/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 Neutron Star Glitches. Occasiona... [FREE SOLUTION] | 91Ó°ÊÓ

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Neutron Star Glitches. Occasionally, a rotating neutron star (see Exercise 10.41 ) undergoes a sudden and unexpected speedup called a glitch. One explanation is that a glitch occurs when the crust of the neutron star settles slightly, decreasing the moment of inertia about the rotation axis. A neutron star with angular speed \(\omega_{0}=70.4 \mathrm{rad} / \mathrm{s}\) underwent such a glitch in October 1975 that increased its angular speed to \(\omega=\omega_{0}+\Delta \omega,\) where \(\Delta \omega / \omega_{0}=2.01 \times 10^{-6} .\) If the radius of the neutron star before the glitch was \(11 \mathrm{km},\) by how much did its radius decrease in the star- quake? Assume that the neutron star is a uniform sphere.

Short Answer

Expert verified
The radius decreased by approximately 11 mm.

Step by step solution

01

Understanding the concept

The glitch phenomenon involves a sudden increase in angular speed due to a change in the moment of inertia of the neutron star. Since the neutron star is approximately spherical, we can relate its moment of inertia to its radius.
02

Determine the Initial and Final Angular Speeds

Given \[ \omega_0 = 70.4\, \text{rad/s} \]and \[ \frac{\Delta \omega}{\omega_0} = 2.01 \times 10^{-6},\]we find \[ \Delta \omega = \omega_0 \times \frac{\Delta \omega}{\omega_0} = 70.4 \times 2.01 \times 10^{-6} \approx 0.000141314 \text{ rad/s}. \]Thus,\[ \omega = \omega_0 + \Delta \omega = 70.400141314 \text{ rad/s}. \]
03

Apply Conservation of Angular Momentum

The angular momentum before and after the glitch is conserved, so \[ I_0 \omega_0 = I \omega, \]where \(I_0 = \frac{2}{5} m R_0^2\) and \(I = \frac{2}{5} m R^2\) are the moments of inertia before and after respectively. Simplifying gives \[ R^2\omega = R_0^2 \omega_0. \]
04

Solve for the New Radius

Replacing \(R\) with \(R_0 - \Delta R\) leads to \[ (R_0 - \Delta R)^2 \omega = R_0^2 \omega_0. \]Simplifying, \[ \omega (R_0^2 - 2R_0 \Delta R + \Delta R^2) = \omega_0 R_0^2. \]Assuming \(\Delta R^2\) is negligible,\[ 2R_0 \Delta R \approx R_0^2(\omega_0/\omega - 1). \]
05

Substitute Known Values and Calculate

Using the values given, \[ R_0 = 11 \times 10^3 \text{ m}, \]\[ \Delta R = \frac{R_0^2 (\omega_0/\omega - 1)}{2R_0}. \]Substituting,\[ \Delta R \approx \frac{11,000^2 (70.4/70.400141314 - 1)}{2 \times 11,000}, \]which simplifies to approximately \( \Delta R \approx 11 \times 10^{-3} \text{ m} \).
06

Conclusion

The radius decreases by approximately \(11\, \text{mm}\) during the starquake.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
The moment of inertia is a fundamental concept in physics that describes an object's resistance to changes in its rotational motion. For a uniform spherical object like a neutron star, the moment of inertia can be calculated using the formula
  • \( I = \frac{2}{5} m R^2 \)
where \( m \) is the mass and \( R \) is the radius. This formula tells us that the moment of inertia is directly related to the radius. Thus, if the radius decreases, the moment of inertia will also decrease, assuming mass remains constant.

In neutron stars, a decrease in the moment of inertia can lead to an increase in rotational speed, if the star's radial shape changes. This is because the total angular momentum must be conserved. Hence, understanding moment of inertia helps us predict how changes in the star’s structure can influence its rotational behavior.
Angular Momentum Conservation
Angular momentum conservation is a critical principle in physics that states the total angular momentum of a closed system remains constant if no external torques act on it. For rotating bodies like neutron stars, this law translates into the equation
  • \( I_0 \omega_0 = I \omega \)
where \( I_0 \) and \( I \) are the initial and final moments of inertia, and \( \omega_0 \) and \( \omega \) are the initial and final angular speeds.

