Chapter 12, Problem 15PQ

(a)

To determine

The rotation rate after the glitch.

Answer to Problem 15PQ

The rotation rate after the glitch is $\underset{\_}{189\text{rad/s}}$.

Explanation of Solution

From Example 12.2, it is found that the original rotation rate is $189\text{rad/s}$. The rotation rate after the glitch (final rotation rate) is the sum of the original rotation rate with the increase in rotation rate.

Write the expression for the rotation rate after the glitch.

  ${\omega }_f={\omega }_{\text{orig}}+\Delta \omega$                                                                                                        (I)

Here, ${\omega }_f$ is the final rotation rate, ${\omega }_{\text{orig}}$ is the original rotation rate, and $\Delta \omega$ is the change (increase) in rotation rate.

It is given that the fractional increase in rotation rate is,

  $\frac{\Delta \omega }{{\omega }_{\text{orig}}}=2.4\times {10}^{-8}$                                                                                                      (II)

Solve equation (II) for $\Delta \omega$ and use it in equation (I).

  $\begin{array}{c}{\omega }_f={\omega }_{\text{orig}}+\left(2.4\times {10}^{-8}\right){\omega }_{\text{orig}}\\ =\left(1+2.4\times {10}^{-8}\right){\omega }_{\text{orig}}\\ \approx {\omega }_{\text{orig}}\end{array}$                                                                                   (IV)

This shows that the rotation rate after the glitch is almost equal to the original rotation rate since the change is negligibly small compared to the original rate. Thus, substituting $189\text{rad/s}$ for ${\omega }_{\text{orig}}$ in equation (IV) yields the rotation rate after the glitch ${\omega }_f$.

  ${\omega }_f\approx 189\text{rad/s}$

Conclusion:

Therefore, the rotation rate after the glitch is $\underset{\_}{189\text{rad/s}}$.

(b)

To determine

The new angular acceleration.

Answer to Problem 15PQ

The new angular acceleration is $\underset{\_}{2.40\times {10}^{-9}{\text{rad/s}}^2}$.

Explanation of Solution

From Example 12.2, it is found that the magnitude of original angular acceleration is $2.39\times {10}^{-9}{\text{rad/s}}^2$. The angular acceleration after the glitch (final angular acceleration) is the sum of the original angular acceleration with the increase in angular acceleration.

Write the expression for the angular acceleration after the glitch.

  ${\alpha }_f={\alpha }_0+\Delta \alpha$                                                                                                         (V)

Here, ${\alpha }_f$ is the final angular acceleration, ${\alpha }_0$ is the original angular acceleration, and $\Delta \alpha$ is the change (increase) in angular acceleration.

It is given that the fractional increase in angular acceleration is,

  $\frac{\Delta \alpha }{{\alpha }_{\text{0}}}=5\times {10}^{-3}$                                                                                                         (VI)

Solve equation (II) for $\Delta \alpha$ and use it in equation (I).

  $\begin{array}{c}{\alpha }_f={\alpha }_{\text{0}}+\left(5\times {10}^{-3}\right){\alpha }_{\text{0}}\\ =\left(1+5\times {10}^{-3}\right){\alpha }_{\text{0}}\\ =1.005{\alpha }_{\text{0}}\end{array}$                                                                                          (VII)

Conclusion:

Substitute $2.39\times {10}^{-9}{\text{rad/s}}^2$ for ${\alpha }_{\text{0}}$ in equation (VII) to find the angular acceleration after the glitch ${\alpha }_f$.

  $\begin{array}{c}{\alpha }_f=1.005\left(2.39\times {10}^{-9}{\text{rad/s}}^2\right)\\ =2.40\times {10}^{-9}{\text{rad/s}}^2\end{array}$

Therefore, the new angular acceleration is $\underset{\_}{2.40\times {10}^{-9}{\text{rad/s}}^2}$.

(c)

To determine

The time taken for the pulsar to return to its pre-glitch rotation rate.

Answer to Problem 15PQ

The time taken for the pulsar to return to its pre-glitch rotation rate is $\underset{\_}{1.9\times {10}^3\text{s}}$.

Explanation of Solution

It is obtained that the angular acceleration after the glitch is $2.40\times {10}^{-9}{\text{rad/s}}^2$.

Write the expression for the angular frequency in term of the angular acceleration.

  $\omega ={\omega }_{\text{orig}}+\alpha t$                                                                                                       (VIII)

Here, $t$ is the time.

Solve equation (VIII) for $t$.

  $t=\frac{\omega -{\omega }_{\text{orig}}}{\alpha }$                                                                                                           (IX)

Multiply and divide the right-hand side of equation (IX) by ${\omega }_{\text{orig}}$.

  $\begin{array}{c}t=\frac{\omega -{\omega }_{\text{orig}}}{\alpha }\left(\frac{{\omega }_{\text{orig}}}{{\omega }_{\text{orig}}}\right)\\ =\left(\frac{\omega -{\omega }_{\text{orig}}}{{\omega }_{\text{orig}}}\right)\left(\frac{{\omega }_{\text{orig}}}{\alpha }\right)\\ =\left(\frac{\Delta \omega }{{\omega }_{\text{orig}}}\right)\left(\frac{{\omega }_{\text{orig}}}{\alpha }\right)\end{array}$                                                                                             (X)

Conclusion:

Substitute $2.4\times {10}^{-8}$ for $\Delta \omega /{\omega }_{\text{orig}}$, $189\text{rad/s}$ for ${\omega }_{\text{orig}}$, and $2.40\times {10}^{-9}{\text{rad/s}}^2$ for $\alpha$ in equation (X) to find $t$.

  $\begin{array}{c}t=\left(2.4\times {10}^{-8}\right)\left(\frac{189\text{rad/s}}{2.40\times {10}^{-9}{\text{rad/s}}^2}\right)\\ =1.9\times {10}^3\text{s}\end{array}$

Therefore, the new angular acceleration is $1.9\times {10}^3\text{s}$.