Chapter 25, Problem 59P

(a)

To determine

The speed of the laser beam when the spot of light on the wall is at distance $x$ from the point $O$.

Answer to Problem 59P

The speed of the laser beam when the spot of light on the wall is at distance $x$ from the point $O$ is $\underset{\_}{v=\left(\frac{4x^2+L^2}{L}\right)\omega }$.

Explanation of Solution

Let $\theta$ be the angle between the dashed line and reflected beam in the Figure P25.59.

From the Figure, write the expression for $\tan \theta$.

    $\tan \theta =\frac{x}{L/2}$        (I)

Here, $x$ distance between the point $O$ and the spot, $L$ is the side of the room.

Solve equation (I) for $x$.

    $x=\frac{1}{2}L\tan \theta$        (II)

Differentiate the equation (II) with respect to time, to find the speed of the laser spot at the wall.

    $\begin{array}{c}v=\frac{dx}{dt}\\ =\frac{d}{dt}\left(\frac{1}{2}L\tan \theta \right)\\ =\frac{1}{2}L{\sec }^2\theta \frac{d\theta }{dt}\end{array}$                                                                                             (III)\

From the Figure, using Pythagorean write the expression for $\sec \theta$.

`    $\begin{array}{c}\sec \theta =\frac{1}{\cos \theta }\\ =\frac{\sqrt{4x^2+L^2}}{L}\end{array}$        (IV)

If the mirror turns through angle $\varphi$, its normal rotates through angle $\varphi$, then the angle of incidence increases by $\varphi$ as does the angle of reflection. Hence the reflected ray rotates through $2\varphi$. As a result the angular speed of the reflected ray is twice that of the mirror.

    $\begin{array}{c}{\omega }_{\text{reflected ray}}=\frac{d\theta }{dt}\\ =2\omega \end{array}$        (V)

Use equation (IV) and (V) in (III).

    $\begin{array}{c}v=\frac{1}{2}L{\left(\frac{\sqrt{4x^2+L^2}}{L}\right)}^{2}2\omega \\ =\left(\frac{4x^2+L^2}{L}\right)\omega \end{array}$        (VI)

Conclusion:

Therefore, the speed of the laser beam when the spot of light on the wall is at distance $x$ from the point $O$ is $\underset{\_}{v=\left(\frac{4x^2+L^2}{L}\right)\omega }$.

(b)

To determine

The value of the $x$ corresponds to the minimum value for the speed.

Answer to Problem 59P

The value of the $x$ corresponds to the minimum value for the speed is $\underset{\_}{0}$.

Explanation of Solution

The speed of the laser beam when the spot of light on the wall is at distance $x$ from the point $O$ is $v=\left(\frac{4x^2+L^2}{L}\right)\omega$. It is clear that the variable in this expression is $x$, so the minimum value of speed will get only when the value of $x$ is zero.

Conclusion:

Therefore, the value of the $x$ corresponds to the minimum value for the speed is $\underset{\_}{0}$.

(c)

To determine

The minimum value for the speed.

Answer to Problem 59P

The minimum value for the speed is $\underset{\_}{v=L\omega }$.

Explanation of Solution

The speed of the laser beam when the spot of light on the wall is at distance $x$ from the point $O$ is $v=\left(\frac{4x^2+L^2}{L}\right)\omega$. It is clear that the variable in this expression is $x$, so the minimum value of speed will get only when the value of $x$ is zero.

Substitute, $0$ for $x$ in the above equation.

    $\begin{array}{c}v=\left(\frac{4\times 0^2+L^2}{L}\right)\omega \\ =L\omega \end{array}$

Conclusion:

Therefore, the minimum value for the speed is $\underset{\_}{v=L\omega }$.

(d)

To determine

The maximum speed of the spot on the wall.

Answer to Problem 59P

The maximum speed of the spot on the wall is $\underset{\_}{2L\omega }$.

Explanation of Solution

The maximum speed occurs when the reflected beam arrives at the corner of the room.

Substitute, $\frac{L}{2}$ for $x$ in the equation $v=\left(\frac{4x^2+L^2}{L}\right)\omega$.

    $\begin{array}{c}v=\left[\frac{4{\left(\frac{L}{2}\right)}^2+L^2}{L}\right]\omega \\ =2L\omega \end{array}$

Conclusion:

Therefore, the maximum speed of the spot on the wall is $\underset{\_}{2L\omega }$.

(e)

To determine

The time interval in which the spot changes from its minimum to maximum speed.

Answer to Problem 59P

The time interval in which the spot changes from its minimum to maximum speed is $\underset{\_}{\frac{\pi }{8\omega }}$.

Explanation of Solution

The reflected laser beam rotates through $\frac{\pi }{4}$ radians between the minimum and maximum speed, then the mirror rotates through $\frac{\pi }{8}$ radians.

The time interval between the minimum and maximum speed can b e calculated as follows,

    $\Delta t=\frac{\Delta \theta }{\omega }$        (VII)

Conclusion:

Substitute, $\frac{\pi }{8}$ for $\Delta \theta$ in the equation (VII), to find $\Delta t$.

    $\Delta t=\frac{\pi }{8\omega }$

Therefore, the time interval in which the spot changes from its minimum to maximum speed is $\underset{\_}{\frac{\pi }{8\omega }}$.