Inverted Pendulum in a Cart: Equations and Simulation

Running head: ASSIGNMENT 1
ASSIGNMENT 1
Name of the Student
Name of the University
Author Note

Question 1:
The schematic diagram of the inverted pendulum in a cart is shown below.
Here, the variables are specified below.
M = mass of the cart
m = mass of the pendulum
x = horizontal displacement of the pivot on the cart.
θ = rotational angle of the pendulum
u = driving force of the carriage.
b = frictional coefficient modelled by a damper.
l = length of the pendulum.
Free body diagram:

Now, force of the cart for its mass M = mass*acceleration ¿ M∗( d x2
d t2 )
Friction acts opposite to its motion. It can be modelled by a damper which has force
b*(dx/dt). The driving force u which is causing the cart to accelerate acting on the direction
of its motion. N is the horizontal component of reaction force by the inverted pendulum
acting on the cart and P is the vertical component.
Now, from onwards the derivatives are represented by dots above the variable names for
simplified views of the equations.
Now, summing the forces by free-body diagram the equation of motion will be
M ̈x +b ̇x + N=u (1)
Now, if θ is the rotational angle then rotational angle change for pendulum’s motion is ̇θ and
its linear change = l ̇θ (as s = r*θ). Hence, linear acceleration of change = l ̈θ.
Now, algebraic sum of all the horizontal forces in the pendulum gives
N=m ̈x +ml ̈θ cos (θ)−ml ( ̇θ2 )∗(sinθ) (2)
P
mg
N N
θu
Friction force
= b*(dx/dt) P
Displacement x