Assume that the function x(t) represents the length of tape that has unwound as a function of time. find θ(t), the angle through which the drum will have rotated, as a function of time. express your answer (in radians) in terms of x(t) and any other given quantities.
We know that arc length (x(t)) is given with the following formula: [tex]x(t)=\theta(t) r[/tex] Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it. If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically: [tex]r(\theta)=r_0-\frac{D\cdot \theta}{2\pi} [/tex] When we plug this back into the first equation we get: [tex]x(t)=\theta(r_0-\frac{D\theta}{2\pi}})\\ \frac{D\theta^2}{2\pi}-\theta r_0+x(t)=0\\[/tex] We must solve this quadratic equation. The final solution is: [tex]\theta=\frac{\pi r_0+\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D},\:\theta=\frac{\pi r_0-\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D}[/tex] It is rather complicated solution. If we asume that the tape has no thickness we get simply: [tex]x(t)=\theta(r_0-\frac{D\theta}{2\pi}});D=0\\
x(t)=\theta r_0\\
\theta(t)=\frac{x(t)}{r_0}[/tex]