When a neutron star experiences a glitch, it is often hypothesized that the star's crust shifts slightly, causing a change in \( I \). Due to angular momentum conservation, a decrease in the moment of inertia \( I \) necessitates an increase in the angular speed \( \omega \), explaining the sudden speedup observed in such glitches.
This relationship is crucial in understanding rotational dynamics not only in stars but also in other celestial and mechanical systems.
Angular Speed Changes
Angular speed, denoted by \( \omega \), is the rate at which an object rotates or revolves relative to another point, such as how fast a neutron star spins around its axis. When the structure of a neutron star changes – for instance, when its outer crust settles – it affects the star's moment of inertia, leading to changes in angular speed.

The formula for the glitch in angular speed can be expressed as
  • \( \omega = \omega_0 + \Delta \omega \)
where \( \Delta \omega \) represents the change in angular speed. For neutron stars, this change is incredibly small but significant.
  • For example, if \( \omega_0 = 70.4 \, \text{rad/s} \) and \( \Delta \omega / \omega_0 = 2.01 \times 10^{-6} \), the resulting \( \Delta \omega \) is a tiny increase of about 0.000141 rad/s.

Such small changes are detectable due to the precise observational techniques used by astronomers, which are essential for studying the dynamics of these fascinating stellar remnants.
Neutron Star Physics
Neutron stars are the remnants of massive stars that have ended their life cycles in spectacular supernovae. These objects are incredibly dense, with masses greater than our Sun but compressed into spheres only about 20 kilometers in diameter. This density results in extraordinary gravitational and magnetic field strengths.

In terms of physics, neutron stars provide extreme environments where theories of gravity and quantum mechanics intersect. Their unique properties, such as high rotational speeds and strong magnetic fields, make neutron stars natural laboratories for testing the limits of physical laws.
  • Glitches, or sudden changes in rotation speeds, are particularly interesting events that suggest a complex internal structure.
  • The crust, superfluid interior, and magnetic fields all interact in ways that are not fully understood, providing rich avenues for research.
Studying these glitches helps scientists unravel the mysteries of nuclear matter under extreme conditions, enhancing our understanding of the universe.

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Most popular questions from this chapter

BIO Gymnastics. We can roughly model a gymnastic tumbler as a uniform solid cylinder of mass 75 \(\mathrm{kg}\) and diameter 1.0 \(\mathrm{m}\) . If this tumbler rolls forward at 0.50 rev/s, (a) how much total kinetic energy does he have, and (b) what percent of his total kinetic energy is rotational?

What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? (a) a uniform solid cylinder; (b) a uniform sphere; ( c) a thin walled, hollow sphere; (d) a hollow cylinder with outer radius \(R\) and inner radius \(R / 2\) .

The solid wood door of a gymnasium is 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high, has total mass \(35.0 \mathrm{kg},\) and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 \(\mathrm{N}\) to the door for 8.00 \(\mathrm{ms} .\) Find the angular speed of the door after the impact. [Hint: Integrating Eq. \((10.29)\) yields \(\Delta L_{z}=\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t=\left(\sum \tau_{z}\right)_{\mathrm{av}} \Delta t .\) The quantity \(\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t\) is called the angular impulse.]

A uniform, \(0.0300-\mathrm{kg}\) rod of length 0.400 \(\mathrm{m}\) rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.0200 \(\mathrm{kg}\) , are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 \(\mathrm{m}\) on each side of the center of the rod, and the system is rotating at 30.0 rev/min. With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. (a) What is the angular speed of the system at the instant when the rings reach the ends of the rod? (b) What is the angular speed of the rod after the rings leave it?

One force acting on a machine part is \(\vec{\boldsymbol{F}}=(-5.00 \mathrm{N}) \hat{\imath}+\) \((4.00 \mathrm{N}) \hat{\boldsymbol{J}}\) . The vector from the origin to the point where the force is applied is \(\vec{r}=(-0.450 \mathrm{m}) \hat{\imath}+(0.150 \mathrm{m}) \hat{\boldsymbol{J}}\) (a) In a sketch, show \(\vec{r}, \vec{\boldsymbol{F}},\) and the origin. (b) Use the right-hand rule to determine the direction of the torque. (c) Calculate the vector torque for an axis at the origin produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).

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