Chapter 20, Problem 97P

(a)

To determine

The rotational speed of the armature of the generator.

Answer to Problem 97P

The rotational speed of the armature of the generator is $\underset{\_}{1.3\times {10}^3\text{rev}/\text{min}}$.

Explanation of Solution

Write the expression for motional emf.

  ${\epsilon }_m=NBA\omega$        (I)

Here, $N$ is the number of turns, $B$ is the magnetic field, $A$ is the area of cross section, $\omega$ is the angular speed.

Substitute, $\pi r^2$ for $A$ in equation (I).

  ${\epsilon }_m=NB\left(\pi r^2\right)\omega$        (II)

Conclusion:

Substitute, $150$ for $N$, $0.20\text{T}$ for $B$, $0.018\text{m}$ for $r$, and $4.2\text{V}$ for ${\epsilon }_m$ in equation (II) to obtain the rotational speed.

  $\begin{array}{c}4.2\text{V}=\left(150\right)\left(0.20\text{T}\right)\pi {\left(0.018\text{m}\right)}^2\omega \\ \omega =\frac{4.2\text{V}}{\left(150\right)\left(0.20\text{T}\right)\pi {\left(0.018\text{m}\right)}^2}\\ =137.5\frac{\text{rad}}{\text{s}}\times \frac{1\text{rev}}{2\pi \text{rad}}\frac{60\text{s}}{1\text{min}}\\ =1.3\times {10}^3\text{rev}/\text{min}\end{array}$

Therefore, the rotational speed of the armature of the generator is $\underset{\_}{1.3\times {10}^3\text{rev}/\text{min}}$.

(b)

To determine

The average torque and instantaneous torque applied by bicycle tire to the generator.

Answer to Problem 97P

The average torque applied by bicycle tire to the generator is $\underset{\_}{0.044\text{N}\cdot \text{m}}$, and the instantaneous torque applied by bicycle tire to the generator is $\underset{\_}{0.087\text{N}\cdot \text{m}}$.

Explanation of Solution

Write the expression for average torque.

  ${\tau }_{av}=\frac{P_{av}}{\omega }$        (III)

Here, $P_{av}$ is the average power, and $\omega$ is the angular frequency.

Conclusion:

Substitute, $60\text{W}$ for $P_{av}$, and $137.5/\text{s}$ for $\omega$ in equation (III) to find the average torque.

  $\begin{array}{c}{\tau }_{av}=\frac{60\text{W}}{137.5/\text{s}}\frac{1\text{J}/\text{s}}{1\text{W}}\frac{1\text{N}\cdot \text{m}}{1\text{J}}\\ =0.044\text{N}\cdot \text{m}\end{array}$

Substitute, $12.0\text{W}$ for $P_m$, and $137.5/\text{s}$ for $\omega$ in equation (III) to find the maximum instantaneous torque.

  $\begin{array}{c}{\tau }_{av}=\frac{12.0\text{W}}{137.5/\text{s}}\frac{1\text{J}/\text{s}}{1\text{W}}\frac{1\text{N}\cdot \text{m}}{1\text{J}}\\ =0.087\text{N}\cdot \text{m}\end{array}$

Therefore, the average torque applied by bicycle tire to the generator is $\underset{\_}{0.044\text{N}\cdot \text{m}}$, and the instantaneous torque applied by bicycle tire to the generator is $\underset{\_}{0.087\text{N}\cdot \text{m}}$.

(c)

To determine

The linear speed of the bicycle to supply emf of $4.2\text{V}$.

Answer to Problem 97P

The linear speed of the bicycle to supply emf of $4.2\text{V}$ is $\underset{\_}{1.4\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for the linear speed of the bicycle.

  $v=r_{shaft}{\omega }_{shaft}$        (IV)

Here, $r_{shaft}$ is the radius of the shaft.

Conclusion:

Substitute, $0.010\text{m}$ for $r_{shaft}$, and $137.5/\text{s}$ for ${\omega }_{shaft}$ in equation (IV) to find the linear speed.

  $\begin{array}{c}v=\left(0.010\text{m}\right)\left(137.5/\text{s}\right)\\ =1.4\text{m}/\text{s}\end{array}$

Therefore, the linear speed of the bicycle to supply emf of $4.2\text{V}$ is $\underset{\_}{1.4\text{m}/\text{s}}$.