Solutions - 0120-02

(1) Integral formula:

$$\begin{array}{c}W=\int dW\gg \gg \int h(y)\rho gdV\\ \\ \gg \gg \int \rho gh(y)A(y)dy\end{array}$$

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Option 1: (2) Setup:

Set $y=0$ at the bottom, increasing upwards.

Radius of the cone with a QLIF:

$$r(y)=1.2+\frac{0-1.2}{4}y\gg \gg 1.2-0.3y$$

Horizontal slice of the cone tower: disk of radius $r$ at height $y$, satisfies:

$$A(y)=\pi r^2=\pi {(1.2-0.3y)}^2$$

The slice at $y$ is raised a distance of $h(y)=y$.

Thus:

$$W\gg \gg {\int }_0^4\rho g\pi y{(1.2-0.3y)}^{2}dy$$

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Option 2: (2) Setup:

Set $y=0$ at the top of the cone, increasing downwards.

Now $h(y)=4-y$ is the distance from the ground up to the height of a slice indexed by $y$.

Radius function:

$$r(y)=0+\frac{1.2-0}{4}y\gg \gg 0.3y$$

Thus:

$$W\gg \gg {\int }_0^4\rho g\pi (4-y){(0.3y)}^{2}dy$$