Chapter 8.1, Problem 10SSC

(a)

To determine

The angular velocity of the disk for the start of the track.

Answer to Problem 10SSC

The angular velocity of the disk for the start of the track is $52\text{rad/s}\text{or 497}\text{rev/min}$ .

Explanation of Solution

Given:

The inner radius $\left(r_{\text{i}}\right)$ of the track is $2.7\text{cm}$ .

The outer radius $\left(r_{\text{f}}\right)$ of the track is $5.5\text{cm}$ .

The linear velocity $\left(v\right)$ of the track is $1.4\text{m/s}$ .

Formula used:

The expression for the angular velocity $\left(\omega \right)$ of a disk of radius $\left(r\right)$ as follows:

  $\omega =\frac{v}{r}$

Here, $v$ is the linear velocity of the disk.

Calculation:

The angular velocity of the disk for the start of the track as follows:

  $\begin{array}{l}{\omega }_{\text{i}}=\frac{v}{r_{\text{i}}}\\ {\omega }_{\text{i}}=\frac{1.4\text{m/s}}{2.7\text{cm}\times \left(\frac{1\text{cm}}{100\text{cm}}\right)}\\ {\omega }_{\text{i}}=\left(\frac{1.4\text{m/s}}{0.027\text{m}}\right)\\ {\omega }_{\text{i}}\approx 52\text{rad/s}\left(\begin{array}{l}\text{rounding off to }\\ \text{two significant figures}\end{array}\right)\end{array}$

  $\begin{array}{l}{\omega }_{\text{i}}\approx 52\left(\frac{\text{rad}}{\text{s}}\right)\times \left(\frac{1\text{rev}}{2\pi \text{rad}}\right)\times \left(\frac{60\text{s}}{1\text{min}}\right)\\ {\omega }_{\text{i}}\approx 497\text{rev/min}\end{array}$

Conclusion:

Therefore, the angular velocity of the disk for the start of the track is $52\text{rad/s}\text{or 497}\text{rev/min}$ .

(b)

To determine

The angular velocity of the disk for the end of the track.

Answer to Problem 10SSC

The angular velocity of the disk for the end of the track is $25\text{rad/s}$ .

Explanation of Solution

Given:

The inner radius $\left(r_{\text{i}}\right)$ of the track is $2.7\text{cm}$ .

The outer radius $\left(r_{\text{f}}\right)$ of the track is $5.5\text{cm}$ .

The linear velocity $\left(v\right)$ of the track is $1.4\text{m/s}$ .

Formula used:

The expression for the angular velocity $\left(\omega \right)$ of a disk of radius $\left(r\right)$ as follows:

  $\omega =\frac{v}{r}$

Here, $v$ is the linear velocity of the disk.

Calculation:

The angular velocity of the disk for the end of the track as follows:

  $\begin{array}{l}{\omega }_{\text{f}}=\frac{v}{r_{\text{f}}}\\ {\omega }_{\text{i}}=\frac{1.4\text{m/s}}{5.5\text{cm}\times \left(\frac{1\text{cm}}{100\text{cm}}\right)}\\ {\omega }_{\text{i}}=\left(\frac{1.4\text{m/s}}{0.055\text{m}}\right)\\ {\omega }_{\text{i}}\approx 25\text{rad/s}\left(\begin{array}{l}\text{rounding off to }\\ \text{two significant figures}\end{array}\right)\end{array}$

Conclusion:

Therefore, the angular velocity of the disk for the end of the track is $25\text{rad/s}$ .

(c)

To determine

The angular acceleration of the disk.

Answer to Problem 10SSC

The angular acceleration of the disk is $-5.9\times {10}^{-3}{\text{rad/s}}^2$ .

Explanation of Solution

Given:

The disk is played for time $t=76\text{min}$ .

The inner radius $\left(r_{\text{i}}\right)$ of the track is $2.7\text{cm}$ .

The outer radius $\left(r_{\text{f}}\right)$ of the track is $5.5\text{cm}$ .

The linear velocity $\left(v\right)$ of the track is $1.4\text{m/s}$ .

Formula used:

The expression for the angular acceleration $\left(\alpha \right)$ of a disk as follows:

  $\alpha =\left(\frac{{\omega }_{\text{f}}-{\omega }_{\text{i}}}{\Delta t}\right)$

Here, ${\omega }_{\text{i}}$ and ${\omega }_{\text{f}}$ is the initial and final angular velocity of the disk, and $\Delta t$ is the time interval.

Calculation:

Frompart (a) and (b), The angular velocity of the disk for the start of the track is ${\omega }_{\text{i}}=52\text{rad/s}$ .

The angular velocity of the disk for the end of the track is ${\omega }_{\text{f}}=25\text{rad/s}$ .

The angular acceleration $\left(\alpha \right)$ of the disk is

  $\begin{array}{l}\alpha =\left(\frac{{\omega }_{\text{f}}-{\omega }_{\text{i}}}{\Delta t}\right)\\ \alpha =\left(\frac{25\text{rad/s}-52\text{rad/s}}{76\text{min}\times \frac{60\text{s}}{1\text{min}}}\right)\\ \alpha =\left(\frac{-27\text{rad/s}}{4560\text{s}}\right)\\ \alpha =-5.9\times {10}^{-3}{\text{rad/s}}^2\end{array}$

Conclusion:

Therefore, the angular acceleration of the disk is $-5.9\times {10}^{-3}{\text{rad/s}}^2$ .