Respuesta :
Answer:
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 = [tex]I_{cm}[/tex] + 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 = [tex]\sqrt{x^2 +y^2 }[/tex]
I_varrilla = 1/12 m L² + m ([tex]\sqrt{ x^2 + y^2 }[/tex])
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}
