Question

To find the total angular displacement during the playing time of the compact disc in part (B) of Example 10.2, the disc was modeled as a rigid object under constant angular acceleration. In reality, the angular acceleration of a disc is not constant. In this problem, let us explore the actual time dependence of the angular acceleration. (a) Assume the track on the disc is a spiral such that adjacent loops of the track are separated by a small distance $h$. Show that the radius $r$of a given portion of the track is given by

$y=r_i+\frac{h\theta }{2\pi }$

where$r_i$ is the radius of the innermost portion of the track and is the angle through which the disc turns to arrive at the location of the track of radius $r$. (b) Show that the rate of change of the angle$\theta$ is given by

$y=r_i+\frac{h\theta }{2\pi }$

where $v$is the constant speed with which the disc surface passes the laser. (c) From the result in part (b), use integration to find an expression for the angle$\theta$ as a function of time. (d) From the result in part (c), use differentiation to find the angular acceleration of the disc as a function of time.

Short answer

The solution is

a) The radius of the given portion of the track is,

$r=r_i+h\left(\frac{\theta }{2\pi }\right)$

b) The rate of the change of the angle is

$\frac{d\theta }{dt}=\frac{v}{r_i+h\left(\frac{\theta }{2\pi }\right)}$

c) The expression for angle using integration $\theta =\frac{\pi }{h}[-r_i+\sqrt{r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)}]$,

d) The disc's angular acceleration using differentiation

$\alpha =\frac{-v}{2{\left(r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)\right)}^{\frac{3}{2}}}$

Step-by-step solution

Angular speed

The relationship between linear speed, radius, and angular speed is

$v=r\left(\frac{d\theta }{dt}\right)\dots \dots \left(1\right)$

The linear speed is$v$ , the radius of the circular path is $r$, and the angular speed is$\frac{d\theta }{dt}$

Model of the disc

The radius of the given portion of the track

(a)

Ifis the angle covered by the spiral, then the increased radius of a particular section of the track is (the distance covered by the spiral).

$r_{\theta }=h\left(\frac{\theta }{2\pi }\right)L\dots \left(2\right)$

The radius of a certain section of the track after spiraling over an angle $\theta$is shown in the diagram.

$\begin{array}{l}r=r_i+r_{\theta }\\ =r_i+h\left(\frac{\theta }{2\pi }\right)\end{array}$

Rate of the change of the angle

(b)

We derive the rate of change of the angle$\theta$ by substituting $r_i+h\left(\frac{\theta }{2\pi }\right)$for the radius of a specific portion of the track (r).

$\begin{array}{l}v=r\left(\frac{d\theta }{dt}\right)\\ v=[r_i+h\left(\frac{\theta }{2\pi }\right)]\left(\frac{d\theta }{dt}\right)\\ \frac{d\theta }{dt}=\frac{v}{r_i+h\left(\frac{\theta }{2\pi }\right)}\mathrm{.......}\dots \left(3\right)\end{array}$

The expression for angle using integration

(c)

By rearrangement and integration of equation (3), we obtain the following expression for the angle theta as a function of time t:

$\begin{array}{l}d\theta =\frac{v}{r_i+h\left(\frac{\theta }{2\pi }\right)}dt\\ [r_i+h\left(\frac{\theta }{2\pi }\right)]d\theta =vdt\end{array}$

Integrate both sides, we get,

$\begin{array}{l}\int r_{i}d\theta +\int h\left(\frac{\theta }{2\pi }\right)d\theta =\int vdt\\ r_i\left(\frac{{\theta }^{0+1}}{0+1}\right)+\frac{h}{2\pi }\left(\frac{{\theta }^{1+1}}{1+1}\right)=v\left(\frac{t^{0+1}}{0+1}\right)\end{array}$

By simplifying the above equation, we get,

$\begin{array}{l}{r}_i\theta +\frac{h}{4\pi }{\theta }^2=vt\\ \frac{h}{4\pi }{\theta }^2+r_i\theta -vt=0\dots \dots \left(4\right)\end{array}$

Equation (4) can be compared to the generic equation for quadratic equation .

$a{x}^2+bx+c=0$

$\begin{array}{l}a=\frac{h}{4\pi }\\ b=r_i\\ c=-vt\end{array}$

Because the magnitude of length and velocity are both positive, the value of $b^2-4ac=r_i^2-4\left(\frac{h}{2\pi }\right)(-vt)$. should also be positive. As a result, these are the genuine answers to equation (4).

$\begin{array}{c}\theta =\frac{-r_i\pm \sqrt{r_i^2-4\left(\frac{h}{2\pi }\right)(-vt)}}{2\left(\frac{h}{2\pi }\right)}\\ =\frac{\pi [-r_i\pm \sqrt{r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)}]}{h}\end{array}$

Since$\left(\frac{2h}{\pi }\right)\left(vt\right)$ is positive, then $r_i\le \sqrt{r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)}$for the positive value of angle, the solution of equation (4) is

$\theta =\frac{\pi }{h}[-r_i+\sqrt{r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)}]$

As a result, the perspective in terms of time t is$\theta =\frac{\pi }{h}[-r_i+\sqrt{r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)}]$

The disc's angular acceleration using differentiation

(d)

We may get the angular acceleration by differentiating the equation for angle theta with respect to time t.

$\begin{array}{l}\frac{d\theta }{dt}=\frac{d}{dt}\left[\frac{\pi }{h}[-r_i+\sqrt{r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)}]\right]\\ =\frac{\pi }{h}[0+\left(\frac{1}{2}\right){\left(r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)\right)}^{\frac{1}{2}-1}\left(\frac{2hv}{\pi }\right)]\\ =v\left[\frac{1}{\sqrt{r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)})}\right]\end{array}$

By differentiating again, we get

$\begin{array}{c}\frac{d^2\theta }{d{t}^2}=\frac{d}{dt}[v\left[\left(\frac{1}{\sqrt{r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)}]]}\right]\right]\\ \alpha =v\left[\frac{-1}{2}{\left(r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)\right)}^{\frac{-1}{2}-1}\right]\\ =\frac{-v}{2{\left(r_i^2+\left(\frac{2h}{\pi }\right)\left(vt\right)\right)}^{\frac{3}{2}}}\end{array}$

The disc's angular acceleration is alpha in this case.