Eq. (1) through (6) represents the original nonlinear equation representing the dynamics of the inverted pendulum. Using ode45 from MATLAB, solve the original nonlinear equations.

Force analysis and system equations Below are the free-body diagrams of the two elements of the inverted pendulum system. Summing the forces in the free-body diagram of the cart in the horizontal direction, you get the following equation of motion. Mi- + bi + N = F (1) Note that you can also sum the forces in the vertical direction for the cart, but no useful information would be gained. Summing the forces in the free-body diagram of the pendulum in the horizontal direction, you get the following expression for the reaction force N. N = ma + mlocos0 - ml02 sin 0 (2) If you substitute this equation into the first equation, you get one of the two governing equations for this system. (M + m)x + bi + mlocos0 - ml0 sin0 = F (3) To get the second equation of motion for this system, sum the forces perpendicular to the pendulum. Solving the system along this axis greatly simplifies the mathematics. You should get the following equation. P sin 0 + N cos0 - mg sin 0 = ml0 + mi cos 0 (4) To get rid of the P and NV terms in the equation above, sum the moments about the centroid of the pendulum to get the following equation. - Plsin 0 - NIcos0 = 10 (5) Combining these last two expressions, you get the second governing equation. (I + mi?)0 + mgl sin 0 = -mli cos 0 (6)(M) mass of the cart 0.5 kg (m) mass of the pendulum 0.2 kg (b) coefficient of friction for cart 0.1 N/m/sec (1) length to pendulum center of mass 0.3 m (I) mass moment of inertia of the pendulum 0.006 kg.m\"2 (F) force applied to the cart (x) cart position coordinate (theta) pendulum angle from vertical (down)