Bike
Problem
Ben has just bought a new bike with 23 inch wheels but without mudguards and he decided to test it despite the fact it was raining. Find the maximum height of the raindrops which were being flicked up from the back wheel if his speed was 18 km/h. Where were the drops when they reached the highest point relative to the bike?
Getting Started
Suppose that a drop detaches from the wheel when it is at angle $\alpha$ with the vertical. The bike is moving horizontally with the velocity $v$ relative to the ground and the points of the wheel is rotating with the velocity $v$ tangentially relative to the wheel. You need to add these velocities vectorially if you want to find the velocity of the drop relative to the ground. Write the equations of motion for the drop in horizontal and vertical directions. Find the expression for $y_{max}$ for an arbitrary angle $\alpha$ and then differentiate to find the angle for which drop reaches its maximum height.
Student Solutions
As suggested in the hint it is useful to choose the coordinate system with the origin at the wheel touching the ground point. The wheel is moving horizontally with the speed $v = 18\mathrm{kmh}^{-1} = 5\mathrm{ms}^{-1}$ and the diameter of the wheel $d = 23 \mathrm{inch} = 0.584\mathrm{m}$. Note $r = d/2$. The wheel is not slipping on the ground, thus the linear speed of the drops on the wheel is $v = 5\mathrm{ms}^{-1}$. Suppose that the drop detaches when it is at an angle $\alpha$ with vertical.
In x direction: $x(t) = -r\sin(\alpha) + vt + vt\cos(\alpha)\;.$
In y direction: $y(t) = r + r\cos(\alpha) + vt\sin(\alpha) - \frac{gt^2}{2}\;.$
We neglect any air resistance then the drop will be at its highest point when its vertical speed is zero.
$$0 = v\sin(\alpha) - gt_{max}$$
Hence, $t_{max} = \frac{v\sin(\alpha)}{g}$. We substitute this expression to the y direction equation to get that
$$y(\alpha) = r(1 + \cos(\alpha))+\frac{v^2\sin(\alpha)^2}{g}-\frac{v^2\sin^2(\alpha)}{2g} = r(1 + \cos(\alpha)) +\frac{v^2\sin^2(\alpha)}{2g}\;.$$
Method 1. Differentiate y with respect to alpha and then solve $\frac{dy}{d\alpha} = 0$. Remember that $\frac{d\sin(\alpha)}{d\alpha} = \cos(\alpha)$ and $\frac{d\cos(\alpha)}{d\alpha} = -\sin(\alpha)$. We can use chain rule to differentiate $\sin^2(\alpha)$: $$\frac{d\sin^2(\alpha)}{d\alpha} = \frac{d\sin^2(\alpha)}{d\sin(\alpha)}\frac{d\sin(\alpha)}{d\alpha} = 2\sin(\alpha)\cos(\alpha)\;.$$ Moreover if C is constant $\frac{dC}{d\alpha} = 0$. Thus, $\frac{dy}{d\alpha} = 0 - r\sin(\alpha) + \frac{v^2}{2g}2\sin(\alpha)\cos(\alpha)$. This means that either $\sin(\alpha) = 0$ or $\cos(\alpha) =\frac{gr}{v^2}$ for $\frac{gr}{v^2} < 1$. If $\sin(\alpha) = 0$ then $\cos(\alpha) = 1$ or $-1$, $y_{max} = 2r = 0.584\mathrm{m}$ or $0$. If $\cos(\alpha) =\frac{gr}{v^2}$ for $\frac{gr}{v^2} < 1$ then $y_{max} = r\left(1+\frac{gr}{v^2}\right) + \frac{v^2}{2g}\left(1 -\frac{g^2r^2}{v^4}\right) = r + \frac{v^2}{2g} + \frac{gr^2}{2v^2} = 1.58\mathrm{m}$. So, the maximum height is $1.58\mathrm{m}$.
Method 2. Use identity $\sin^2(\alpha) = 1 - \cos^2(\alpha)$ and note $X = \cos(\alpha)$. Then $y(X) = r + rX + \frac{v^2}{2g} - \frac{v^2}{2g}X^2$ which is a quadratic and the critical point can be found. Try it!
Find the $x$ coordinate of the drops when they are at the highest point.
$$\begin{eqnarray} x_{max} &=& -r\sin(\alpha) + vt(1 + \cos(\alpha)) = -r\sin(\alpha) + v\frac{v\sin(\alpha)}{g}(1 + \cos(\alpha))\\ &=& \sin(\alpha)\left(\frac{v^2}{g}(1 + \cos(\alpha)) - r\right)\\
&=& \sqrt{1 - \frac{g^2r^2}{v^4}}\left(\frac{v^2}{g}\left(1 + \frac{gr}{v^2}\right) -r\right) =\frac{v^2}{g} \sqrt{1 - \frac{g^2r^2}{v^4}}
\end{eqnarray}$$
The coordinate of the wheel axle is $x_{axle} = vt = v\frac{v\sin(\alpha)}{g} = \frac{v^2}{g} \sqrt{1 - \frac{g^2r^2}{v^4}}$. Thus, $x_{axle} = x_{max}$. The drops are just above the wheel axle.