Consider the following rotational system that consists of a

Consider the following rotational system that consists of a rotating bar B_1 with moment of inertia J_1 = 20 Kg m^2 and length L_1 = 2 m supported by a spring-damper suspension as shown in Fig. 1 where k = 1200 N/m and b = 50 N sec/m. A rotating machine results in the excitation force f(t) = 20sin omega t at a distance alpha = 1.2 m from the rotating point O. a) Compute the transfer function G(s) = Theta_1(s)/F(s) where theta_i(t) is the angular displacement of the bar B_1. b) Compute the worst-case excitation frequency omega* that results in the maximum steady-state angular displacement of the bar B_1. c) Compute the steady state amplitude of the angular displacement theta_1(t) at this excitation frequency omega *.

Solution

solution:

1)for given system as spring connected in serieshence resultant K

K=.5k=600

b=50

2)equation of motion is

angle is m

Jm\'\'+bl^2m\'+KL^2m=f(t)

on applying laplace transform we get

m(s)/f(s)=a/(Js^2+bL^2s+KL^2)

0n puuting value we get

m(s)/f(s)=1.2/(20s^2+200s+2400)

3)for F(t)=20sinwt

on applying laplace transform we get

F(s)=20w/(s^2+w^2)

4)on putting value in m(s) we get

m(s)=20wa/(s^2+w^2)(s+5)^2

0n differentiating with respect w and equating to zero,to get maximum rotation m we get that

w=s=-5 rad/s

w=5 rad/s

6)on applying reverse laplace transform we get solution as

for s=w

m(s)=12(1/(s)(s+5)^2)

on solving we get

(1/(s)(s+5)^2)=A/s+B/(s+5)+C/(s+5)^2

for s=-5,5,0

we get A=1/25,B=-1/25,c=-1/5

on solving finally we get

m(t)=9.6(1-e^-5t-5te^-5t)sinwt

and w=5 rad/s

we get

m(t)=9.6(1-e^-5t-5te^-5t)sin5t

and amplitude is

amplitude is=9.6(1-e^-5t-5te^-5t)