Chapter 8, Problem 102P

(a)

To determine

Find the ratio of the angular momentum for rotating star.

Answer to Problem 102P

The ratio of the angular momentum for rotating star is $1$.

Explanation of Solution

Write the equation for angular momentum of a rigid object.

$L=I\omega$ (I)

Here, $L$ is the angular momentum of a rigid object, $I$ is the rotational inertia of the object, and $\omega$ is the angular velocity.

For a rigid object rotating around a fixed axis, angular momentum does not tell us anything, because the rotational inertia is constant for such an object since the distance $r_{\text{n}}$ between every point on the object and axis stays the same.

Conservation of angular momentum: If the net external torque acting on a system is zero, then the angular momentum of the system cannot change. This is the law of conservation of momentum.

The law of conservation of momentum is expressed as,

$\begin{array}{c}{\displaystyle \sum \tau }=0\\ L_{\text{i}}=L_{\text{f}}\end{array}$ (II)

Here, $L_{\text{i}}$ is the initial angular momentum and $L_{\text{f}}$ is the final angular momentum.

Conclusion:

Since, from the above explanation the rotating star is converted into the pulsar under the influence of gravity. This pulsar does not have any shape so it is can be considered as rigid object. The rotational inertia is constant for such an object.

Since the angular momentum is conserved, the ratio is 1.

(b)

To determine

Find the ratio of the angular velocity for rotating star.

Answer to Problem 102P

The ratio of the angular momentum for rotating star is $1.0\times {10}^8$.

Explanation of Solution

From the equation (I), write the equation for angular velocity of the object.

$\omega =\frac{L}{I}$ (III)

Write the equation for the inertia of the rotating body.

$I=m{r}^2$ (IV)

Here, $m$ is the mass of the object and $r$ is the radius of the object.

Rewrite equation (III) by using equation (IV).

$\omega =\frac{L}{m{r}^2}$ (V)

Since, from the equation (III) and (V) the rotational inertia is proportional to the square of the radius.

Conclusion:

The ratio of the angular velocities is.

$\frac{{\omega }_{\text{f}}}{{\omega }_{\text{i}}}=\frac{r_{\text{i}}^2}{r_{\text{f}}^2}$

Here, ${\omega }_{\text{i}}$ is the angular velocity of the star before collapse and ${\omega }_{\text{f}}$ is the angular velocity of the star after collapse.

The radius of the pulsar is $1.0\times {10}^{-4}$ times the radius of the star before collapse. There is no change in mass in both the cases mass is evenly distributed in a spherical shaped star.

$\begin{array}{c}\frac{{\omega }_{\text{f}}}{{\omega }_{\text{i}}}={\left(\frac{1}{1.0\times {10}^{-4}}\right)}^2\\ =1.0\times {10}^8\end{array}$

Therefore, the ratio of the angular momentum for rotating star is $1.0\times {10}^8$.

(c)

To determine

Find the ratio of the rotational kinetic energy of the star after collapse to the values of before collapse.

Answer to Problem 102P

The ratio of the rotational kinetic energy of the star is $1.0\times {10}^8$.

Explanation of Solution

Write the equation for the rotational kinetic energy.

$K_{\text{rot}}=\frac{1}{2}I{\omega }^2$ (VI)

Here, $K_{\text{rot}}$ is the rotational kinetic energy of the star.

Rewrite the equation (I) for $I$.

$I=\frac{L}{\omega }$ (VII)

Rewrite equation (VI) by using equation (VII).

$\begin{array}{c}{K}_{\text{rot}}=\frac{1}{2}\left(\frac{L}{\omega }\right){\omega }^2\\ =\frac{1}{2}L\omega \end{array}$ (VIII)

Conclusion:

The ratio of the rotational kinetic energies of the star is.

$\frac{K_{\text{f}}}{K_{\text{i}}}=\frac{{\omega }_{\text{f}}}{{\omega }_{\text{i}}}$

The radius of the pulsar is $1.0\times {10}^{-4}$ times the radius of the star before collapse. There is no change in mass in both the cases mass is evenly distributed in a spherical shaped star from part (b).

$\frac{K_{\text{f}}}{K_{\text{i}}}=1.0\times {10}^8$

Therefore, the ratio of the rotational kinetic energy of the star is $1.0\times {10}^8$.

(d)

To determine

The period of the star’s rotation after collapse.

Answer to Problem 102P

The period of the star’s rotation is $0.10\text{s}$.

Explanation of Solution

Write the equation for the period related to the angular velocity.

$T=\frac{2\pi }{\omega }$ (IX)

Here, $T$ is the period of rotation of the star.

Conclusion:

The period of the star after collapse is.

$\frac{T_{\text{f}}}{T_{\text{i}}}=\frac{{\omega }_{\text{i}}}{{\omega }_{\text{f}}}$

Here, $T_{\text{f}}$ is the period of star after collapse and $T_{\text{i}}$ is the period of star before collapse.

Rewrite the above equation for $T_{\text{f}}$.

$T_{\text{f}}=\frac{{\omega }_{\text{i}}}{{\omega }_{\text{f}}}{T}_{\text{i}}$ (X)

The radius of the pulsar is $1.0\times {10}^{-4}$ times the radius of the star before collapse. There is no change in mass in both the cases mass is evenly distributed in a spherical shaped star.

Substitute $1.0\times {10}^{-8}$ for $\left({\omega }_{\text{i}}/{\omega }_{\text{f}}\right)$ and $1.0\times {10}^7\text{s}$ for $T_{\text{i}}$ in equation (X).

$\begin{array}{c}{T}_{\text{f}}=\left(1.0\times {10}^{-8}\right)\left(1.0\times {10}^7\text{s}\right)\\ =1.0\times {10}^{-1}\text{s}\\ =0.10\text{s}\end{array}$

Therefore, period of the star’s rotation after collapse is $0.10\text{s}$.