Acompound disk of outside diameter 128 cm is made up of a uniform solid disk of radius 37.0 cm and area density 4.60 g/cm2 surrounded by a concentric ring of inner radius 37.0 cm, outer radius 64.0 cm, and area density 1.60 g/cm2find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center. in kg/m^2.

I total = 5,101 kg m²

Explanation:

The moment of inertia of a body is given by the expression

     I = ∫ r² dm

For some bodies with symmetry they are tabulated, for an axis that passes through its center

Disc                I = ½ M R2

hollow ring    I = ½ M (R12 + R22)

let's apply this equation our case.

The body is formed by two consecutive bodies a disc plus a hollow ring, with the same center of rotation, so the moment of inertia that is a scalar we can add

      I total = I disk + I ring

To find the mass let's use the definition of density

       ρ = m / A

The area of a circle is

        A = π R²

Disk

        A₁ = π 37²

        A₁ = 4.30 10³ cm2

Ring

       A₂ = π (R₂² - r₁²)

       A₂ = π (64² - 37²)

       A₂ = 8.57 10³ cm2

We cleared the dough

       m = ρ A

Disk

       m₁ = 4.60 4.30 10³

       m₁ = 19.78 10³ g = 19.78 kg

Ring

      m₂ = ρ₂ A₂

      m₂ = 1.60 8.57 10³

      m₂ = 13,712 10³ g = 13,712 kg

Now we can calculate the moments of inertia of each body

     I disk = ½ 19.78 0.37²

     I disk = 1,354 kg m²

     I ring = = ½ 13,712 (0.37² + 0.64²)

     I ring = 3.747 Kg m²

The moment of total inertia

     I total = I disk + I ring

     I total = 1,354 + 3,747

     I total = 5,101 kg m²

Note that the total radius of the system is 101 cm

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