/*! 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 43 Under some circumstances, a star... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 – 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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An engine delivers 175 hp to an aircraft propeller at 2400 rev \(/\) min. (a) How much torque does the aircraft engine provide? (b) How much work does the engine do in one revolution of the propeller?

A woman with mass \(50 \mathrm{~kg}\) is standing on the rim of a large horizontal disk that is rotating at \(0.80 \mathrm{rev} / \mathrm{s}\) about an axis through its center. The disk has mass \(110 \mathrm{~kg}\) and radius \(4.0 \mathrm{~m} .\) Calculate the magnitude of the total angular momentum of the woman-disk system. (Assume that you can treat the woman as a point.)

A block with mass \(m\) is revolving with linear speed \(v_{1}\) in a circle of radius \(r_{1}\) on a frictionless horizontal surface (see Fig. E10.42). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to \(r_{2}\). (a) Calculate the tension \(T\) in the string as a function of \(r\), the distance of the block from the hole. Your answer will be in terms of the initial velocity \(v_{1}\) and the radius \(r_{1} .\) (b) Use \(W=\int_{r_{1}}^{r_{2}} \overrightarrow{\boldsymbol{T}}(r) \cdot d \overrightarrow{\boldsymbol{r}}\) to calculate the work done by \(\vec{T}\) when \(r\) changes from \(r_{1}\) to \(r_{2}\). (c) Compare the results of part (b) to the change in the kinetic energy of the block.

A machine part has the shape of a solid uniform sphere of mass \(225 \mathrm{~g}\) and diameter \(3.00 \mathrm{~cm}\). It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of \(0.0200 \mathrm{~N}\) at that point. (a) Find its angular acceleration. (b) How long will it take to decrease its rotational speed by \(22.5 \mathrm{rad} / \mathrm{s} ?\)

A 42.0-cm-diameter wheel, consisting of a rim and six spokes, is constructed from a thin, rigid plastic material having a linear mass density of \(25.0 \mathrm{~g} / \mathrm{cm} .\) This wheel is released from rest at the top of a hill \(58.0 \mathrm{~m}\) high. (a) How fast is it rolling when it reaches the bottom of the hill? (b) How would your answer change if the linear mass density and the diameter of the wheel were each doubled?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.