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When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of the star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius \(R\) that is initially rotating at 2.0 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is \(4.0 R\). Assume that all of the star's original mass is contained in the shell.

Short Answer

Expert verified
The angular velocity after the explosion is 0.05 revolutions per day.

Step by step solution

01

Understand the conservation of angular momentum

In this problem, we use the principle of conservation of angular momentum. Before the explosion, the star is a solid sphere of radius \( R \). After the explosion, the star's mass spreads out into a spherical shell of radius \( 4R \). Angular momentum is conserved since there is no external torque.
02

Define initial angular momentum

Before the explosion, the star is a solid sphere. The moment of inertia, \( I_1 \), for a solid sphere is given by \( \frac{2}{5}mR^2 \), where \( m \) is the mass of the star, and \( R \) is its radius. The initial angular velocity, \( \omega_1 \), is 2 revolutions per day. Thus, the initial angular momentum, \( L_1 \), of the star is \( I_1 \times \omega_1 = \frac{2}{5}mR^2 \times 2.0 \).
03

Define final angular momentum

After the explosion, the star becomes a thin shell with the radius expanding to \( 4R \) and with the same mass \( m \). The moment of inertia, \( I_2 \), for a shell is \( m(4R)^2 = 16mR^2 \). Therefore, the final angular momentum, \( L_2 \), is \( I_2 \times \omega_2 = 16mR^2 \times \omega_2 \), where \( \omega_2 \) is the angular velocity we need to find.
04

Set initial angular momentum equal to final angular momentum

Using conservation of angular momentum, we have \( L_1 = L_2 \). Substitute \( L_1 = \frac{2}{5}mR^2 \times 2.0 \) and \( L_2 = 16mR^2 \times \omega_2 \). Thus, \( \frac{4}{5}mR^2 = 16mR^2 \times \omega_2 \).
05

Solve for the final angular velocity \( \omega_2 \)

Simplify \( \frac{4}{5}mR^2 = 16mR^2 \times \omega_2 \) by canceling \( mR^2 \) from both sides, yielding \( \frac{4}{5} = 16 \times \omega_2 \). Solve for \( \omega_2 \) to get \( \omega_2 = \frac{4}{5 \times 16} = \frac{1}{20} \).
06

Express the final angular velocity in revolutions per day

The final angular velocity \( \omega_2 = \frac{1}{20} \) is already given in terms of revolutions per unit time, where the unit time corresponds to one day. Therefore, \( \omega_2 = 0.05 \) revolutions per day.

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Key Concepts

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

Angular Momentum
Angular momentum is a key concept in physics, representing the rotational analog of linear momentum. It is essentially the quantity of rotation that an object possesses. Angular momentum is particularly crucial in situations where rotational movements are involved, such as celestial events like supernovas. It is determined by three factors:
  • The mass of the object
  • The shape or distribution of that mass (or moment of inertia)
  • Angular velocity (how fast the object is spinning)
The formula for angular momentum ( \( L \) ) is given by: \[ L = I \times \omega \] where \( I \) represents the moment of inertia and \( \omega \) is the angular velocity. When stars explode in a supernova, although their mass disperses into a shell, the angular momentum of the system before the explosion remains equal to that after the explosion, assuming no external torques are acting.
Moment of Inertia
The moment of inertia is one of the most important concepts when it comes to understanding rotational motion. It tells us how difficult it is to change the rotational speed of an object. Think of it as the rotational "mass" in Newton's laws of motion. The moment of inertia is highly dependent on how the mass of an object is distributed relative to the axis of rotation.
For simple objects, moments of inertia can be calculated easily. For a solid sphere, like the initial state of a star in our supernova problem, the moment of inertia \( I_1 \) is: \[ I_1 = \frac{2}{5}mR^2 \] after the explosion, the star decompresses into a thin shell. In this case, the moment of inertia \( I_2 \) for a spherical shell is: \[ I_2 = m(4R)^2 = 16mR^2 \] The differences in these formulas highlight how the moment of inertia changes substantially with the change in shape and mass distribution.
Angular Velocity
Angular velocity connects the rotation of an object to time. Essentially, it describes how fast something is spinning. In everyday terms, it is like the speedometer for rotating bodies, indicating how quickly rotation occurs over time. Angular velocity is expressed in units like revolutions per day or radians per second.
In the given problem, the star starts with an angular velocity \( \omega_1 \) of 2 revolutions per day. After a supernova explosion, the star's mass expands into a larger shell of radius \( 4R \) . By applying the conservation of angular momentum, we can determine the new angular velocity \( \omega_2 \) . We find \( \omega_2 = 0.05 \) revolutions per day, showing how slower the star spins as a result of the expansion.
Conservation of Angular Momentum
The principle of conservation of angular momentum is foundational in rotational dynamics and crucial in the analysis of systems like collapsing stars. According to this principle, the total angular momentum of a closed system remains constant in the absence of external torques. This means that any change in the distribution of mass or moment of inertia must be compensated by a change in angular velocity.
In our supernova example, initially, the star spins as a solid sphere, and after its explosion, becomes a thin sphere of radius \( 4R \). No external forces impact the system, so the angular momentum remains constant: \[ L_1 = L_2 \] With the initial and final moments of inertia known, and using conservation laws, we can equate and solve for the new angular velocity, showcasing the profound effect of this conservation principle.

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Most popular questions from this chapter

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a $$30.0^{\circ}$$ angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.

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