Solve the chain rule in lagrange equation

Hai guys,

I would like to ask your opinion about this matter. I have a code as shown below which basically, is the Lagrange equation. I have reached to the Lagrange equation and next, I would like to differentiate the Lagrange equation.

Basically, the Lagrange eqaution is as :

 L = KE - PE

As shown in the pic, now, I would like to differentiate the L to the r_dot. Could anyone advice me? I have try several ways, but it does not give me the correct answer. Thanks in advance

Regards,

Siti

syms x1(t) x2(t) x3(t) x4(t)
syms a la 
syms m1 m2 m3 m4  I1 I2 I3 I4 
syms g h1 h2 h3 h4

T0_1=[cos(x1(t)) 0  sin(x1(t)) 0 ;
      sin(x1(t)) 0 -cos(x1(t)) 0;
      0  1  0 0;
      0  0 0 1;];

   T1_2=[cos(x2(t)) 0 sin(x2(t)) 0;
         sin(x2(t)) 0 -cos(x2(t)) 0;
         0  1  0  0;
           0  0 0 1;];

   T2_3=[cos(x3(t)) 0 sin(x3(t)) 0;
         sin(x3(t)) 0 -cos(x3(t)) 0;
         0  1 0 -a;
           0  0 0 1;];

   T3_4=[cos(x4(t)) 0 -sin(x4(t)) -la*cos(x4(t));
         sin(x4(t)) 0 cos(x4(t))  -la*sin(x4(t));
         0  -1  0 0;
         0  0  0 1;];

     T1 = T0_1;
     T2 = T1*T1_2;
     T3 = T2*T2_3;
     T4 = T3*T3_4;

   xT1 = T1(1,4);
   xT1_dot(t) = diff(xT1);
   yT1 = T1(2,4);
   yT1_dot(t) = diff(yT1);

   xT2 = T2(1,4);
   xT2_dot(t) = diff(xT2);
   yT2 = T2(2,4);
   yT2_dot(t) = diff(yT2);

   xT3 = T3(1,4)
   xT3_dot(t) = diff(xT3);
   yT3 = T3(2,4);
   yT3_dot(t) = diff(yT3);

   xT4 = T4(1,4);
   xT4_dot(t) = diff(xT4);
   yT4 = T4(2,4);
   yT4_dot(t) = diff(yT4);

%  v1 = simplify((xT1_dot(t) + yT1_dot(t))^2);
%  v2 = simplify((xT2_dot(t) + yT2_dot(t))^2);
%  v3 = simplify((xT3_dot(t) + yT3_dot(t))^2);
%  v4 = simplify((xT4_dot(t) + yT4_dot(t))^2);

   v1 = ((xT1_dot(t) + yT1_dot(t))^2);
   v2 = ((xT2_dot(t) + yT2_dot(t))^2);
   v3 = ((xT3_dot(t) + yT3_dot(t))^2);
   v4 = ((xT4_dot(t) + yT4_dot(t))^2);

   KE = (1/2)*m1*v1 + (1/2)*m2*v2 + (1/2)*m3*v3 + (1/2)*m4*v4 + ...
       (1/2)*I1*diff(x1(t), t) + (1/2)*I2*diff(x2(t), t) + (1/2)*I3*diff(x3(t), t) + (1/2)*I4*diff(x4(t), t) ; 

   PE = m1*g*h1 + m2*g*h2 + m3*g*h3 + m4*g*h4 ;

   Lag = KE - PE;

Answer 1

[Mischa Kim]

Siti, the Euler-Lagrange tool package on File Exchange does exactly what you are looking for.

[siti khadijah]

Thanks Mischa Kim :)