How much work must be done to build a conical pile

How much work must be done to build a conical pile of sand with base radius $R=1.2$ m and height $h=1$ m, if all the sand is lifted from ground level and the specific gravity of sand is $2$?

Step 1: Radius as a function of height

With the base at $y=0$ (radius $R$) and the apex at $y=h$ (radius $0$), similar triangles give

\[r(y)=R\left(1-\frac{y}{h}\right)=\frac{R(h-y)}{h}.\]

Step 2: Mass of a thin disk

With sand density $\rho=2000\ \text{kg/m}^3$ (specific gravity $2$ relative to water):

\[dm=\rho\,\pi r^2(y)\,dy=\frac{2000\pi R^2(h-y)^2}{h^2}\,dy.\]

Step 3: Work to lift the disk from the ground

\[dW=(dm)gy=\frac{2000\pi R^2g}{h^2}\,y(h-y)^2\,dy.\]

Step 4: Integrate over the cone

Expanding $y(h-y)^2=h^2y-2hy^2+y^3$ and integrating from $0$ to $h$ gives $\int_0^hy(h-y)^2\,dy=h^4/12$ (a standard computation, equivalent to a Beta-function value). So

\[W=\frac{2000\pi R^2g}{h^2}\cdot\frac{h^4}{12}=\frac{500\pi R^2gh^2}{3}.\]

\[\boxed{W=\frac{500\pi R^2gh^2}{3}}\]

Step 5: Numerical value

With $R=1.2$ m, $h=1$ m, $g=9.8\ \text{m/s}^2$:

\[W=\frac{500\pi(1.2)^2(9.8)(1)^2}{3}=\frac{7056\pi}{3}\approx7390\ \text{J}\approx7.39\ \text{kJ}.\]

Step 6: Center-of-mass check

The cone's volume is $V=\tfrac13\pi R^2h$, so $M=\tfrac23000\pi R^2h$; comparing, $W=(500\pi R^2gh^2)/3=Mg(h/4)$, so the effective lift is exactly $h/4$ — the known centroid height (from the base) of a solid cone, confirming the calculation.