On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant linear speed of $\mathit{v}\mathbf{=}\mathbf{1}\mathbf{.25}\mathbf{\text{\hspace{0.17em}m/s}}$. Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the CD is played. (See Exercise 9.20.) Let’s see what angular acceleration is required to keep constant. The equation of a spiral is $\mathit{r}\left(\theta \right)\mathbf{=}{\mathit{v}}_{\mathbf{0}}\mathbf{+}\mathit{\beta }\mathit{\theta }$, where ${\mathit{r}}_{\mathbf{0}}$ is the radius of the spiral at $\mathit{u}\mathbf{=}\mathbf{0}$and $\mathit{\beta }$is a constant. On a CD, ${\mathit{r}}_{\mathbf{0}}$is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, $\mathit{\beta }$must be positive so that $\mathit{r}$increases as the disc turns and $\mathit{\theta }$ increases.

1. When the disc rotates through a small angle $\mathit{d}\mathit{\theta }$, the distance scanned along the track is $\mathit{d}\mathit{s}\mathbf{=}\mathit{r}\mathit{d}\mathit{\theta }$. Using the above expression for $\mathit{r}\left(\theta \right)$, integrate $\mathit{d}\mathit{s}$to find the total distance $\mathit{s}$scanned along the track as a function of the total angle $\mathit{\theta }$through which the disc has rotated.
2. Since the track is scanned at a constant linear speed $\mathit{v}$, the distance found in part (a) is equal to. Use this to find $\mathit{\theta }$as a function of time. There will be two solutions for $\mathit{\theta }$; choose the positive one, and explain why this is the solution to choose.
3. Use your expression for $\mathit{r}\left(\theta \right)$to find the angular velocity ${\mathit{\omega }}_{\mathbf{z}}$ and the angular acceleration${\mathit{\alpha }}_{\mathbf{z}}$as functions of time. Is ${\mathit{\alpha }}_{\mathbf{z}}$constant?
4. On a CD, the inner radius of the track is $\mathbf{25}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mm}$, the track radius increases by 1.55µm per revolution, and the playing time is 74.00 min . Find r₀, β, and the total number of revolutions made during the playing time.
5. Using your results from parts (c) and (d), make graphs of ω_z(in rad/s) versust andα_z(in rad/s²) versus t between t=0and t=74.00 min.

Step by step solution

01

Identification of given data:

Here we have $v=1.25\text{\hspace{0.17em}m/s}$

Constant linear speed of scanning $r\left(\theta \right)=r_0+\beta \theta$

$Theequationofaspiral{r}_0=theradiusofthespiralatthe\theta =0$The radius of the spiral at the

$\beta$is the constant

02

Angular velocity and angular acceleration

(a) Angular velocity: The pace at which an object's location angle with respect to time changes is known as its angular velocity.

${\mathit{\omega }}_{\mathbf{z}}\mathbf{=}\frac{\mathbf{d}\mathbf{\theta }}{\mathbf{d}\mathbf{t}}$

Where${\omega }_z$is an angular velocity and$\theta$ is angle of rotation.

(b) Angular acceleration: The time rate of change of angular velocity is known as angular acceleration

${\alpha }_z=\frac{d{\omega }_z}{dt}$

Where ${\omega }_z$is an angular velocity and${\alpha }_z$ is angular acceleration.

03

Finding the total distance s scanned along the track as a function of the total angle θ through which the disc has rotated.

Here we have$dS=rd\theta$,.

From above equation, we can determine the total distances $S=f\left(\theta \right)$.

$\begin{array}{rcl}ds& =& rd\theta \\ ds& =& \left(r_0+\beta \theta \right)d\theta \\ {\int }_0^{s}ds& =& {\int }_0^{\theta }\left(r_0+\beta \theta \right)d\theta \\ S-0& =& {\int }_0^{\theta }{r}_{0}d\theta +{\int }_0^{\theta }\beta \theta d\theta \\ S& =& r_0+\beta \frac{{\theta }^2}{2}\end{array}$

04

Step 4: Finding θ as a function of time by using s found in part (a)

In the case of uniform rectilinear motion, the distance traveled can be calculated as:

$S=vt$
Using the equation above, we can determine the form of function$\theta =f\left(t\right)$.

$$\begin{gathered} S=vt \\ {r}_0\theta +{\beta }^{{\theta }^2}=vt \\ \frac{\beta }{2}{\theta }^2+r_0\theta -vt=0 \end{gathered}$$

The equation above is a quadratic equation so, its solution can be written in the following form:

$\begin{array}{c}\theta =\frac{-r_0\pm \sqrt{r_0^2+4\frac{\beta }{2}vt}}{\beta }\\ =\frac{-r_0\pm \sqrt{r_0^2+2\beta vt}}{\beta }\end{array}$

In the previous equation, we will consider only positive solutions. So,

$\theta =\frac{-r_0+\sqrt{r_0^2+2\beta vt}}{\beta }$

05

 Step 5: To find the angular velocity and angular acceleration as functions of time. 

