First find the moment of inertia of the slender rod about a line L passing through its center of mass.

"Slender" means the length is much greater than the width or depth.

Consider a differential element of length dr at a distance r from the center of mass. The mass of the element is equal to the product of its density and volume:

     dm = ρ dv = ρ A dr

Substitute this into equation for the moment of inertia of the rod about L,

     

Note that the mass of the rod is given by

     m_r = ρ V_rod = ρ A L_r

thus

     I_rod = 1/12 m_r L_r²

This is the same result listed in the Sections Appendix. for a slender rod.

Now the parallel axis theorem can be used to calculate the moment of inertia of the rod about the line L_o passing through the point 0,

     I_o-rod = I_rod + d_r² m_r

              = 1/12 m_r L_r² + (1/2 L_r)² m_r

             = 1/3 m_r L_r² = 0.3333 (0.5 slugs) (1.2 ft)²

             = 0.2400 slugs-ft²

Next, determine the moment of inertia of the disk about a line L passing through its center of mass, perpendicular to the flat face of the disk. Consider a differential element of width dr at a radius r from the center of mass. The mass of the element is equal to the product of its density and volume:

     dm = ρ dv = ρ t_d 2 π r dr

Substitute this into the moment of inertia equation to get I of the disk about L,

           = 0.5 ρ t_d π r_d⁴

The mass of the element is equal to the product of its density and volume,

     m_d = ρ V_d = ρ π r_d² t_d

     I_disk = 1/2 m_d r_d²

Again, parallel axis theorem can be used to find the moment of inertia of the disk about the line Lo passing through the point 0,

     I_o-disk = I_disk + d_d² m_d

           = 1/2 m_d r_d² + (L_r + r_d)² m_d

           = m_d [1/2 r_d² + (L_r + r_d)²]

          = 2.67 [1/2 (0.4)² + (1.2 + 0.4)²]

         = 7.049 slugs-ft²

The sum of the moment of inertia of the disk and rod to will give the total moment of inertia of the pendulum about L_o as

     I_o-pendulum = I_o-rod + I_o-disk

         = 7.289 slugs-ft²