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(III) An asteroid of mass \(1.0 \times 10^5 kg\), traveling at a speed of 35 km/s relative to the Earth, hits the Earth at the equator tangentially, in the direction of Earth's rotation, and is embedded there. Use angular momentum to estimate the percent change in the angular speed of the Earth as a result of the collision.

Short Answer

Expert verified
The percent change in Earth's angular speed is negligible due to the small mass of the asteroid relative to Earth.

Step by step solution

01

Determine Initial Angular Momentum of Earth

The Earth can be approximated as a solid sphere rotating around its axis. The formula for Earth's initial angular momentum is \[ L_{ ext{initial}} = I_{ ext{Earth}} \omega_{ ext{Earth}} \] where \[ I_{ ext{Earth}} = \frac{2}{5} M_{ ext{Earth}} R_{ ext{Earth}}^2 \] is Earth's moment of inertia, with \( M_{ ext{Earth}} = 5.972 \times 10^{24} \text{ kg} \) and \( R_{ ext{Earth}} = 6.371 \times 10^6 \text{ m} \).The initial angular speed of Earth \( \omega_{\text{Earth}} = \frac{2 \pi}{T} \), where \( T = 86400 \text{ s} \) (seconds in a day).Calculate \( I_{\text{Earth}} \omega_{\text{Earth}} \) to find Earth's initial angular momentum.
02

Determine Angular Momentum of Asteroid

The angular momentum of the asteroid about the center of the Earth when hitting the Earth is given by \[ L_{ ext{asteroid}} = m_{ ext{asteroid}} v_{ ext{asteroid}} R_{ ext{Earth}} \]where \( m_{\text{asteroid}} = 1.0 \times 10^5 \text{ kg} \) and \( v_{\text{asteroid}} = 35 \times 10^3 \text{ m/s} \).Compute \( L_{\text{asteroid}} \) using these values.
03

Calculate Total Angular Momentum After Collision

The total angular momentum after the collision is the sum of the Earth's initial angular momentum and the angular momentum of the asteroid. \[ L_{ ext{total}} = L_{ ext{initial}} + L_{ ext{asteroid}} \] This value represents the new total angular momentum of the Earth after the asteroid is embedded.
04

Calculate New Angular Speed of Earth

Using the conservation of angular momentum, set the final total angular momentum equal to the new angular momentum of the Earth. \[ L_{ ext{total}} = I_{ ext{Earth}} \omega_{ ext{new}} \]Solve for \( \omega_{\text{new}} \), the new angular speed of the Earth. \[ \omega_{\text{new}} = \frac{L_{ ext{total}}}{I_{ ext{Earth}}} \]
05

Determine Percent Change in Angular Speed

Calculate the percent change in angular speed using the formula: \[ \text{Percent Change} = \left( \frac{\omega_{\text{new}} - \omega_{\text{Earth}}}{\omega_{\text{Earth}}}\right) \times 100\% \]Substitute the values obtained for \( \omega_{\text{new}} \) and \( \omega_{\text{Earth}} \) to find this change.

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Key Concepts

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

moment of inertia
The moment of inertia is a crucial concept when studying rotary systems. Think of it as the rotational counterpart of mass in linear motion. For a body rotating around an axis, the moment of inertia measures how difficult it is to change its rotational speed. This depends both on the mass of the object and how that mass is distributed relative to the axis of rotation.

For Earth, which can be approximated as a solid sphere, the moment of inertia is calculated using the formula: \[ I = \frac{2}{5} M R^2 \]where:
  • \(M\) is the mass of the Earth.
  • \(R\) is the radius of the Earth.
This formula shows that objects with more mass and those whose mass is farther from the axis have a larger moment of inertia. In our problem, this formula allows us to calculate how resistant Earth is to changes in its rotation when influenced by an external factor, such as an incoming asteroid.
angular speed
Angular speed defines how fast something spins around a central point or axis. It is usually denoted by the symbol \( \omega \) and expressed in radians per second rad/s. For the Earth, which rotates once a day, angular speed can be calculated using the period of its rotation as follows:

\[ \omega = \frac{2 \pi}{T} \]where \( T \) is the time it takes for one complete rotation. In Earth's case, this period is one day or 86400 seconds. Thus, the initial angular speed can be found, which gives us a baseline against which any changes are measured.

When an asteroid impacts Earth, it imparts its own angular momentum, affecting this speed. By calculating the new angular speed using the conservation of momentum, we can assess the impact of such cosmic events on our planet's rotation.
conservation of angular momentum
The principle of conservation of angular momentum is fundamental in physics. It states that if no external torques act on a system, the total angular momentum remains constant. This principle applies to various systems, making it an essential tool in analyzing rotational dynamics.

In our exercise, both the Earth and the asteroid act as a closed system. When the asteroid collides with Earth and becomes embedded, the system's total angular momentum before and after the collision remains the same. This allows us to set up the equation:\[ L_{\text{initial}} + L_{\text{asteroid}} = L_{\text{total}} \]Here:
  • \( L_{\text{initial}} \) is the Earth's initial angular momentum.
  • \( L_{\text{asteroid}} \) is the asteroid's angular momentum at impact.
  • \( L_{\text{total}} \) is the total angular momentum after the collision.
By using this equation, we can solve for the new angular speed. This helps scientists understand how external influences modify a planet's rotation without additional forces or energy added to or removed from the system.

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Most popular questions from this chapter

(II) A cooling fan is turned off when it is running at 850 rev/min. It turns 1250 revolutions before it comes to a stop. \((a)\) What was the fan's angular acceleration, assumed constant? \((b)\) How long did it take the fan to come to a complete stop?

(II) A centrifuge rotor rotating at 9200 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 m\(\cdot\)N. If the mass of the rotor is 3.10 kg and it can be approximated as a solid cylinder of radius 0.0710 m, through how many revolutions will the rotor turn before coming to rest, and how long will it take?

(II) Two masses, \(m_A = 32.0 kg\) and \(m_B = 38.0 kg\), are connected by a rope that hangs over a pulley (as in Fig. 8-54). The pulley is a uniform cylinder of radius \(R = 0.311 m\) and \(mass 3.1 kg\). Initially \(m_A\) is on the ground and \(m_B\) rests 2.5 m above the ground. If the system is released, use conservation of energy to determine the speed of just before it strikes the \(m_B\) ground. Assume the pulley bearing is frictionless.

(II) A child rolls a ball on a level floor 3.5 m to another child. If the ball makes 12.0 revolutions, what is its diameter?

(II) A potter's wheel is rotating around a vertical axis through its center at a frequency of 1.5 rev/s. The wheel can be considered a uniform disk of mass 5.0 kg and diameter 0.40 m. The potter then throws a 2.6-kg chunk of clay, approximately shaped as a flat disk of radius 7.0 cm, onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it? Ignore friction.

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