The wheels on a bicycle have $r$-inch radii. After the front wheel picks up a tack, the bike rolls for another $d$ feet and stops. How far above the ground is the tack?
I've been thinking about this problem for a couple of days and I keep coming up against the same issue. I can do the problem fine as long as I have actual numbers for $r$ and $d$, but in the general case I'm having trouble figuring out how to find the angle of rotation. Dividing $d$ by the circumference gives the number of times that the circle rotates, and then to find the angle of rotation I just need multiply the fractional part of this number by 360. In the general case, I keep getting stuck on how to convey the fractional part of a number. I've found that the total number of rotations made is $\dfrac{6d}{r\pi}$; is there a way to notate the fractional part of this number? Is there a way to do this so that I don't need that?
I could write it as $\dfrac{6d}{r\pi}-\lfloor \dfrac{6d}{r\pi}\rfloor$ but I'd like to avoid introducing floor notation if possible.
The number of rotations can be found by $\frac{d}{2\pi r}$, or the distance divided by the circumference of the wheel.
Multiplying by $2\pi$ gets you the total number of radians: $\theta = 2\pi \cdot \frac{d}{2\pi i} = \frac{d}{r}$
The vertical coordinate of the tack with respect to the center of the wheel is generally given by $r \sin \left(\theta \pm \frac{\pi}{2}\right).$ The $\pm \frac{\pi}{2}$ is a phase shift to describe the fact that the tack is on the ground when it gets picked up. The $\pm$ part depends on whether you consider positive angles to go counterclockwise (i.e. bike traveling to the left) or clockwise (i.e. bike traveling to the right)
To get the distance from the ground simply add the wheel's radius.