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Ch 12: Rotation of a Rigid Body
Chapter 12, Problem 12

During most of its lifetime, a star maintains an equilibrium size in which the inward force of gravity on each atom is balanced by an outward pressure force due to the heat of the nuclear reactions in the core. But after all the hydrogen 'fuel' is consumed by nuclear fusion, the pressure force drops and the star undergoes a gravitational collapse until it becomes a neutron star. In a neutron star, the electrons and protons of the atoms are squeezed together by gravity until they fuse into neutrons. Neutron stars spin very rapidly and emit intense pulses of radio and light waves, one pulse per rotation. These 'pulsing stars' were discovered in the 1960s and are called pulsars. a. A star with the mass (M = 2.0 X 10^30 kg) and size (R = 7.0 x 10^8 m) of our sun rotates once every 30 days. After undergoing gravitational collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.10 s. By treating the neutron star as a solid sphere, deduce its radius.

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Step 1: Calculate the initial angular velocity of the star. Use the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of rotation in seconds.
Step 2: Apply the conservation of angular momentum, assuming no external torques act on the system. The initial angular momentum \( L_i \) can be calculated using \( L_i = I_i \omega_i \), where \( I_i = \frac{2}{5}MR^2 \) is the moment of inertia of the initial star.
Step 3: Calculate the final angular velocity \( \omega_f \) of the neutron star using the observed pulse period \( T_f = 0.10 \) s, again using \( \omega_f = \frac{2\pi}{T_f} \).
Step 4: Set the initial angular momentum equal to the final angular momentum \( L_f = I_f \omega_f \), where \( I_f = \frac{2}{5}MR_f^2 \) is the moment of inertia of the neutron star. Solve for \( R_f \) using the equation \( \frac{2}{5}MR^2 \omega_i = \frac{2}{5}MR_f^2 \omega_f \).
Step 5: Simplify the equation to find \( R_f \) by cancelling out common terms and solving for \( R_f \), resulting in \( R_f = R \sqrt{\frac{\omega_i}{\omega_f}} \).

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Key Concepts

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

Gravitational Equilibrium

Gravitational equilibrium in a star occurs when the inward gravitational force pulling matter toward the center is balanced by the outward pressure generated from nuclear fusion reactions in the core. This balance allows the star to maintain a stable size and prevents it from collapsing under its own gravity. When the nuclear fuel is depleted, this equilibrium is disrupted, leading to gravitational collapse.
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Neutron Star Formation

A neutron star forms when a massive star exhausts its nuclear fuel and undergoes gravitational collapse, compressing electrons and protons to form neutrons. This process results in an extremely dense object, where the gravitational forces are so strong that normal atomic structures cannot exist. Neutron stars are typically about 1.4 times the mass of the sun but compressed into a radius of only about 10 kilometers.
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Pulsars and Rotation

Pulsars are a type of neutron star that emit beams of electromagnetic radiation, including radio waves, due to their rapid rotation and strong magnetic fields. As the neutron star spins, these beams sweep across space, and if aligned with Earth, they can be detected as regular pulses. The rotation period of a pulsar can be extremely short, often just milliseconds, which is a key characteristic that distinguishes them from other celestial objects.
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Related Practice
Textbook Question
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Textbook Question
A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs tangent to the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg. What is the merry-go-round's angular velocity, in rpm, after John jumps on?
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Textbook Question
FIGURE P12.82 shows a cube of mass m sliding without friction at speed v₀. It undergoes a perfectly elastic collision with the bottom tip of a rod of length d and mass M = 2m. The rod is pivoted about a frictionless axle through its center, and initially it hangs straight down and is at rest. What is the cube's velocity—both speed and direction—after the collision?
404
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Textbook Question
The bunchberry flower has the fastest-moving parts ever observed in a plant. Initially, the stamens are held by the petals in a bent position, storing elastic energy like a coiled spring. When the petals release, the tips of the stamen act like medieval catapults, flipping through a 60° angle in just .30 ms to launch pollen from anther sacs at their ends. The human eye just sees a burst of pollen; only high-speed photography reveals the details. As FIGURE CP12.91 shows, we can model the stamen tip as a 1.0-mm-long, 10 μg rigid rod with a 10 μg anther sac at the end. Although oversimplifying, we'll assume a constant angular acceleration. b. What is the speed of the anther sac as it releases its pollen?
427
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Textbook Question
Determine the moment of inertia about the axis of the object shown in FIGURE P12.52.
544
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Textbook Question
A rod of length L and mass M has a nonuniform mass distribution. The linear mass density (mass per length) is λ = cx^2 , where x is measured from the center of the rod and c is a constant. b. Find an expression for c in terms of L and M.
399
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