Here we have to consider $\theta =f\left(t\right)$calculated in the part b), we can determine the change of the angular velocity over time ${\omega }_z=f\left(t\right)$, using the equation (1):

$\begin{array}{rcl}& & \begin{array}{c}{\omega }_z=\frac{d\theta }{dt}\\ =\frac{d}{dt}\left(\frac{-r_0+\sqrt{r_0^2+2\beta vt}}{\beta }\right)\end{array}\\ & & \begin{array}{c}=\frac{1}{\beta }\frac{d}{dt}\left(\sqrt{r_0^2+2\beta vt}\right)\\ =\frac{2\beta v1}{\beta }{\left(r_0^2+2\beta vt\right)}^{-\frac{1}{2}}\\ =\frac{v}{\sqrt{r_0^2+2\beta vt}}\end{array}\end{array}$

Using the equation 2 we can calculate ${\alpha }_z=f\left(t\right)$ as

$\begin{array}{c}{\alpha }_z=\frac{d{\omega }_z}{dt}\\ =\frac{d}{dt}\left(\frac{v}{\sqrt{r_0^2+2\beta vt}}\right)\\ =v\frac{d}{dt}{\left(r_0^2+2\beta vt\right)}^{-\frac{1}{2}}\\ =v\left(-\frac{2\beta v}{2}\right){\left(r_0^2+2\beta vt\right)}^{-\frac{3}{2}}\\ =-\beta v^2{\left(r_0^2+2\beta vt\right)}^{-\frac{3}{2}}\\ =-\frac{\beta v^2}{\sqrt{{\left(r_0^2+2\beta vt\right)}^3}}\end{array}$

06

Step 6: Finding r0,β and the total number of revolutions made during the playing time.

Using the parameters given from part d) we can determine the angle$\theta$, Before, it is necessary to show $\beta$in units of$\frac{\text{m}}{\text{rad}}$.

$\begin{array}{c}\beta =1.55\frac{\mu \text{m}}{\text{revolution}}\\ =1.55\cdot \times {10}^{-6}\frac{\text{m}}{\text{revolution}}\cdot \frac{1\text{revoltuon}}{2\pi \text{rad}}\\ =0.247\times {10}^{-6}\text{\hspace{0.17em}m/rad}\end{array}$

Now, using the equation from part b) we can determine as follows.

$\begin{array}{c}\theta =\frac{-r_0+\sqrt{r_0^2+2\beta vt}}{\beta }\\ =\frac{-25\cdot {10}^{-3}\text{\hspace{0.17em}m}+\sqrt{{\left(25\cdot {10}^{-3}\text{\hspace{0.17em}m}\right)}^2+2(0.247\times {10}^{-6}\text{\hspace{0.17em}m/rad})\left(1.25\text{\hspace{0.17em}m/s}\right)\left(4440\text{\hspace{0.17em}s}\right)}}{0.247\cdot {10}^{-6}\text{\hspace{0.17em}m/rad}}\\ =133697.45\text{\hspace{0.17em}revolution}\end{array}$

07

Make graphs of ωz  (in rad/s ) versus  t and  αz (in rad/s2) versus   t between t=0  and  t=74.00 min from the result of part C and D.

Based on the laws calculated in the part under c ). And by applying the parameters from the part under d), it is possible to show graphs of the dependence of ${\omega }_z=f\left(t\right)$

So, we have ${\omega }_z=\frac{v}{\sqrt{r_0^2+2\beta vt}}$

So, graph of above function is which is given by,

As well as ${\alpha }_z=f\left(t\right)$

Here we have,${\alpha }_z=-\frac{\beta v^2}{\sqrt{{\left(r_0^2+2\beta vt\right)}^3}}$

So, graph of above function is given as

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