Question

Xosin(ot) Shown in the figure below is a rigid pendulum bar of I= 1 m and mass m, 1 kg attached to a roller, of mass m, = 0.2 kg, whose motion is described by X, sin(@t), Xo = 1 [m]. Model the pendulum bar as a uniform slender rod which has a moment of inertia with respect to its mass centre Ig given by m1/12. Use tangential-normal coordinate system to analyze the dynamics of the pendulum bar and use MATLAB to simulate O(t) for the following questions. А m 0 (1) [21%] Show two free-body-diagrams, one for the roller and one for the pendulum bar. (2) [35%] Derive the equation of motion of 0 in the tangential and normal directions using the Newton's second law. (3) [10%] Show the first-order-form moment equation used in the pendulum simulation. (4) [28%] Given the initial conditions (0) = 60, (0) = 0, let 0,max be the maximum 0that the pendulum will not make a full revolution about the support A in the first 50 seconds. Consider w in (0.5, 3.0] rad/s and carry out the simulation to find the o versus 0,max graph. (5) [20%] Consider the pendulum of the same mass and length but with a different geometrical shape. The moment of inertia Ig is therefore different from the pendulum bar considered above. Consider two geometrical shapes: one is a square of length a = 1/72 m and one is a circle of radius of r=0.5 m; see Figure below. Carry out the simulations using MATLAB to find their respective o versus Oo,max graphs for o in [0.5, 3.0] rad/s. Discuss your results.

Answer 1

DOM5 Page No. DOMS Page to Date Date 7 2 free body diagram roller pendulum bar 120 d42 g sino L txosincet ST tangential direct Dorma direction O=goo d²o d²a mg dt2 dt2 Now we will solve equation a to get time perpod reduce the second order differential equation in first order do dog sino =0 Equation of motion Fa-mgsino ma= - mosino a = -g sino mg -g sino dt² arc length of a pendulum (an be thought as . position of system * + dt dt2 dt L masing 2 do 1 d dt (40) acoso] Integrate to get differential eanath de 2- 29 CoSD = (at are = LB acceleration of a system will now be d20 a = L dt2 with initial conda o'c01=0 B (Ol=00 we can solve forc C= -29 costo L (dzo gsino = 0 de LIJ-100

CORE 13 DOM5 Page No. BOMS Page No Date 10 0.5 2 3 rad/sec Ido 00) - 29/6050-coso.) (dt take o Square root on both sides and ignore negative sign because we are Solving for time LHS is rate of change of angle with respect to time and we are going to find time como period. so we need time with hespect to anglo: so because of this we are going to inverse entire equation and inte - grate from o to Do T=4 2 1 9 V2 coso-Losoo todo dt W2 - 29 (Loso-osobesar) 29.91) (COSCO - COS C00) L 015 1 O = 9.15 maa W = 2 W = 3 similarly to = 18137 Oo = 37.23 Do = 57.22 LIJ-180