A rigid body consists of four bodies joined together, as drawn below to scale. The point is at one corner of the rectangle and the component bodies are: A uniform disk of radius and mass . A uniform rod of length and mass . A uniform rectangle with side lengths and , and mass . A point mass at with mass . What is the moment of inertia about the axis through the point

I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) ^ {3/2}

Explanation:

The moment of inertia is a scalar quantity, therefore additive, therefore we can find the moments of inertia of each body with respect to the point and add them.

Let's use the parallel axis theorem for the moment of inertia.

          I = + m d²

the moments and inertia of the bodies are

disk                    Icm = ½ m R²

rod                     Icm = 1/12 m L²

rectangle           Icm = 1/12 m (a² + b²)

where a and b are the sides of the rectangle

Let's fix a reference frame on the point body, the length of the rectangle is x and its height y, the total moment of inertia is

          I_total = I_point + I_disco + I_rod + I_rectangular

the moment of inertia of the point is

           I = m r² = m 0

           I_point = 0

disk moment of inertia

suppose it is on the y-axis with x = 0

           I_disco = ½ m R² + m y²

moment inertia rod

located in the opposite corner

The distance from the point to the center of the mass of the rod is

           R =

           I_varrilla = 1/12 m L² + m ()

rectangle moment inertia

located on the x axis

            I_rectangle = ½ m (a² + b²) + m x²

we substitute

         I_total = 0 + ½ m R² + m y² + 1/12 m L² + m (Ra x ^ 2 + y²) + ½ m (a² + b²) + m x²

         I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) + m √(x^2 + y²)

         I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) ^ {3/2}

probably not if net force is zero