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Ch 10: Dynamics of Rotational Motion
Chapter 10, Problem 10

Asteroid Collision! 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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Identify the initial angular momentum of the Earth. The angular momentum (L) can be calculated using the formula L = I \omega, where I is the moment of inertia of the Earth and \omega is the angular velocity of the Earth.
Calculate the moment of inertia (I) of the Earth assuming it is a uniform sphere. The formula for the moment of inertia of a sphere is I = \frac{2}{5} M R^2, where M is the mass of the Earth and R is its radius.
Determine the new angular velocity (\omega') of the Earth after the collision, given that the length of the day is to increase by 25%. Since the length of the day is inversely proportional to the angular velocity, the new angular velocity will be \omega' = \frac{\omega}{1.25}.
Calculate the angular momentum of the asteroid just before impact. Assume the asteroid travels with a velocity v and has a mass m. The angular momentum of the asteroid can be approximated by L_a = m R v, where R is the radius of the Earth.
Set up the conservation of angular momentum equation to solve for the mass of the asteroid. The total angular momentum before the collision (Earth + asteroid) must equal the total angular momentum after the collision. Thus, \frac{2}{5} M R^2 \omega + m R v = \frac{2}{5} M R^2 \frac{\omega}{1.25}. Solve this equation for m in terms of M.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Angular Momentum

Angular momentum is a measure of the rotational motion of an object and is conserved in a closed system. In the context of the Earth and an asteroid collision, the total angular momentum before and after the collision must remain constant. This principle helps us understand how the collision affects the Earth's rotation and, consequently, the length of a day.
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Moment of Inertia

The moment of inertia is a property that quantifies how mass is distributed relative to an axis of rotation. For the Earth, this distribution affects how easily it can change its rotational speed when an external mass, like an asteroid, impacts it. A change in the moment of inertia due to the asteroid's mass will influence the Earth's rotation period, thereby altering the length of a day.
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Conservation of Energy

The conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another. In the case of the asteroid collision, the kinetic energy of the asteroid will be converted into other forms of energy, such as heat and potential energy, affecting the Earth's rotational dynamics. Understanding this concept is crucial for calculating the necessary mass of the asteroid to achieve the specified change in day length.
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Related Practice
Textbook Question
The Spinning Figure Skater. The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center (Fig. E10.43). When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thinwalled, hollow cylinder. His hands and arms have a combined mass of 8.0 kg. When outstretched, they span 1.8 m; when wrapped, they form a cylinder of radius 25 cm. The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40 kg•m2 . If his original angular speed is 0.40 rev/s, what is his final angular speed?

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Textbook Question
CP A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.40). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. (c) Find the change in kinetic energy of the block.
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Textbook Question
CP A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.40). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad>s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. (d) How much work was done in pulling the cord?
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Textbook Question
A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500 kg, traveling perpendicular to the door at 12.0 m/s just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?
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Textbook Question
A uniform, 4.5-kg, square, solid wooden gate 1.5 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.1-kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.0 m/s in the opposite direction. (a) What is the angular speed of the gate just after it is struck by the unfortunate raven?
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Textbook Question
A uniform, 4.5-kg, square, solid wooden gate 1.5 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.1-kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.0 m/s in the opposite direction. (b) During the collision, why is the angular momentum conserved but not the linear momentum?
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