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Chapter 10: Problem 64

Most stars maintain an equilibrium size by balancing two forces - an inward gravitational force and an outward force resulting from the star's nuclear reactions. When the star's fuel is spent, there is no counterbalance to the gravitational force. Whatever material is remaining collapses in on itself. Stars about the same size as the Sun become white dwarfs, which glow from leftover heat. Stars that have about three times the mass of the Sun compact into neutron stars. And a star with a mass greater than three times the Sun's mass collapses into a single point, called a black hole. In most cases, protons and electrons are fused together to form neutrons-this is the reason for the name neutron star. Neutron stars rotate very fast because of the conservation of angular momentum. Imagine a star of mass \(5.00 \cdot 10^{30} \mathrm{~kg}\) and radius \(9.50 \cdot 10^{8} \mathrm{~m}\) that rotates once in 30.0 days. Suppose this star undergoes gravitational collapse to form a neutron star of radius \(10.0 \mathrm{~km} .\) Determine its rotation period.

Short Answer

Expert verified
Answer: The rotation period of the neutron star after gravitational collapse is approximately \(_\Box_\) days.

Step by step solution

01

Recall the equation of angular momentum

The equation for angular momentum (L) is given by: $$L = Iω$$ where I is the moment of inertia and ω is the angular velocity.
02

Calculate the initial angular momentum

Before the collapse, the star has mass \(M_1 = 5.00 \cdot 10^{30} \mathrm{~kg}\), radius \(R_1 = 9.50 \cdot 10^{8} \mathrm{~m}\), and rotation period of \(P_1 = 30.0 \mathrm{~days}\). The moment of inertia for a sphere is given by: $$I_1 = \dfrac{2}{5}M_1R_1^2$$ To find the initial angular velocity, we can use the formula: $$ω_1 = \dfrac{2π}{P_1}$$ Note that we need to convert the period from days to seconds. The initial angular momentum, \(L_1\), is given by: $$L_1 = I_1ω_1$$
03

Calculate the final moment of inertia and angular momentum

After the collapse, the star has a new radius, \(R_2 = 10.0 \mathrm{~km}\). We assume that the mass remains the same during the collapse, so \(M_2 = M_1\). The final moment of inertia, \(I_2\), is given by: $$I_2 = \dfrac{2}{5}M_2R_2^2$$ Using the conservation of angular momentum, we know that the final angular momentum, \(L_2\), is equal to the initial angular momentum, \(L_1\) i.e. $$L_2 = L_1$$
04

Calculate the final angular velocity and rotation period

Now, we can find the final angular velocity, \(ω_2\), using the final angular momentum and moment of inertia: $$ω_2 = \dfrac{L_2}{I_2}$$ To find the final rotation period, \(P_2\), we use the equation: $$P_2 = \dfrac{2π}{ω_2}$$ Finally, we can convert the rotation period from seconds back to days.

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Most popular questions from this chapter

A circular platform of radius \(R_{p}=4.00 \mathrm{~m}\) and mass \(M_{\mathrm{p}}=400 .\) kg rotates on frictionless air bearings about its vertical axis at 6.00 rpm. An 80.0 -kg man standing at the very center of the platform starts walking \((\) at \(t=0)\) radially outward at a speed of \(0.500 \mathrm{~m} / \mathrm{s}\) with respect to the platform. Approximating the man by a vertical cylinder of radius \(R_{\mathrm{m}}=0.200 \mathrm{~m}\) determine an equation (specific expression) for the angular velocity of the platform as a function of time. What is the angular velocity when the man reaches the edge of the platform?

An ice skater spins with her arms extended and then pulls her arms in and spins faster. Which statement is correct? a) Her kinetic energy of rotation does not change because, by conservation of angular momentum, the fraction by which her angular velocity increases is the same as the fraction by which her rotational inertia decreases. b) Her kinetic energy of rotation increases because of the work she does to pull her arms in. c) Her kinetic energy of rotation decreases because of the decrease in her rotational inertia; she loses energy because she gradually gets tired.

In a tire-throwing competition, a man holding a \(23.5-\mathrm{kg}\) car tire quickly swings the tire through three full turns and releases it, much like a discus thrower. The tire starts from rest and is then accelerated in a circular path. The orbital radius \(r\) for the tire's center of mass is \(1.10 \mathrm{~m},\) and the path is horizontal to the ground. The figure shows a top view of the tire's circular path, and the dot at the center marks the rotation axis. The man applies a constant torque of \(20.0 \mathrm{~N} \mathrm{~m}\) to accelerate the tire at a constant angular acceleration. Assume that all of the tire's mass is at a radius \(R=0.35 \mathrm{~m}\) from its center. a) What is the time, \(t_{\text {throw }}\) required for the tire to complete three full revolutions? b) What is the final linear speed of the tire's center of mass (after three full revolutions)? c) If, instead of assuming that all of the mass of the tire is at a distance \(0.35 \mathrm{~m}\) from its center, you treat the tire as a hollow disk of inner radius \(0.30 \mathrm{~m}\) and outer radius \(0.40 \mathrm{~m}\), how does this change your answers to parts (a) and (b)?

A 25.0 -kg boy stands \(2.00 \mathrm{~m}\) from the center of a frictionless playground merry-go-round, which has a moment of inertia of \(200 . \mathrm{kg} \mathrm{m}^{2} .\) The boy begins to run in a circular path with a speed of \(0.600 \mathrm{~m} / \mathrm{s}\) relative to the ground. a) Calculate the angular velocity of the merry-go-round. b) Calculate the speed of the boy relative to the surface of the merry-go- round.

A thin uniform rod (length \(=1.00 \mathrm{~m},\) mass \(=2.00 \mathrm{~kg})\) is pivoted about a horizontal frictionless pin through one of its ends. The moment of inertia of the rod through this axis is \(\frac{1}{3} m L^{2} .\) The rod is released when it is \(60.0^{\circ}\) below the horizontal. What is the angular acceleration of the rod at the instant it is released?
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