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Chapter 9: Problem 86

The Crab Nebula is a cloud of glowing gas about 10 light-years across, located about 6500 light-years from the earth (Fig. P9.86). It is the remnant of a star that underwent a supernova \(e x\) plosion, seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about \(5 \times 10^{31} \mathrm{~W},\) about \(10^{5}\) times the rate at which the sun radiates energy. The Crab Nebula obtains its energy from the rotational kinetic energy of a rapidly spinning neutron star at its center. This object rotates once every \(0.0331 \mathrm{~s},\) and this period is increasing by \(4.22 \times 10^{-13} \mathrm{~s}\) for each second of time that elapses. (a) If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star. (b) Theories of supernovae predict that the neutron star in the Crab Nebula has a mass about 1.4 times that of the sun. Modeling the neutron star as a solid uniform sphere, calculate its radius in kilometers. (c) What is the linear speed of a point on the equator of the neutron star? Compare to the speed of light. (d) Assume that the neutron star is uniform and calculate its density. Compare to the density of ordinary rock \(\left(3000 \mathrm{~kg} / \mathrm{m}^{3}\right)\) and to the density of an atomic nucleus (about \(\left.10^{17} \mathrm{~kg} / \mathrm{m}^{3}\right) .\) Justify the statement that a neutron star is essentially a large atomic nucleus.

Short Answer

Expert verified
The short answers to the questions are: (a) The moment of inertia of the neutron star can be found using the formula for rotational kinetic energy and the given data about rate of energy distribution and rotation time. (b) The radius of the neutron star, modelled as a solid uniform sphere, can be calculated using the formula for moment of inertia and the given data about moment of inertia and mass. (c) The linear speed of a point on the neutron star's equator can be calculated using the formula for tangential speed, applying the known radius and angular speed. (d) The density of the neutron star can be calculated by dividing the mass of the star by its volume, modelled as a sphere.

Step by step solution

01

Finding the Moment of Inertia

The rate at which energy is released by the nebula is given as \(5 \times 10^{31} \mathrm{~W}\), and it is mentioned that this is equal to the rate at which energy is lost by the neutron star. The rate of energy loss is equal to the change in energy over change in time, using the fact that rotational kinetic energy is given by \(\frac{1}{2} I \omega^{2}\), where I is the moment of inertia and 蠅 is the angular velocity. Furthermore, given that initial time for the rotation is \(0.0331 s\) and the increase in time per second is \(4.22 \times 10^{-13} s\). The moment of inertia can hence be found using the formula \(dE/dt=-d\left(\frac{1}{2} I \omega^{2}\right)/dt\).
02

Calculating the Radius of the Neutron Star

The neutron star's mass is given as 1.4 times the mass of the sun. Let's represent the sun's mass as \(M_{sun}\), and the neutron star's mass as \(M_{ns} = 1.4 \times M_{sun}\). The neutron star is modelled as a solid uniform sphere, thus its moment of inertia is given by \(I=\frac{2}{5} M_{ns} R^{2}\), where R is the radius of the sphere. Using the moment of inertia found in step 1, the radius of the neutron star can be obtained.
03

Finding the linear speed of a point on the Neutron Star's equator

This is basically the tangential speed of a point on neutron star, which can be calculated using the formula \(v = R \times \omega\). Here, \(\omega\) is the angular speed and can be obtained from \(1/\) rotation time and R is the radius obtained from step 2.
04

Calculating the Density of the Neutron Star

Density is mass per unit volume. For a sphere the volume is given by \((4/3) \pi R^{3}\), where R is the radius of the sphere. Substituting the values of the mass of the neutron star and the radius calculated in step 2 into this formula will give the density.

