Civil Engineering - Applied Mechanics - Discussion

Centre of gravity of a thin hollow cone = 1/3
Centre of gravity of a solid cone = 1/4.

Hollow cone = h/3.
Solid cone = h/4.

Cone:h/4.
Hollow cone : 2h/3.

The correct Answer must be option A.

It Should be h/3.

Hollow: h/3.
Solid: h/4.
From base.

Why not h/4 is there any differencee in C.G. of hollow and solid circular cone?

When the cone is positioned upright, standing on its tip we can find the center of mass by the formulas,
x' = ∫(x*da)/∫(da).
y' = ∫(y*da)/∫(da).
---with the limits from,
x = 0 to x = b(base).
y = 0 to y = h(height).

By intuition center of mass along x = 0.

We can represent the equation of the cone as y = mx, where m is the slope.

So finding the center of mass along y we find,
∫yda = ∫y*2πxdy, since da=2πxdy from infinitesimal area A = 2πrh.
= 2π∫(y^2)dy/m.
= 2π(y^3)/3m evaluated from 0 to h.
= 2πh^3/3m, we can find m from y = mx ---> h = mb so m=h/b.
= 2πb(h^2)/3.

Then we have to find ∫da,
∫da = ∫2πxdy.
= ∫2πydy/m.
= π(h^2)/m.
= πbh.

so y' = ∫(y*da)/∫(da) = 2πb(h^2)/3πbh = 2h/3.
This means the center of mass along the vertical axis of the cone is 2h/3 when the cone is STANDING ON ITS TIP.
y = 2h/3 and x = 0.

From top its 2h/3 and from base it is h/3.

How come 2h/3, It should be h/3?