Question

1 m Mm F Sl. In the mechanism shown the circular body with the mass moment of inertia I about 0. i.e. the center of gravity of the body, rotates about O. The T shaped body with the mass m is attached to that circular k body through the joint A such that OA = r. and it translates along the horizontal direction. The linear spring with stiffness k is placed between the T shaped body and the ground such that its deformation is zero when =0. Moreover the circular body is subject to the motor moment Mm while the translating body is exerted the force F. (a) Find the equivalent mass moment of inertia (les) of the whole mechanism about 0. (b) Obtain the equation of motion of the mechanism using leg and the force field of the mechanism. (c) Find the angular acceleration of the rotating body and linear acceleration of the translating body by exploiting the equation of motion of the mechanism for the following values: Мm 100 Nm, k = 5000 Nm-7, r = 20 cm, 0 = 90F = 200 N, w = 20 rads-?, m = 2 kg, I = 1 kgm.

Answer 1

* Solution :-), i.e m Ko 5000 Nm? Mm os 20cm r k O905 I position of the T-shaped body from the pivot o. PAGE- Considering the parameters , conditions from the question Mm 100 Nm m = 2 kg I = 1 kg m Hele considhing an the centre of the slot of the T-shaped RAO RBA body and rigidly attached to it. R,30 -> The rectos Bizo represents the Fa 2001 w = zorad st 2 4 imaginary point B et f B Equation +RRA - RAO * Loop closure RBOR The equation can be decomposed into too scoll equatis follows as

RBO = RAO COSO = scose PAGE-2 19 RBA - rsino Rao sino - Differentiating eq Da wireline we get ÅBo=- rosino 2 Er@ gives the lined relocity of the Ta shaped body. Diffese, tioting eq@ with gespect to time >(3 Reo E-sësing - récoso Equation gives the lined orelesetion of T-shaped Let La BO á R.B0- E-80'sino. The Kinetic energy of the system K.Ea (50+ mé mai) that the spring It is = (It mo sinzo) 0² is unstretched when Oro given the deformation in spring can be with as 0030 --- Then

PAGES defined as here sae IX so; a gcosa (Itma sinto)ő tmshoco The potential energy of system will be PE = 'a xo? (1050-1)? The lagrangian is L: KEPE The vistool Look done by the extesmal forces is W = Mm So of Sal [5 +ms?sino) Ó?_Kx?((050-132] =(Mm +Frsino ) so Genela, lized force F = Mm t fasino Lengrange's Equation given es (10) Substituting LF we get d dt al do cf = Mnt Frsino

PAGE" Equation is the governing differential equation of motion The Co-efficient of ö is equivalent of inestia of system mess moment i e Ies - It me? sino Read ronging © Y mo ócos Osino ö Mm of Fysino +kt+[Cos 0 - )sino It mg sino Substituting the valves. ö 100 +200(0-25 in 6) +500 x(0.234[cos gyvi] sin 96 ~2x0.24 202 Cos 30 sielo 1 + 2 23sin?901 2x0.2 0 --ss.5556 sad 1² From Equation ③ on 3 we caluciate lineal accellation de T- shaped body å card sind od caso =-0.2% (655.5550 sin 90 -0.2x20 x20?cos 90 B = 11:1171 m2 // PLEASE UPVOTE x اد