Question

ructions: To receive full credit, you must answer all questions and show all derivation of your final answer. 1) For the system shown below the following data is given: k = 1000N/m, k2 = 500 N/m, c-500 N-s/m, m=10 kg, r= 5cm, Jo=1 kg-m2, Fo = 50 N, a = 20 rad/s !3! !3! Pulley, mass moment of inertia J Fsin er Figure 1: springs-mass-pulley system Find the following: a) Equation of motion (EOM) w.r.t the hinge (0) b) Steady-state response of the system, 0(t)

Answer 1

Pulley, mans moment of inertia, so om Fo sink wt 2 Given 1000 NI K Kq = Soo nim, C- 500N-SIM m 10 kg To Set To tym Fo= W ON, 20 rad 1 2) The equation of motion for the moment of force about point o. . -K2 (ro)r - TÖ - K; 4(er) -ci (20)+(Fo sinwt) (ev)- mäler) = 0 --K2 {"(?)}r - Yo Te - Ky() c2r) - ci(x) & Fosinut (2) Märzy)=0 - IT 9Y - 2 K,72 - Y + 2r Fo sinut - mano

[1th Jo term] i + qorin [ 7 kar + 2 Kir] x = Ev to sinut o from given values and putting those values in above Eqreation we get Tulä +508 +119.51- ssin 20t) - Eq" @ The relation for steady response is Tr(t) = x sin (wt-o) - Et © . The marimum aplitude t ī (kamer te weg . [ (12.5-112] + [ 50 x 207]k 1 0.01135699m The phase angle o $ = tant c om $- fant Liais- Cucos) SO x 20 1 - 0,229 13 rad,

Hence the equation of motion of system is * (t) = x sin (wto) = 0.0 135699) sin [wt-(-0.22913) X(t): 0.01135699 sin [wt +0.22913)

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