1Velocity and Acceleration AnalysisDerivative MethodThe angular velocity of link 1 is constant and has the valuen = 50 rpm and ω =˙φ =πn30= 5.235 rad/s,The velocity is obtained taking the derivative of the position with respect totime, t. The symbolic variable t is introduced in Matlab with the statementsymt = sym(’t’,’real’);The coordinates of the joint B arexB(t) = AB cos φ(t) and yB(t) = AB sin φ(t),and the position vector of B is rB= xBı + xB.To calculate symbolically the position of the joint B, the following Matlabcommands are usedxB = AB*cos(sym(’phi(t)’)); yB = AB*sin(sym(’phi(t)’));rB = [ xB yB 0 ];The statement sym(’phi(t)’) represents the mathematical function φ(t)and is introduced with the command sym that constructs symbolic numbers,variables and objects. The function phi has one argument, the time t.To calculate numerically the position of the joint B, the symbolic variablesneed to be substituted with the input data. To apply a transformation ruleto a particular expression expr, type subs(expr,lhs,rhs). The statementsubs(expr,lhs,rhs) replaces lhs with rhs in the symbolic expression expr.For the mechanism, the numerical values for the joint B arexBn = subs(xB, ’phi(t)’, pi/6); yBn = subs(yB, ’phi(t)’, pi/6);rBn = subs(rB, ’phi(t)’, pi/6);The linear velocity vector of B1= B2isvB1= vB2= ˙xBı + ˙yB,where˙xB=dxBdt= −AB˙φ sin φ and ˙yB=dyBdt= AB˙φ cos φ,2are the components of the velocity vector of B1= B2.To calculate symbolically the components of the velocity vector using theMatlab the command diff(f,t) is used, which gives the derivative of fwith respect to t.The symbolical expression of the velocity vector of B1= B2is obtain withthe statementvB = diff(rB,t);The numerical values for the components of the velocity of B1= B2are˙xB= −0.140 (5.235) sin 30◦= −0.366 m/s,˙yB= 0.140 (5.235) cos 30◦= 0.634 m/s.To obtain the numerical values in Matlab first diff(’phi(t)’,t) is re-placed with omega and then phi(t) is replaced with pi/6vBnn = subs(vB,diff(’phi(t)’,t),omega); vBn = subs(vBnn,’phi(t)’,pi/6);Instead of replacing diff(’phi(t)’,t) with omega and then replac-ing ’phi(t)’ with pi/6, a list with the symbolical variable ’phi(t)’,diff(’phi(t)’,t), and diff(’phi(t)’,t,2) is createdslist = {’phi(t)’, diff(’phi(t)’,t), diff(’phi(t)’,t,2)};Next a list with the numerical values for slist is introducednlist = {pi/6, omega, 0}; % numerical values for slistThe velocities and accelerations need to be calculated at the momentwhen driver link makes and angle φ(t) = π/6 with the horizontal and˙φ(t) =ω and¨φ(t) = ˙ω = 0. To obtain the numerical value for the symbolic vectorrB the following statements are introducedvBn = subs(vB,slist,nlist); VB = double(vBn);The statement double(S) converts the symbolic object S to a numeric object.3The linear acceleration vector of B1= B2isaB1= aB2= ¨xBı + ¨yB,where¨xB=d ˙xBdt= −AB˙φ2cos φ − AB¨φ sin φ,¨yB=d ˙yBdt= −AB˙φ2sin φ + AB¨φ cos φ.For the considered mechanism the angular acceleration of the link 1 is¨φ =˙ω = 0. The numerical values of the acceleration of B are¨xB= −0.140 (5.235)2cos 30◦= −3.323 m/s2,¨yB= −0.140 (5.235)2sin 30◦= −1.919 m/s2.The Matlab command used to calculate symbolically the acceleration vec-tor isaB = diff(vB,t);The numerical value for the vector aB is obtained withaBn = double(subs(aB,slist,nlist));The coordinates of the joint D are xDand yD. The Matlab commands usedto calculate the position of D areeqnD1 = ’( xDsol - xC )^2 + ( yDsol - yC )^2 = CD^2 ’;eqnD2 = ’(yB - yC) / (xB - xC) = (yDsol - yC) / (xDsol - xC)’;solD = solve(eqnD1, eqnD2, ’xDsol, yDsol’);Two sets of solutions are found for the position of the joint D that arefunctions of the angle φ(t) (i.e., functions of time):xDpositions = eval(solD.xDsol); yDpositions = eval(solD.yDsol);xD1 = xDpositions(1); xD2 = xDpositions(2);yD1 = yDpositions(1); yD2 = yDpositions(2);4To determine the correct position of the joint D for the mechanism, anadditional condition is needed. For the first quadrant, 0 ≤ φ ≤ 90◦, thecondition is xD≤ xC.This condition using the Matlab command is:xD1n = subs(xD1,’phi(t)’,pi/6); % xD1 for phi(t)=pi/6if xD1n < xCxD = xD1; yD = yD1;elsexD = xD2; yD = yD2;endrD = [ xD yD 0 ]; % position vector of D in term of phi(t)The linear velocity vector of the joint D3= D4(on link 3 or link 4) isvD3= vD4= ˙xDı + ˙yD,where˙xD=dxDdtand ˙yD=dyDdt,are the components of the velocity vector of the joint D, respectively, on thex-axis and the y-axis.To calculate symbolically the components of this velocity vector the followingMatlab commands are usedvD = diff(rD,t);The numerical solutions using Matlab isvDn = double(subs(vD,slist,nlist));For the considered mechanism the numerical values are˙xD= 0.067 m/s and ˙yD= −0.814 m/s.The linear acceleration vector of D3= D4isaD3= aD4= ¨xDı + ¨yD,5where¨xD=d ˙xDdtand ¨yD=d ˙yDdt.To calculate symbolically the components of the acceleration vector the fol-lowing Matlab commands are used:aD = diff(vD,t);The numerical values of the acceleration of D3= D4are¨xD= 4.617 m/s2and ¨yD= −1.811 m/s2,and can be printed using MatlabaDn = double(subs(aD,slist,nlist)); % numerical value for aDfprintf(’aD3 = aD4 = [ %g, %g, %g ] (m/s^2) \n’, aDn);The angle φ2(t) = φ3(t) is determined as a function of time t from theequation of the slope of the line BC:tan φ2(t) = tan φ3(t) =yB(t) − yCxB(t) − xC.The Matlab function atan(z) gives the arc tangent of the number zand the angle φ2is calculated symbolicallyphi2 = atan((yB-yC)/(xB-xC));The numerical value is given byphi2n = subs(phi2,’phi(t)’,pi/6);The angular velocity ω2(t) = ω3(t) is the derivative with respect to timeof the angle φ2(t)ω2=dφ2(t)dt.Symbolically, the angular velocity ω2= ω3is calculated using Matlab6dphi2 = diff(phi2,t);and is the numerical values are printed using the Matlab statementsdphi2nn = subs(dphi2,diff(’phi(t)’,t),omega);dphi2n = subs(dphi2nn,’phi(t)’,pi/6);fprintf(’omega2 = omega3 = %g (rad/s) \n’, dphi2n);The angular acceleration α2(t) = α3(t) is the derivative with respect totime of the angular velocity ω2(t):α2(t) =dω2(t)dt.Symbolically, using Matlab, the angular acceleration α2isddphi2 = diff(dphi2,t);The numerical values of the angles, angular velocities, and angular accel-erations for the links 2 and 3