Answer:
I₁ / I₂ = 1.43
Explanation:
To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended
Before starting let's reduce all units to the SI system
d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m
d₂ = 38 in = 96.52 10⁻² m
The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm
I₁ = I_man + 2 I_ arm
Man indicates that we can approximate them to a cylinder where the average diameter is
d = (d₁ + d₂) / 2
d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m
The average radius is
r = d / 2 = 50.8 10⁻² m = 0.508 m
The mass of the trunk is the mass of man minus the masses of each arm.
M = M_man - 0.2 M_man = 80 (1-0.2)
M = 64 kg
The moments of inertia are:
A cylinder with respect to a vertical axis: Ic = ½ M r²
A rod that rotates at the end: I_arm = 1/3 m L²
Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.
I1 = I_arm + m D²
Where D is the distance from the axis of rotation of the arm to the axis of the body
D = d / 2 = 101.6 10⁻² /2 = 0.508 m
Let's replace
I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]
Let's calculate
I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]
I₁ = 8.258 + 5.33 + 4.129
I₁ = 17,717 Kg m² / s²
Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,
I₂ = ½ M r² + 2 m D²
I₂ = ½ 64 0.508² + 2 8 0.508²
I₂ = 8,258 + 4,129
I₂ = 12,387 kg m² / s²
The relationship between these two magnitudes is
I₁ / I₂ = 17,717 /12,387
I₁ / I₂ = 1.43
