Chapter 16, Problem 110P

To determine

The angular speed of the dipole when it reaches $\theta =0$, if it was released from rest at $\theta =90.0°$.

Answer to Problem 110P

The angular speed of the dipole when it reaches $\theta =0$, if it was released from rest at $\theta =90.0°$, is $20\text{rad}/\text{s}$.

Explanation of Solution

Write the expression for the torque acting on the dipole.

$\tau =qEd\sin \theta$ (I)

Here, $\tau$ is the torque on the dipole, $q$ is the charge, $E$ is the electric field acting, $d$ is dipole length and $\theta$ is the angle the dipole makes with electric field

Write the expression for the angular acceleration.

$\alpha =\frac{\tau }{I}$

Here, $\alpha$ is the angular acceleration and $I$ is the moment of inertia of the dipole

Put equation (I) in the above equation.

$\alpha =\left(\frac{qEd}{I}\right)\sin \theta$ (II)

Since the expression for the angular acceleration contains $\sin \theta$ , it is a variable function. This implies Newton’s constant acceleration equations of motion cannot be applied here.

The best way to calculate angular speed is to use right average angular acceleration. This average acceleration must lie between $qEd/I$ and $0$ within the range of $\theta =0$ to $0=\frac{\pi }{2}$.

Substitute $\frac{\pi }{2}$ for $\theta$ to get ${\alpha }_{\text{max}}$ and $0$ for $\theta$ to get ${\alpha }_{\text{min}}$ in equation (II).

$\begin{array}{c}{\alpha }_{\text{max}}=\left(\frac{qEd}{I}\right)\sin \frac{\pi }{2}\\ =\frac{qEd}{I}\end{array}$

$\begin{array}{c}{\alpha }_{\text{min}}=\left(\frac{qEd}{I}\right)\sin 0\\ =0\end{array}$

Here, ${\alpha }_{\text{min}}$ is the minimum angular acceleration and ${\alpha }_{\text{max}}$ is the maximum angular acceleration.

The following figure gives the graph of $\sin \theta$ for the interval $\theta =0\text{to }\theta =\frac{\pi }{2}$.

Above figure indicates that average value of $\sin \theta$ in this range is larger than one half of maximum value, because of the shape of the graph.

Consider the area of rectangle having height equal to right average acceleration and width $\frac{\pi }{2}$.This area has same value as area below the graph of $\sin \theta$ versus $\theta$ in the limit $\theta =0$ to $\theta =\frac{\pi }{2}$.

From more advanced mathematics, area of the sine function is the equal to $1$, when $\theta$ is in radians.

Write the expression for the area of rectangle.

$\begin{array}{c}\sin {\theta }_{\text{avg}}\left(\frac{\pi }{2}\right)=1\\ \sin {\theta }_{\text{avg}}=\frac{2}{\pi }\end{array}$

Write the expression for the angular velocity.

${\omega }_{\text{f}}^2-{\omega }_{\text{i}}^2=2\alpha \left({\theta }_{\text{f}}-{\theta }_{\text{i}}\right)$ (III)

Here, ${\omega }_{\text{f}}$ is the final angular velocity, ${\omega }_{\text{i}}$ is the initial angular velocity, ${\theta }_{\text{f}}$ is the final angular position and ${\theta }_{\text{i}}$ is the initial angular position.

The dipole system consists of charges at end of uniform rod.

Write the expression for moment of inertia of dipole system.

$I=m{r}^2+\left(\frac{1}{12}\right)M{L}^2+m{r}^2$ (IV)

Here, $m$ is the mass of charge, $r$ is the distance between center to end of rod, $M$ is the mass of rod and $L$ is the length of rod.

Conclusion:

Substitute $5.0\text{g}$ for $m$ , $20.0\text{g}$ for $M$ , $3.5\text{cm}$ for $r$ and $7.0\text{cm}$ for $L$ in equation (IV) to get $I$.

$\begin{array}{c}I=\left(5.0\text{g}\times \frac{1\text{kg}}{1000\text{g}}\right){\left(3.5\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2+\left(\frac{1}{12}\right)\left(20.0\text{g}\frac{1\text{kg}}{1000\text{g}}\right){\left(7.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2+\left(5.0\text{g}\times \frac{1\text{kg}}{1000\text{g}}\right){\left(3.5\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =2.0\times {10}^{-5}\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0\text{rad}/\text{s}$ for ${\omega }_{\text{i}}$ , $\frac{qEd}{I}\sin {\theta }_{\text{avg}}$ for $\alpha$ , $\frac{\pi }{2}$ for ${\theta }_{\text{f}}$ and $0$ for ${\theta }_{\text{i}}$ in equation(III) to get ${\omega }_{\text{f}}$.

$\begin{array}{c}{\omega }_{\text{f}}^2-{\left(0\text{rad}/\text{s}\right)}^2=2\times \frac{qEd}{I}\sin {\theta }_{\text{avg}}\left(\frac{\pi }{2}-0\right)\\ {\omega }_{\text{f}}^2=\frac{q\pi Ed}{I}\sin {\theta }_{\text{avg}}\end{array}$

Substitute $\frac{2}{\pi }$ for $\sin {\theta }_{\text{avg}}$ in above equation to get the final expression for ${\omega }_{\text{f}}$.

$\begin{array}{c}{\omega }_{\text{f}}^2=\frac{q\pi Ed}{I}\frac{2}{\pi }\\ {\omega }_{\text{f}}=\sqrt{\frac{2qEd}{I}}\end{array}$

Substitute $3.0\text{μC}$ for $q$ , $2.0\times {10}^4\text{N}/\text{C}$ for $E$ , $7.0\text{cm}$ for $d$ and $2.0\times {10}^{-5}\text{kg}\cdot {\text{m}}^2$ for $I$ in above equation to get ${\omega }_{\text{f}}$.

$\begin{array}{c}{\omega }_{\text{f}}=\sqrt{\frac{2\left(3.0\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}\right)\left(2.0\times {10}^4\text{N}/\text{C}\right)\left(7.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{2.0\times {10}^{-5}\text{kg}\cdot {\text{m}}^2}\frac{1\text{kg}\cdot \text{m}}{1\text{N}\cdot {\text{s}}^2}\frac{1\text{rad}}{1}}\\ =20\text{rad}/\text{s}\end{array}$

Therefore, the angular speed of the dipole when it reaches $\theta =0$, if it was released from rest at $\theta =90.0°$, is $20\text{rad}/\text{s}$.