Answer:
a) I = 1,75 10-² kg m² and b) I = 1.49 10⁻² kg m²
Explanation:
The expression for the moment of inertia is
I = ∫ r² dm
The moment of inertia is a scalar by which an additive magnitude, we can add the moments of inertia of each part of the system, taking into account the axis of rotation.
I = I core + I shell
The moment of inertia of a solid sphere is
I sphere = 2/5 MR²
The moment of inertia of a thin spherical shell is
I shell = 2/3 M R²
a) Let's apply to our system, first to the core of weight 1.6 kg and diameter 0.196m, the radius is half the diameter
R = d / 2
R= 0.196 m / 2 = 0.098 m
I core = 2/5 1.6 0.098²
I core = 6.147 10-3 kg m²
Let's calculate the moment of inertia of the shell of mass 1.6 kg with a diameter of 0.206 m
R = 0.206 / 2
R = 0.103 m
I shell = 2/3 1.6 0.103²
I shell = 1,132 10-2 kg m²
The moment of inertia of the ball is the sum of these moments of inertia,
I = I core + I shell
I = 6,147 10⁻³ + 1,132 10⁻² = 6,147 10⁻³ + 11.32 10⁻³
I = 17.47 10⁻³ kg m²
I = 1,747 10-² kg m²
b) Now the ball is report with mass 3.2kg and diameter 0.216 m
R = 0.216 / 2
R = 0.108 m
It is a uniform sphere
I = 2/5 M R²
I = 2/5 3.2 0.108²
I = 1.49 10⁻² kg m²
