# Civil Engineering - Applied Mechanics - Discussion

Discussion Forum : Applied Mechanics - Section 1 (Q.No. 46)

46.

The c.g. of a thin hollow cone of height

*h*, above its base lies on the axis, at a height ofDiscussion:

10 comments Page 1 of 1.
Mali alpesh k said:
2 years ago

Centre of gravity of a thin hollow cone = 1/3

Centre of gravity of a solid cone = 1/4.

Centre of gravity of a solid cone = 1/4.

Onkar k said:
4 years ago

Hollow cone = h/3.

Solid cone = h/4.

Solid cone = h/4.

Tirth shete said:
5 years ago

Cone:h/4.

Hollow cone : 2h/3.

Hollow cone : 2h/3.

Shashu said:
5 years ago

The correct Answer must be option A.

Ashwini said:
6 years ago

It Should be h/3.

Venkat said:
6 years ago

Hollow: h/3.

Solid: h/4.

From base.

Solid: h/4.

From base.

Samprit said:
7 years ago

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

Krp said:
8 years ago

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.

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.

Danesh said:
1 decade ago

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

Neetha said:
1 decade ago

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

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