Question

Locate the centre of gravity of the right circular cone of base radius r and height h shown in Fig.

Answer 1

Taking origin at the vertex of the cone and selecting the axis as shown in Fig. it can be observed that due to symmetry the coordinates of centre of gravity \(\bar y\) and \(\bar z\) are equal to zero, i.e. the centre of gravity lies on the axis of rotation of the cone. To find its distance \(\bar x\) from the vertex, consider an elemental plate at a distance x. Let the thickness of the elemental plate be dx. From the similar triangles OAB and OCD, the radius of elemental plate z is given by

z = \(\frac xhr\)

∴ Volume of the elemental plate dv

dv = πz² dx = πx² \(\frac {r^2}{h^2}\) dx

If γ is the unit weight of the material of the cone, then weight of the elemental plate is given by:

Now, substituting the value of dW in (i), above, we get:

Thus, in a right circular cone, centre of gravity lies at a distance 3/4 h from vertex along the axis of rotation i.e., at a distance h/4 from the base.