CALCULATION OF MOMENT OF INERTIA OF CERTAIN SYMMETRIC OBJECTS

There are many calculations in which the following are very important.

( a ) Moment of inertia of a thin uniform rod about an axis, passing through its centre and perpendicular to its length:

To calculate moment of inertia of a thin rod of length l and mass M about an axis yy’ passing through its center O and perpendicular to its length, consider O as origin and X-axis along the length of the rod. A small element of length dx of the rod is at a distance x from O. 

The moment of inertia of this element about yy’ is

d I = M dx · x² because moment of inertia of the rod,

         ℓ

    +ℓ/2

I =   ∫  [ M dx · x²]

  -ℓ/2         ℓ

   =  M [ℓ³ + ℓ³]

       3ℓ [8  +  8 ]

    = Mℓ²

       12

 ( b ) Moment of inertia of a thin ring or a thin walled hollow cylinder or a thin walled hollow sphere:

As the entire mass, M, of a thin ring is at the same distance, equal to the radius R of the ring from its centre, the moment of inertia of a thin ring about an axis passing through its centre and perpendicular to its plane is MR2. Similarly, the moments of inertia of a thin walled cylinder about its geometric axis or of a thin walled hollow sphere about its center are also given by MR2, where M represents their mass and R their radii.

( c ) Moment of inertia of a disc or a solid cylinder:

To calculate moment of inertia of a disc of uniform thickness t, radius R and mass M about an axis passing through its centre and perpendicular to its plane, consider an element of the disc in the form of a thin ring of thickness dx at a distance x from its centre. Mass of this ring is 2 π x dx ⋅ t ⋅ ρ, where ρ is the density of the material of the ring. Therefore, the moment of inertia of the ring about an axis passing through the centre, O, of the disc and perpendicular to its plane is 

d I = ( 2 π x dx . t . ρ ) x² = ( 2 π t ρ ) x³ dx

Therefore moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane is

     R

Ι = ∫( 2 t ρ ) x3 dx = ( 2 π t ρ ) [x⁴ /4]  = 2 π t ρR⁴ / 4

     0

   = 1/2 M R²

so therefore moment of inertia of a solid cylinder about its axis is also  1/2 M R² 

 ( d ) Moment of inertia of a thin walled hollow sphere about its diameter:

Moment of inertia of a thin walled hollow sphere about its center is I ₀ = MR².

By perpendicular axes theorem for three dimensional bodies, 2 I ₀ = I _x + I _y + I _z 

Now I _x = I _y = I _z  = moment of inertia, I , of the hollow sphere about its diameter.

Therefore I = 2 I ₀ = 2 MR²

                      ^{3           3}