Question

Olt) 1422LLA 1. The system at the right is subject to the harmonic + x(t) force f(t) = Fo sin ot as shown, with amplitude 50 N and a forcing frequency due to a motor (not m shown) with speed = 191 rotations per minute (RPM). Mass m can only translate horizontally and the rod is pinned at point O. The parameters are: r = 5 cm, m= 10 kg, Jo = 1 kg m-, kı = 1000 N/m, ka = 500 N/m, and V kz - Jo, 1 C= 500 N-s/m. pivot a. Convert the motor speed to rad/sec to get o) - point o for the forcing function. b. Confirm for yourself that this is a 1 - DOF system. Determine the relationship between the rod rotation and its translation x? c. What are the equivalent mass and stiffness of the system as seen at the x(t) coordinate system? d. What are the damping ratio and the natural frequency of the system? e. What is the steady-state response of the system?

Answer 1

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Answer's Given data, The harmonic force PCt)= Fosirrot. Amplitude =SON Speed - 191 rotatione per minute CRPM) the parameters are: 8=5cm , m=lokg, Jo = 1 Kg. m²; K1=1000N/m, K2 = Soon/m & C=S00N-5 Ing W is the angel od velocity. then, we z W= 2X*X191 W=20.000 rad/s Degree of freedom of a system is defined as the number of independent inputs needed for the systems to work effectiely.hese, these are two inputs faesent in this system, eithor we can give a rotation to the rod or we can gie a translation to the mars. But these inputs ade dependent on each otheya is if we give an angular motion, it wil alfomatica sexults in the translation of the mass and vice versa.

So this is a single Dof system * Relation between oft) and xCt);- let we give a small rotation do, which will results in translatón by die amount. Then, considering O as the centre of a circle with radius ar upon which we are making do totalion, For lim dd 28 I doo To = 28 for for when doo, when doo de - da then, lim do-so de los ar dx=28.do. mtegrating, we get, X(t)= 20.0thtc at t=0, 0 = 0 and x = 0 >)C=0. (264) = 28. 06t))

© let, t be the tension in the string connecting rod and mass m Fee body dia grain of sod is, Esping = K2xXxo {'Mo=0 = (1280) X-T428 = Jowo (2=w² t = K2820 - Jowo - 28. writing the ervation of motion for the mass for a small displacement oiwe get, mätatk x = 50 Sinut -t. on) mű + Gi+kix+(*27°0-Jows) -5o3irest mx totkat I 28 x 28x0 ?? mx + (atklat /K28² - Jow? - ) X2 = 508 not

00 [4; + kos te test) x+ó = sesinat. equivalement mass malokg. equivalent stilom oss , keq = 4, + Kxd? Jo W2 From the given data, 482 Veq= 1000+ [(500x 018 29-(wa coolz)? Kenq 51000 +120.99 keq = 1120.99N/m @ whakea - 11120.99 I met 1 20 = 10.58 radle critical daming coefficient = 25 km 6 = 2x 51120.99x/ <t - ul: 75 3 4 5 » Damping ratio, & = c m - - 500 211.753 3.2.363 X-