ (1- (0.8)^2 ) \ + \ (2 \times 0.1306 \times 0.8)^2}}$

$\theta_{st} = 0.064$rad

$M_o$ = 76 7. 36 N-m

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