Position of centre of gravity of the frustrum of a cone:

situate the position of C. G. of the frustrum of a cone with height H = 12 cm and with  diameter 8 cm and 12 cm at top and bottom, respectively, of the frustrum of the cone as shown in Figure (a).

Solution

Generators B₁ T₁ & B₂ T₂ is extended to meet at A.

The volume V of the frustrum T₁ T₂ B₂ B₁ of a cone may be assumed as (V₁ - V₂),

Where V₁ = volume of cone B₁ B₂ A and V₂ = volume of cone T₁ T₂ A as illustrated in Figure (b).

Let similar triangles ATT₂ and ABB₂

T T2 / B B2 = AT/ A B = 4 / 6

∴ 4/6 = H ₂ / (H + H ₂)

4 (12 +  H ₂ ) = 6 H ₂

∴ H ₂ = 24 cm

∴          H₁ = Height of full cone = 24 + 12 = 36 cm

∴ V₁ =1/3 π r₁² H₁ = (π × 36 × 36)/3 = 432 π cm³

Likewise,

V₂  =(1/3) π 4² × 24 = 128 π cm³

∴ V =  (432 - 128) π = 304 π cm³

Considering moments about the axis B₁ B B₂,

304 π × z¯ = 432 π( 36/4) -  128 π (6 + 12)

∴ z¯ =  5.21 cm