Chapter 3, Problem 74PQ

Problems 74 and 75 are paired.

74. N A classroom clock has a small magnifying glass embedded near the end of the minute hand. The magnifying glass may be modeled as a particle. Class begins at 7:55 and ends at 8:50. The length of the minute hand is 0.300 m. a. Find the average velocity of the magnifying glass at the end of the minute hand using the coordinate system shown in Figure P3.74. Give your answer in component form. b. Find the magnitude and direction of the average velocity. c. Find the average speed and in the CHECK and THINK step, compare to the average velocity.

To determine

The average velocity of the magnifying glass at the end of the minute hand using the coordinate system.

Answer to Problem 74PQ

The average velocity of the magnifying glass is $\underset{\_}{{\overrightarrow{v}}_{\text{av}}=-3.33\times {10}^{-5}\left(\widehat{\text{i}}+\widehat{\text{j}}\right)\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for the average velocity.

    ${\overrightarrow{\text{v}}}_{\text{av}}=\frac{\Delta \overrightarrow{\text{r}}}{\Delta t}$                                                                                                                  (I)

Here, ${\overrightarrow{\text{v}}}_{\text{av}}$ is the average velocity, $\Delta \overrightarrow{\text{r}}$ is the change in position, $\Delta t$ is the change in time.

The initial position of the minute hand is given by,

    ${\overrightarrow{\text{r}}}_i=\left(0.300\text{m}\right)\left[\cos \left(120°\right)\widehat{\text{i}}\text{+}\text{sin}\left(120°\right)\widehat{\text{j}}\right]$                                                     (II)

The final position of the minute hand is given by,

    ${\overrightarrow{\text{r}}}_f=\left(0.300\text{m}\right)\left[\cos \left(150°\right)\widehat{\text{i}}\text{+}\text{sin}\left(150°\right)\widehat{\text{j}}\right]$                                                   (III)

Conclusion:

It takes $55\text{min}$ to change the position, substitute $55\text{min}$ for $\Delta t$ and equation (II) and equation (III) in equation (I) to find ${\overrightarrow{v}}_{\text{av}}$.    $\begin{array}{c}{\overrightarrow{\text{v}}}_{\text{av}}=\frac{\left(0.300\text{m}\right)\left[\cos \left(150°\right)\widehat{\text{i}}\text{+}\text{sin}\left(150°\right)\widehat{\text{j}}\right]-\left(0.300\text{m}\right)\left[\cos \left(120°\right)\widehat{\text{i}}\text{+}\text{sin}\left(120°\right)\widehat{\text{j}}\right]}{55\text{min}\times \frac{60\text{s}}{1\text{min}}}\\ =-3.33\times {10}^{-5}\left(\widehat{\text{i}}\text{+}\widehat{\text{j}}\right)\text{m}/\text{s}\end{array}$

Therefore, the average velocity of the magnifying glass is $\underset{\_}{{\overrightarrow{v}}_{\text{av}}=-3.33\times {10}^{-5}\left(\widehat{\text{i}}+\widehat{\text{j}}\right)\text{m}/\text{s}}$.

To determine

The magnitude and the direction of the average velocity.

Answer to Problem 74PQ

The magnitude and the direction of the average velocity is $\underset{\_}{4.71\times {10}^{-5}\text{m}/\text{s}}$ at $\underset{\_}{45°\text{below the }-x\text{axis}}$.

Explanation of Solution

The magnitude of the average velocity is given by taking the square root of the sum of the squares of the $i^{\text{th}}$ and $j^{\text{th}}$ components.

  ${\overrightarrow{\text{v}}}_{\text{av}}=-3.33\times {10}^{-5}\left(\widehat{\text{i}}+\widehat{\text{j}}\right)\text{m}/\text{s}$

The magnitude of the average velocity is,

    $\begin{array}{c}\left|{\overrightarrow{v}}_{\text{av}}\right|=3.33\times {10}^{-5}\sqrt{{\left(1\right)}^2+{\left(1\right)}^2}\text{m}/\text{s}\\ =3.33\times {10}^{-5}\sqrt{2}\text{m}/\text{s}\\ =4.71\times {10}^{-5}\text{m}/\text{s}\end{array}$

The direction of the average velocity is given by,

    $\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$                                                                                                     (IV)

Conclusion:

Substitute $-3.33\times {10}^{-5}$ for $v_x$ and $-3.33\times {10}^{-5}$ for $v_y$ in equation (IV) to find $\theta$.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-3.33\times {10}^{-5}}{-3.33\times {10}^{-5}}\right)\\ ={\tan }^{-1}\left(1\right)\\ =45°\text{below the }-x\text{axis}\text{.}\end{array}$

Therefore, the magnitude and the direction of the average velocity is $\underset{\_}{4.71\times {10}^{-5}\text{m}/\text{s}}$ at $\underset{\_}{45°\text{below the }-x\text{axis}}$.

To determine

The comparison of average speed with the average velocity.

Answer to Problem 74PQ

The speed is about an order of magnitude larger because the distance traveled is much larger than the displacement at that time.

Explanation of Solution

The average speed is the total distance traveled in the time the clock hand moves from the initial to the final position.  The distance traveled is 55 out of the 60 of the circumference traced out by the end of the minute hand which can be expressed as,

    $d=\frac{55}{60}\left(2\pi r\right)$                                                                                                  (V)

Write the expression for the displacement.

    $s=\frac{d}{\Delta t}$                                                                                                            (VI)

Here, $s$ is the displacement, $\Delta t$ is the change in time.

Conclusion:

Substitute $0.300\text{m}$ for $r$ in equation (V) to find $d$.

    $\begin{array}{c}d=\frac{55}{60}\left(2\pi \times 0.300\text{m}\right)\\ =1.73\text{m}\end{array}$

Substitute $1.73\text{m}$ for $d$ and $55\text{min}$ for $\Delta t$ in equation (VI) to find $s$.

    $\begin{array}{c}s=\frac{1.73\text{m}}{55\text{min}\left(\frac{60\text{s}}{1\text{min}}\right)}\\ =\frac{1.73\text{m}}{3300\text{s}}\\ =5.24\times {10}^{-4}\text{m}\end{array}$

Comparing the value of displacement and distance the magnitude is larger for the distance so that the magnitude of the speed will be higher than the magnitude of the average velocity.

Therefore, the speed is about an order of magnitude larger because the distance traveled is much larger than the displacement at that time.