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Chapter 10: Problem 43

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly \(10^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was \(7.0 \times 10^{5} \mathrm{~km}\) (comparable to our sun); its final radius is \(16 \mathrm{~km}\). If the original star rotated once in 30 days, find the angular speed of the neutron star.

Short Answer

Expert verified
The final angular speed \(\omega_2\) of the neutron star can be calculated using the law of conservation of angular momentum and the given values for initial and final radius and the initial angular speed.

Step by step solution

01

Calculating Initial Angular Momentum

Compute the initial angular momentum before the collapse. Use the formula for the moment of inertia of a sphere \(I = \frac{2}{5}mr^2\) and the formula of angular momentum \(L = I\omega\). Here, m is the mass of the star, r is its initial radius, and \(\omega\) is its initial angular speed. We could denote them as m, \(r_1\) and \(\omega_1\). However, we don't know the mass of the star, so we can only write the initial angular momentum \(L_1\) as a function of m: \(L_1 = I_1\omega_1= \frac{2}{5}m r_1^{2}\omega_1\).
02

Calculating Final Angular Momentum

Following the same logic, the final angular momentum \(L_2\) after the collapse will be \(L_2 = I_2\omega_2=\frac{2}{5}m r_2^{2}\omega_2\), where \(r_2\) is the final radius and \(\omega_2\) is the final angular speed of the star.
03

Using Conservation of Angular Momentum

According to the law of conservation of angular momentum, the initial and final angular momentum should be equal, hence \(L_1 = L_2\). Meaning \(\frac{2}{5}m r_1^{2}\omega_1 = \frac{2}{5}m r_2^{2}\omega_2\). As we can see, the mass of the star 'm' and the fraction \(\frac{2}{5}\) are present on both sides of the equation, so they can be cancelled out. That leaves us with the equation \(r_1^{2}\omega_1 = r_2^{2}\omega_2\) to calculate the final angular speed.
04

Solving for the Final Angular Speed

Solving the equation for \(\omega_2\), the final angular speed, we get \(\omega_2 = \frac{r_1^{2}\omega_1} {r_2^{2}}\). By substituting the given values into the equation, the final angular speed \(\omega_2\) can be found.

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Key Concepts

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

Conservation of Angular Momentum
Conservation of angular momentum is a fundamental principle in physics, stating that if no external torque acts on a system, the total angular momentum of that system remains constant. In simpler terms, an object's spin will not change unless something causes it to. This principle is crucial in understanding the behavior of objects in rotational motion, such as stars, planets, and even galaxies.

For instance, in the exercise provided, a star collapses into a neutron star. Despite its significant change in size, if the system is isolated (meaning no external forces or moments act upon it), the star's angular momentum before and after the collapse must be identical. This conservation underpins the calculation to find the angular speed of the neutron star post-collapse. The process takes into account the star's initial rotation rate and adjusts it based on changes in the star's moment of inertia due to the collapse.

Understanding the law of conservation of angular momentum helps us comprehend how different bodies in space can change their rotational speeds after events such as supernovae or collisions, which can alter their distribution of mass.
Neutron Star Density
Neutron stars are incredibly dense, the density being roughly a colossal amount of times greater than that of ordinary solid matter. This extreme density means that a tremendous amount of mass is packed into a very small volume. A neutron star's density arises from its formation process; when a massive star ends its life cycle, it can go supernova, ejecting its outer layers. What remains collapses under gravity to an almost point-like object, leaving behind a neutron star.

The density of a neutron star is such that a teaspoonful of its material would weigh about a billion tons on Earth. This supreme compactness drastically increases the star's mass per unit volume (density), which, along with its radius, affects the star's moment of inertia 鈥� a key factor when calculating changes in angular speed as seen in the textbook exercise. The exercise involves using the concept of density to understand the scale of the physical changes a star undergoes when it collapses into a neutron star.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotation about an axis. It is an important concept, especially when dealing with rotational motion in physics. The moment of inertia is dependent on the object's mass and how that mass is distributed relative to the axis of rotation.

In our cosmic example involving a neutron star, we use the formula for the moment of inertia of a sphere, given by \( I = \frac{2}{5}mr^2 \) where \( m \) is the mass and \( r \) is the radius. Before and after the collapse of the star into a neutron star, the moments of inertia change dramatically due to the vast differences in their radii.

This change in the moment of inertia is key to solving for the angular speed after the collapse. The moment of inertia is inversely proportional to the angular speed, following the conservation of angular momentum. Therefore, as a star collapses and its moment of inertia decreases (due to a decrease in radius), its angular speed increases, resulting in a faster spinning neutron star, as calculated in the exercise.

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Most popular questions from this chapter

Example 10.7 calculates the friction force needed for a uniform sphere to roll down an incline without slipping. The incline is at an angle \(\beta\) above the horizontal. And the example discusses that the friction is static. (a) If the maximum friction force is given by \(f=\mu_{\mathrm{s}} n,\) where \(n\) is the normal force that the ramp exerts on the sphere, in terms of \(\beta\) what is the minimum coefficient of static friction needed if the sphere is to roll without slipping? (b) Based on your result in part (a), what does the minimum required \(\mu_{\mathrm{s}}\) become in the limits \(\beta \rightarrow 90^{\circ}\) and \(\beta \rightarrow 0^{\circ} ?\)

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 \(R / 2\).

Stabilization of the Hubble Space Telescope. The Hubble Space Telescope is stabilized to within an angle of about 2 -millionths of a degree by means of a series of gyroscopes that spin at 19,200 rpm. Although the structure of these gyroscopes is actually quite complex, we can model each of the gyroscopes as a thin-walled cylinder of mass \(2.0 \mathrm{~kg}\) and diameter \(5.0 \mathrm{~cm},\) spinning about its central axis. How large a torque would it take to cause these gyroscopes to precess through an angle of \(1.0 \times 10^{-6}\) degree during a 5.0 hour exposure of a galaxy?

A uniform solid disk made of wood is horizontal and rotates freely about a vertical axle at its center. The disk has radius \(0.600 \mathrm{~m}\) and mass \(1.60 \mathrm{~kg}\) and is initially at rest. A bullet with mass \(0.0200 \mathrm{~kg}\) is fired horizontally at the disk, strikes the rim of the disk at a point perpendicular to the radius of the disk, and becomes embedded in its rim, a distance of \(0.600 \mathrm{~m}\) from the axle. After being struck by the bullet, the disk rotates at \(4.00 \mathrm{rad} / \mathrm{s}\). What is the horizontal velocity of the bullet just before it strikes the disk?

A thin-walled, hollow spherical shell of mass \(m\) and radius \(r\) starts from rest and rolls without slipping down a track (Fig. \(\mathbf{P 1 0 . 7 2}\) ). Points \(A\) and \(B\) are on a circular part of the track having radius \(R\). The diameter of the shell is very small compared to \(h_{0}\) and \(R,\) and the work done by rolling friction is negligible. (a) What is the minimum height \(h_{0}\) for which this shell will make a complete loop-the-loop on the circular part of the track? (b) How hard does the track push on the shell at point \(B,\) which is at the same level as the center of the circle? (c) Suppose that the track had no friction and the shell was released from the same height \(h_{0}\) you found in part (a). Would it make a complete loop-theloop? How do you know? (d) In part (c), how hard does the track push on the shell at point \(A,\) the top of the circle? How hard did it push on the shell in part (a)?
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