Analysis Of Engineering Systems - Homework 3

David Clark MAE 488 Homework #3

4.17) Input program: %4.17 % Analysis of the wheel determines: I* (ThetaDD) = Fw * R % Rearranging eq 1 and replacing ThetaDD with XDD % Fw=(I*XDD)/R^2 % Analysis of the spring determines: fs = k*(y-x) % Sum of the forces for the system is % m*XDD=fs-Fw % Substituting everything, you get the final form % (m+I/R^2) * XDD + k*x = k*y % You know L(y) is 1/s since y(t) is a step function % Plug and chug to get x(t) = 1-cos(10*(2/3)^0.5*t m = 10; R=0.3; k=1000; sys=tf(5,[3,21,30]) step(sys) disp('x(t)=1-cos(10*(2/3)^0.5 *t)') Output: Transfer function: 5 ----------------3 s^2 + 21 s + 30 x(t)=1-cos(10*(2/3)^0.5 *t) >>

David Clark MAE 488 Homework #3

4.33) Input program: % 4.33 % sum of the forces (assume the plane is the x axis) % m*xDD=-m*g*sin(Theta)-c*xD % divide by m and rearrage to get % xDD+(c/m)xD=g*sin(Theta) % plug in the knows and you get % xDD+(1/6)xD=4.905 or 6*xDD+xD=29.43 % -- PART B --% Take the laplace of the whole equation from part A % 6*(s*V-4)+V=29.43/s % Rearrage and solve to get % V = 29.43/s - 29.43/(s+(1/6)) + 4/(s+(1/6)) % Inverse laplace and combine the two right terms to get % v(t) = 29.43 - 25.43 exp(-t/6) v=inline('29.43-25.43*exp(-t/6)') % To find steady state, take the derivative and solve for the root % vD=(25.43/6)*exp(-t/6) % t is approximately 103 seconds disp('The velocity at 103 seconds is: '),v(103)

David Clark MAE 488 Homework #3

Output: v= Inline function: v(t) = 29.43-25.43*exp(-t/6) The velocity at 103 seconds is: ans = 29.4300

4.34) Input Program: % 4.34 % After the laplace transfer, the right hand size % of the equation can be solved using residue [r,p,K]=residue( 6000, [40 680 1200 0]) % given the output of residue, the equation is x=inline('0.7692*exp(-15*t)-5.7692*exp(-2*t)+5') % using the same technique for part B [r,p,K]=residue( 6000, [40 400 1200 0]) % given the output of residue, the equation is x=inline('-5*5^5*exp(-5*t)*sin((5^5*t)-5*5^5*exp(-5*t)*cos(5^5*t)') Output: r= 0.7692 -5.7692 5.0000 p= -15 -2 0 K= [] x= Inline function: x(t) = 0.7692*exp(-15*t)-5.7692*exp(-2*t)+5 r= -2.5000 + 5.5902i -2.5000 - 5.5902i

David Clark MAE 488 Homework #3

5.0000 p= -5.0000 + 2.2361i -5.0000 - 2.2361i 0 K= [] x= Inline function: x(t) = -5*5^5*exp(-5*t)*sin((5^5*t)-5*5^5*exp(-5*t)*cos(5^5*t)

4.51 Input %4.51 % FBD of mass 1 yields % m1*x1DD=f(t)-k1*x1-k2*x1+k2*x2 % FBD of mass 2 yields % m2*x2DD=k2*x1-k2*x2 % Laplace these equations (and do some substitution) per work attached % the equations become % X1/F(s)= (s^2+1e3)/(20*s^4+1.1e3s^2+3e7) % X2/F(s)= (1e3)/(20*s^4+1.1e5*s^2+3e7) sys=tf([1,0,1e3],[20,0,1.1e5,0,3e7]) step(sys) Output Transfer function: s^2 + 1000 --------------------------20 s^4 + 110000 s^2 + 3e007

David Clark MAE 488 Homework #3