Question

Consider your moment of inertia about a vertical axis through the center of your body, both when you are standing straight up with your arms flat against your sides, and when you are standing straight up holding your arms straight out to your sides. Estimate the ratio of the moment of inertia with your arms straight out to the moment of inertia with your arms flat against your sides. (Assume that the mass of an average adult male is about 80 kg, and that we can model his body when he is standing straight up with his arms at his sides as a cylinder. From experience in men's clothing stores, a man's average waist circumference seems to be about 34 inches, and the average chest circumference about 42 inches, from which an average circumference can be calculated. We'll also assume that about 20% of your body's mass is in your two arms, and that each has a length L = 1 m, so that each arm has a mass of about m = 8 kg.)

Answer 1

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