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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Intro to Angular Collisions
Problem 52
Textbook Question
(III) A thin uniform rod of mass M and length ℓ rests on a frictionless table and is struck at a point ℓ/4 from its cm by a clay ball of mass m moving at speed υ (Fig. 11–40). The ball sticks to the rod. Determine the translational and rotational motion of the rod after the collision.


1
Identify the principles involved: This is a collision problem involving both linear momentum and angular momentum conservation. Since the table is frictionless, no external forces or torques act on the system, so both linear and angular momentum are conserved.
Apply conservation of linear momentum: The total linear momentum before the collision is the momentum of the clay ball, which is \( p_{initial} = m \cdot \nu \). After the collision, the total linear momentum is shared between the rod and the clay ball (now stuck together). Write the equation for conservation of linear momentum: \( m \cdot \nu = (M + m) \cdot V \), where \( V \) is the final velocity of the center of mass of the system.
Apply conservation of angular momentum: The angular momentum before the collision is due to the clay ball, calculated about the center of mass of the rod. The initial angular momentum is \( L_{initial} = m \cdot \nu \cdot (\ell/4) \). After the collision, the angular momentum is shared between the rotational motion of the rod and the clay ball. Write the equation for conservation of angular momentum: \( m \cdot \nu \cdot (\ell/4) = I_{total} \cdot \omega \), where \( I_{total} \) is the moment of inertia of the rod-ball system about the center of mass, and \( \omega \) is the angular velocity after the collision.
Determine the moment of inertia of the system: The moment of inertia of the rod about its center is \( I_{rod} = \frac{1}{12} M \ell^2 \). The clay ball contributes \( I_{ball} = m \cdot (\ell/4)^2 \) to the total moment of inertia. Combine these to find \( I_{total} = \frac{1}{12} M \ell^2 + m \cdot (\ell/4)^2 \).
Solve for the final translational and rotational motion: Use the linear momentum equation to solve for \( V \), the velocity of the center of mass. Then, use the angular momentum equation to solve for \( \omega \), the angular velocity. These two quantities describe the translational and rotational motion of the rod after the collision.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Conservation of Momentum
The principle of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on it. In this scenario, the momentum before the collision (the clay ball's momentum) must equal the momentum after the collision (the combined system of the rod and the ball). This concept is crucial for determining the translational motion of the rod after the impact.
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Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion, depending on the mass distribution relative to the axis of rotation. For the rod, the moment of inertia can be calculated using the formula I = (1/3)ML² for a uniform rod about its center of mass. Understanding this concept is essential for analyzing the rotational motion of the rod after the collision.
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Angular Momentum
Angular momentum is the rotational equivalent of linear momentum and is defined as the product of an object's moment of inertia and its angular velocity. The conservation of angular momentum states that if no external torque acts on a system, the total angular momentum remains constant. This principle is vital for determining how the rod will rotate after the clay ball sticks to it, especially since the collision occurs off-center.
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