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Key Concepts

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

Neutron Star
A neutron star is the incredibly dense core left behind after a massive star undergoes a supernova explosion. These stars are made mostly of neutrons and are incredibly compact.
  • They have a mass about 1.4 times that of the sun, but with a diameter of only about 20 kilometers.
  • The extreme density means that their gravitational pull is incredibly strong, which causes them to rotate rapidly.
As a result, neutron stars are often observed emitting beams of electromagnetic radiation as pulsars.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotation. For a solid sphere, it is calculated using the formula:
\[ I = \frac{2}{5} M R^2 \]
where \(M\) is the mass and \(R\) is the radius of the sphere. In the context of the Crab Nebula's neutron star:
  • We use the rotational kinetic energy and the rate of energy loss to determine \(I\).
  • Knowing the spin rate helps us calculate how energy is distributed in the rotation.
This is crucial to understanding how the star's energy powers the surrounding nebula.
Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its rotation. It is given by:
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]
where \(\omega\) is the angular velocity. For the Crab Nebula's neutron star:
  • The energy released comes from the star's rotational kinetic energy.
  • The change in rotational speed over time indicates a transfer of energy from the star to the nebula.
This loss of kinetic energy is observable as a slow down in rotation frequency.
Supernova
A supernova is a powerful explosion that occurs at the end of a massive star's life cycle.
  • During this event, most of the star's material is expelled into space, forming objects like the Crab Nebula.
  • The core that remains may become a neutron star, packed with nuclear-density material.
These explosions are crucial in dispersing elements into the universe and play a significant role in cosmic evolution.
Linear Speed
The linear speed of a point on the neutron star's equator is its tangential speed. It can be calculated using:
\[ v = R \times \omega \]
where \(R\) is the radius found earlier and \(\omega\) is the angular speed. For neutron stars, this speed:
  • Can be extremely high due to their rapid rotation.
  • Although fast, remains well below the speed of light.
Such high speeds illustrate the energetic nature of these celestial bodies.
Nuclear Density
Neutron stars are packed with matter at nuclear density, around \(10^{17} \text{ kg/m}^3\).
  • This density is immensely greater than ordinary rock, which has a density of about \(3000 \text{ kg/m}^3\).
  • It is comparable to that found within an atomic nucleus.
This remarkable density results in a gravitational field that can warp space-time, showcasing the extraordinary nature of neutron stars.
Neutron Star Radius
The radius of a neutron star can be determined by its mass and moment of inertia, using the formula for a sphere's inertia.
  • It is affected by the mass, which is typically 1.4 times that of the sun.
  • Using geometry and physics, we can calculate that these stars, despite their mass, are only about the size of a city.
Understanding the radius helps in grasping just how compact these stars are, containing vast amounts of mass in a tiny volume.

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Most popular questions from this chapter

A thin, light wire is wrapped around the rim of a wheel as shown in Fig. E9.49. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius \(0.180 \mathrm{~m}\) and moment of inertia for rotation about the axle of \(I=0.480 \mathrm{~kg} \cdot \mathrm{m}^{2}\). A small block with mass \(0.340 \mathrm{~kg}\) is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-9.00 \mathrm{~J}\) of work as the block descends \(3.00 \mathrm{~m}\). What is the magnitude of the angular velocity of the wheel after the block has descended \(3.00 \mathrm{~m} ?\)

A physics student of mass \(43.0 \mathrm{~kg}\) is standing at the edge of the flat roof of a building, \(12.0 \mathrm{~m}\) above the sidewalk. An unfriendly \(\operatorname{dog}\) is running across the roof toward her. Next to her is a large wheel mounted on a horizontal axle at its center. The wheel, used to lift objects from the ground to the roof, has a light crank attached to it and a light rope wrapped around it; the free end of the rope hangs over the edge of the roof. The student grabs the end of the rope and steps off the roof. If the wheel has radius \(0.300 \mathrm{~m}\) and a moment of inertia of \(9.60 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for rotation about the axle, how long does it take her to reach the sidewalk, and how fast will she be moving just before she lands? Ignore friction in the axle.

Spin cycles of washing machines remove water from clothes by producing a large radial acceleration at the rim of the cylindrical tub that holds the water and clothes. Estimate the diameter of the tub in a typical home washing machine. (a) What is the rotation rate, in rev/min, of the tub during the spin cycle if the radial acceleration of points on the tub wall is \(3 g ?\) (b) At this rotation rate, what is the tangential speed in \(\mathrm{m} / \mathrm{s}\) of a point on the tub wall?

A disk of radius \(25.0 \mathrm{~cm}\) is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. \(\mathbf{P 9 . 6 1}\) ). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is \(1.80 \mathrm{~m} / \mathrm{s}^{2}\). (a) Find \(A\). (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of \(15.0 \mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches \(15.0 \mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .)\)

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant linear speed of \(v=1.25 \mathrm{~m} / \mathrm{s} .\) Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the \(\mathrm{CD}\) is played. (See Exercise \(9.20 .\) ) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta)=r_{0}+\beta \theta,\) where \(r_{0}\) is the radius of the spiral at \(\theta=0\) and \(\beta\) is a constant. On a \(\mathrm{CD}, r_{0}\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d \theta,\) the distance scanned along the track is \(d s=r d \theta .\) Using the above expression for \(r(\theta),\) integrate \(d s\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) since the track is scanned at a constant linear speed \(v,\) the distance \(s\) found in part (a) is equal to vi. Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta ;\) choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_{z}\) and the angular acceleration \(\alpha_{z}\) as functions of time. Is \(\alpha_{z}\) constant? (d) On a CD, the inner radius of the track is \(25.0 \mathrm{~mm}\), the track radius increases by \(1.55 \mu \mathrm{m}\) per revolution, and the playing time is \(74.0 \mathrm{~min} .\) Find \(r_{0}, \beta,\) and the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_{z}\) (in rad/s) versus \(t\) and \(\alpha_{z}\) (in rad/s \(^{2}\) ) versus \(t\) between \(t=0\) and \(t=74.0 \mathrm{~min}\)
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