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 B_{1} T_{1} & B_{2} T_{2} is extended to meet at A.
The volume V of the frustrum T_{1} T_{2} B_{2} B_{1} of a cone may be assumed as (V_{1} - V_{2}),
Where V_{1} = volume of cone B_{1} B_{2} A and V_{2} = volume of cone T_{1} T_{2} A as illustrated in Figure (b).
Let similar triangles ATT_{2} and ABB_{2}
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
∴ H_{1} = Height of full cone = 24 + 12 = 36 cm
∴ V_{1} =1/3 π r_{1}^{2} H_{1} = (π × 36 × 36)/3 = 432 π cm^{3}
Likewise,
V_{2} =(1/3) π 4^{2} × 24 = 128 π cm^{3}
∴ V = (432 - 128) π = 304 π cm^{3}
Considering moments about the axis B_{1} B B_{2},
304 π × z¯ = 432 π( 36/4) - 128 π (6 + 12)
∴ z¯ = 5.21 cm