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Ch 10: Dynamics of Rotational Motion
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 14th EditionCh 10: Dynamics of Rotational MotionProblema 46
Capitolo 10, Problema 46

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth's mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

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Understand the problem: We need to find the mass of the asteroid that would cause the Earth's rotation period to increase by 25%. This involves the conservation of angular momentum, as the collision is inelastic and the asteroid embeds itself into the Earth.
Identify the initial and final states: Initially, the Earth is rotating with an angular velocity \( \omega_i \) and has a moment of inertia \( I_i = \frac{2}{5}MR^2 \), where \( M \) is the Earth's mass and \( R \) is its radius. After the collision, the new angular velocity is \( \omega_f \) and the moment of inertia becomes \( I_f = \frac{2}{5}MR^2 + mR^2 \), where \( m \) is the mass of the asteroid.
Apply the conservation of angular momentum: The initial angular momentum \( L_i = I_i \omega_i \) must equal the final angular momentum \( L_f = I_f \omega_f \). Therefore, \( \frac{2}{5}MR^2 \omega_i = \left( \frac{2}{5}MR^2 + mR^2 \right) \omega_f \).
Relate the change in the Earth's rotation period: The period of rotation \( T \) is inversely proportional to the angular velocity, so \( T_f = 1.25T_i \) implies \( \omega_f = \frac{\omega_i}{1.25} \). Substitute this into the angular momentum equation.
Solve for the mass of the asteroid \( m \): Rearrange the equation from step 3 to solve for \( m \) in terms of \( M \). This will give you the mass of the asteroid needed to increase the Earth's day by 25%.

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Conservation of Angular Momentum

The conservation of angular momentum states that if no external torques act on a system, the total angular momentum remains constant. In this scenario, the earth-asteroid system's angular momentum before and after the collision must be equal, allowing us to relate the asteroid's mass to the change in the earth's rotational period.
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Conservation of Angular Momentum

Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotation. For a uniform sphere like the earth, it is calculated as I = (2/5)MR², where M is the mass and R is the radius. Understanding how the asteroid's mass affects the earth's moment of inertia is crucial for determining the change in rotational speed.
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Intro to Moment of Inertia

Rotational Kinematics

Rotational kinematics involves the study of rotational motion without considering the forces that cause it. The earth's rotational period is related to its angular velocity, and a change in this period due to the collision can be analyzed using the relationship between angular velocity and moment of inertia, helping to find the asteroid's required mass.
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Kinematics Equations
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