Bowl Over Solution

The following is a solution to Puzzle #12 – Bowl Over. Take a look if you haven’t already.

 

The angle \theta is defined to be that which the bowl’s axis of symmetry subtends with the vertical direction. This angle measures the extent to which the bowl is tipped. When \theta is equal to 0, the bowl is untipped; when \theta is \frac{\pi}{2}, the bowl has been tipped completely. If \theta were allowed to go to \pi, the bowl would be upside-down, but we won’t need to worry about this case.

bowlBy the law of ‘those angles look the same therefore they are’, the angle \theta is also that which the plane of the liquid’s surface makes with the rim of the bowl.

Trigonometry allows us to work out h:

h=r\sin\theta

How does the distance h help? We can work out the volume of the liquid remaining in the bowl using integration. The shape filled by the liquid is a spherical cap.

A differential element of volume occupied by a thin slice of the liquid is

dV=\pi x^2 dy

where the y axis is assumed to run vertically and the x axis in the plane of the liquid’s surface. The first factor is the cross-sectional area of the slice, the second is its thickness. To find the total volume of the liquid, we integrate:

\displaystyle V=\pi\int_h^r x^2 dy

So h is one of the limits of integration. The surface of the spherical cap is described  by the Cartesian equation

x^2+y^2=r^2

Hence

\displaystyle V=\pi\int_h^r (r^2-y^2)dy

This integral can be solved using the substitution

y=r\sin\varphi

dy=r\cos\varphi\ d\varphi

\displaystyle V=\pi r^3\int_{\varphi(h)}^{\varphi(r)} (\cos\varphi-\cos\varphi\sin^2\varphi)d\varphi

\displaystyle V=\pi r^3 \Bigg[\sin\varphi-\frac{1}{3}\sin^3\varphi\Bigg]_{\varphi(h)}^{\varphi(r)}

\displaystyle V=\pi r^3 \Bigg[\frac{y}{r}-\frac{1}{3}\Big(\frac{y}{r}\Big)^3\Bigg]_h^r

\displaystyle V=\frac{1}{3}\pi r^3\Bigg(2+\frac{h}{r}\Bigg[\Big(\frac{h}{r}\Big)^2-3\Bigg]\Bigg)

and given that h=r\sin\theta,

V(\theta)=\frac{1}{3}\pi r^3(2+\sin\theta[\sin^2\theta-3])

 

Given this expression for the volume of liquid, we can work out how fast liquid is poured from the bowl:

\displaystyle \frac{dV}{dt}=\frac{dV}{d\theta}\frac{d\theta}{dt}

\displaystyle \frac{dV}{dt}=\frac{1}{3}\pi r^3 \frac{d}{d\theta}(2+\sin\theta [\sin^2\theta -3]) \frac{d\theta}{dt}

\displaystyle \frac{dV}{dt}=\frac{1}{3}\pi r^3 \frac{d}{d\theta}(\cos\theta [\sin^2\theta -3]+\sin\theta [2\sin\theta\cos\theta]) \frac{d\theta}{dt}

\displaystyle \frac{dV}{dt}=\pi r^3(\sin^2\theta - 1)\cos\theta \frac{d\theta}{dt}

\displaystyle \frac{dV}{dt}=-\pi(r\cos\theta)^3 \frac{d\theta}{dt}

We assumed for simplicity that the angle \theta changes at a constant rate \omega as the bowl is tipped. Hence

\displaystyle \frac{dV(t)}{dt}=-\pi\omega(r\cos\theta)^3

Neat!

 


bowlThe flow rate has a maximum when t=0; tipping the bowl only a small distance from its equilibrium position spills a large amount of liquid.

The flow rate goes to 0 as the last of the liquid is poured from the bowl.

It can be shown that the flow rate is monotonically decreasing – it is at all times decreasing.

The animation to the right gives a visualisation of the flow rate. All hail GeoGebra!

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