Answer:
maximum isolator stiffness k =1764 kN-m
Explanation:
mean speed of rotation [tex]=\frac{N_1 +N_2}{2}[/tex]
[tex]Nm = \frac{500+750}{2} = 625 rpm[/tex]
[tex]w =\frac{2\pi Nm}{60}[/tex]
=65.44 rad/sec
[tex]F_T = mw^2 e[/tex]
[tex]F_T = mew^2[/tex]
= 0.1*(65.44)^2
F_T =428.36 N
Transmission ratio [tex]=\frac{300}{428.36} = 0.7[/tex]
also
transmission ratio [tex]= \frac{1}{[\frac{w}{w_n}]^{2} -1}[/tex]
[tex]0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}[/tex]
SOLVING FOR Wn
Wn = 42 rad/sec
[tex]Wn = \sqrt {\frac{k}{m}[/tex]
k = m*W^2_n
k = 1000*42^2 = 1764 kN-m
k =1764 kN-m
