Chapter 12, Problem 60PQ

(a)

To determine

The average angular velocity of the pin wheel if it rotates from $\theta =0°$ to $\theta =180°$ in a time of $0.150\text{s}$.

Answer to Problem 60PQ

The average angular velocity of the pin wheel if it rotates from $\theta =0°$ to $\theta =90°$ in a time of $0.150\text{s}$ is $\underset{\_}{10.5\text{rad/s}}$.

Explanation of Solution

Given that the pin wheel rotates from $\theta =0°$ to $\theta =90°$ in a time of $0.150\text{s}$.

Write the expression for the average angular velocity of the pin wheel.

  ${\omega }_{\text{av}}=\frac{\Delta \theta }{\Delta t}$                                                                                                                   (I)

Here, ${\omega }_{\text{av}}$ is the average angular velocity, and $\Delta \theta$ is the angular displacement, and $\Delta t$ is the time duration.

Write the expression for the angular displacement of the pin wheel.

  $\Delta \theta ={\theta }_f-{\theta }_i$                                                                                                              (II)

Here, ${\theta }_f$ is the final angular distance, and ${\theta }_i$ is the initial angular distance.

Use equation (II) in (I).

  ${\omega }_{\text{av}}=\frac{{\theta }_f-{\theta }_i}{\Delta t}$                                                                                                            (III)

Conclusion:

Substitute $90°$ for ${\theta }_f$, $0°$ for ${\theta }_i$, and $0.150\text{s}$ for $\Delta t$ in equation (III) to find the average angular velocity of the pin wheel ${\omega }_{\text{av}}$.

  $\begin{array}{c}{\omega }_{\text{av}}=\frac{90°-0°}{0.150\text{s}}\\ =\frac{\left(90°-0°\right)\times \frac{2\pi \text{rad}}{360°}}{0.150\text{s}}\\ =10.5\text{rad/s}\end{array}$

Therefore, the average angular velocity of the pin wheel if it rotates from $\theta =0°$ to $\theta =90°$ in a time of $0.150\text{s}$ is $\underset{\_}{10.5\text{rad/s}}$.

(b)

To determine

The average angular velocity of the pin wheel if it rotates from $\theta =0°$ to $\theta =180°$ in a time of $0.300\text{s}$.

Answer to Problem 60PQ

The average angular velocity of the pin wheel if it rotates from $\theta =0°$ to $\theta =180°$ in a time of $0.300\text{s}$ is $\underset{\_}{10.5\text{rad/s}}$.

Explanation of Solution

Given that the pin wheel rotates from $\theta =0°$ to $\theta =180°$ in a time of $0.300\text{s}$.

Equation (III) gives the average angular velocity of the pin wheel.

  ${\omega }_{\text{av}}=\frac{{\theta }_f-{\theta }_i}{\Delta t}$

Conclusion:

Substitute $180°$ for ${\theta }_f$, $0°$ for ${\theta }_i$, and $0.300\text{s}$ for $\Delta t$ in equation (III) to find the average angular velocity of the pin wheel ${\omega }_{\text{av}}$.

  $\begin{array}{c}{\omega }_{\text{av}}=\frac{180°-0°}{0.300\text{s}}\\ =\frac{\left(180°-0°\right)\times \frac{2\pi \text{rad}}{360°}}{0.300\text{s}}\\ =10.5\text{rad/s}\end{array}$

Therefore, the average angular velocity of the pin wheel if it rotates from $\theta =0°$ to $\theta =180°$ in a time of $0.300\text{s}$ is $\underset{\_}{10.5\text{rad/s}}$.

(c)

To determine

The average angular velocity of the pin wheel if it rotates from $\theta =0°$ to $\theta =270°$ in a time of $0.450\text{s}$.

Answer to Problem 60PQ

The average angular velocity of the pin wheel if it rotates from $\theta =0°$ to $\theta =270°$ in a time of $0.450\text{s}$ is $\underset{\_}{10.5\text{rad/s}}$ or $\underset{\_}{-3.5\text{rad/s}}$.

Explanation of Solution

Given that the pin wheel rotates from $\theta =0°$ to $\theta =270°$ in a time of $0.450\text{s}$.

Equation (III) gives the average angular velocity of the pin wheel.

  ${\omega }_{\text{av}}=\frac{{\theta }_f-{\theta }_i}{\Delta t}$

Conclusion:

Substitute $270°$ for ${\theta }_f$, $0°$ for ${\theta }_i$, and $0.450\text{s}$ for $\Delta t$ in equation (III) to find the average angular velocity of the pin wheel ${\omega }_{\text{av}}$.

  $\begin{array}{c}{\omega }_{\text{av}}=\frac{270°-0°}{0.450\text{s}}\\ =\frac{\left(270°-0°\right)\times \frac{2\pi \text{rad}}{360°}}{0.450\text{s}}\\ =10.5\text{rad/s}\end{array}$

Since the direction in which the pin wheel rotates is not mentioned, the net displacement corresponding to $270°$ is equivalent to a net angular displacement corresponding to $90°$ clockwise rotation ($-90°$). Hence, substituting $-90°$ for ${\theta }_f$, $0°$ for ${\theta }_i$, and $0.450\text{s}$ for $\Delta t$ in equation (III) yields,

  $\begin{array}{c}{\omega }_{\text{av}}=\frac{-90°-0°}{0.450\text{s}}\\ =\frac{\left(-90°-0°\right)\times \frac{2\pi \text{rad}}{360°}}{0.450\text{s}}\\ =-3.5\text{rad/s}\end{array}$

Therefore, the average angular velocity of the pin wheel if it rotates from $\theta =0°$ to $\theta =270°$ in a time of $0.450\text{s}$ is $\underset{\_}{10.5\text{rad/s}}$ or $\underset{\_}{-3.5\text{rad/s}}$.

(d)

To determine

The average angular velocity of the pin wheel if it rotates from $\theta =0°$ through one revolution to $\theta =360°$ in a time of $0.600\text{s}$.

Answer to Problem 60PQ

The average angular velocity of the pin wheel if it rotates from $\theta =0°$ through one revolution to $\theta =360°$ in a time of $0.600\text{s}$ is $\underset{\_}{0\text{rad/s}}$.

Explanation of Solution

Given that the pin wheel from $\theta =0°$ through one revolution to $\theta =360°$ in a time of $0.600\text{s}$

Equation (III) gives the average angular velocity of the pin wheel.

Equation (I) gives the average angular velocity of the pin wheel.

  ${\omega }_{\text{av}}=\frac{\Delta \theta }{\Delta t}$

Conclusion:

During one revolution, the pin wheel covers $360°$ and reaches the same point from which it started its rotation, that is $0°$. Since the final position and the initial position of the pin wheel’s rotation, the angular displacement is zero ($\Delta \theta =0$). This leads equation (I) to yield the average angular velocity of the pinwheel to be zero.

  $\begin{array}{c}{\omega }_{\text{av}}=\frac{0\text{rad}}{0.600\text{s}}\\ =0\text{rad/s}\end{array}$

Therefore, the average angular velocity of the pin wheel if it rotates from $\theta =0°$ through one revolution to $\theta =360°$ in a time of $0.600\text{s}$ is $\underset{\_}{0\text{rad/s}}$.