Question

centre of gravity of a hollow cone?

Answer 1

Answer on Question #38576, Physics, Mechanics | Kinematics | Dynamics

Centre of gravity of a hollow cone?

Solution.

Take the apex of the cone shown in Figure as the origin. The cone is symmetric about $x$ -axis, therefore $y$ -coordinate of centre of gravity is given $y_{c} = 0$ .

Consider the cone as split into an infinite number of rings.

I mean rings.

When we calculate center of gravity of a uniform solid cone we consider the cone as split into an infinite number of disks. When we calculate center of gravity of a hollow cone we consider the cone as split into an infinite number of rings.

Consider one such ring of thickness $dx$ at $x$ .

Radius of ring $y = \frac{r}{h} x$

Mass of ring $dm = \rho \cdot dV = \rho (2\pi \cdot y\cdot ds)t$

$$d m = \rho \cdot \left(2 \pi \frac {r}{h \cos \alpha} x d x\right) t,$$

where $\rho$ - density and $\frac{dx}{ds} = \cos \alpha$ , $t$ - thickness of cone side.

ds - height of the ring's side

t - thickness of a cone side (thickness of a side of the ring).

The $x$ -coordinate of centre of gravity is given by

$$x _ {c} = \frac {\int x d m}{\int d m}$$

or

$$x _ {c} = \frac {\int_ {0} ^ {h} x \rho \cdot \left(2 \pi \frac {r}{h \cos \alpha} x d x\right) t}{\int_ {0} ^ {h} \rho \cdot \left(2 \pi \frac {r}{h \cos \alpha} x d x\right) t} = \frac {\int_ {0} ^ {h} x ^ {2} d x}{\int_ {0} ^ {h} x d x} = \frac {\left. \frac {x ^ {3}}{3} \right| _ {0} ^ {h}}{\left. \frac {x ^ {2}}{2} \right| _ {0} ^ {h}} = \frac {2}{3} h.$$

The distance of centre of gravity of hollow cone from the vertex is $(2/3)h$ and from the base is $(1/3)h$ .

Answer. $y_{c} = 0, x_{c} = \frac{2}{3} h$ .