/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 An oxygen molecule \(\left(\math... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 67

An oxygen molecule \(\left(\mathrm{O}_{2}\right)\) rotates in the \(x y\) -plane about the \(z\) -axis. The axis of rotation passes through the center of the molecule, perpendicular to its length. The mass of each oxygen atom is \(2.66 \cdot 10^{-26} \mathrm{~kg},\) and the average separation between the two atoms is \(d=1.21 \cdot 10^{-10} \mathrm{~m}\) a) Calculate the moment of inertia of the molecule about the \(z\) -axis. b) If the angular speed of the molecule about the \(z\) -axis is \(4.60 \cdot 10^{12} \mathrm{rad} / \mathrm{s},\) what is its rotational kinetic energy?

Short Answer

Expert verified
Answer: The moment of inertia of an oxygen molecule about the z-axis is approximately 4.93 x 10^-46 kg路m^2, and its rotational kinetic energy is approximately 5.20 x 10^-20 J.

Step by step solution

01

a) Calculate the moment of inertia of the molecule about the z-axis

First, let's consider the molecule as a system of two point masses. Each oxygen atom can be considered a point mass with mass \(m=2.66 \times 10^{-26} kg\). The moment of inertia for a point mass rotating around an axis is given by the formula \(I = md^2\), where \(m\) denotes mass, and \(d\) denotes distance from the axis of rotation. Both oxygen atoms rotate at distance \(d/2\). Now, we need to calculate the moment of inertia for both atoms and add them together. \(I_{total} = I_{1} + I_{2} = m\left(\frac{d}{2}\right)^2 + m\left(\frac{d}{2}\right)^2 = 2m\left(\frac{d}{2}\right)^2\) Given the mass, \(m\), and distance, \(d\), we can find \(I_{total}\): \(I_{total} = 2\left(2.66 \cdot 10^{-26}\:\mathrm{kg}\right)\left(\frac{1.21 \cdot 10^{-10}\:\mathrm{m}}{2}\right)^2\) Now, calculate the numerical value. \(I_{total} \approx 4.93 \times 10^{-46}\: \mathrm{kg\:m^2}\) The moment of inertia of the molecule about the z-axis is \(4.93 \times 10^{-46}\: \mathrm{kg\:m^2}\).
02

b) Calculate the rotational kinetic energy of the molecule

To find the rotational kinetic energy, we use the following formula: \(K_{rot} = \frac{1}{2}I\omega^2\) We have the moment of inertia, \(I_{total} = 4.93 \times 10^{-46}\: \mathrm{kg\:m^2}\), and the angular velocity, \(\omega = 4.60 \times 10^{12}\: \mathrm{rad/s}\). Plugging these values into the equation: \(K_{rot} = \frac{1}{2}\left(4.93 \times 10^{-46}\: \mathrm{kg\:m^2}\right)\left(4.60 \times 10^{12}\: \mathrm{rad/s}\right)^2\) Now, calculate the numerical value. \(K_{rot} \approx 5.20 \times 10^{-20}\: \mathrm{J}\) The rotational kinetic energy of the molecule is approximately \(5.20 \times 10^{-20}\: \mathrm{J}\).

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Key Concepts

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

Moment of Inertia
Understanding the moment of inertia is crucial when dealing with rotational motion. It's the rotational equivalent of mass for linear motion. This property indicates how much torque is required to change the object鈥檚 rotational speed. For a single point mass, it depends only on the mass of that point and the square of its perpendicular distance from the axis of rotation. The formula is as simple as I = md^2, with I representing the moment of inertia, m the mass, and d the distance from the axis.

In more complex systems, like the oxygen molecule in our problem, composed of two point masses, each mass contributes to the total moment of inertia. We calculate each point mass's moment of inertia and sum them. Recognizing that both atoms are equally distanced from the rotation axis simplifies our work: we double the moment of inertia of one atom. Therefore, the formula becomes I_total = 2m(d/2)^2, enabling us to calculate the molecule's moment of inertia accurately.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or revolves relative to another point, in this case, the axis of rotation. The faster the object spins, the greater its angular velocity, commonly denoted by the Greek letter \(\omega\). The unit of angular velocity is radians per second (rad/s). Radian is a measure of angle, equivalent to the angle at the center of a circle that subtends an arc equal in length to the radius.

To comprehend angular velocity鈥檚 impact, consider holding a string with a weight at one end and spinning it. If you speed up the spinning, the force you feel pulling on the string increases due to the higher angular velocity. In our molecule example, the angular velocity given is exceptionally high, which implies a very rapid spinning motion around the axis.
Rotational Motion
Rotational motion is movement around an axis or point. It's foundational in understanding how various objects move in our universe鈥攆rom planets orbiting stars to the electron clouds within an atom. In our daily lives, we see rotational motion in wheels, fans, and even merry-go-rounds. What makes rotational motion interesting and sometimes complex is the interplay between the moment of inertia and the angular velocity.

The rotational kinetic energy is a form of energy an object possesses due to its rotation, and it's influenced by both the object's moment of inertia and its angular velocity. The formula K_{rot} = \(\frac{1}{2}I\omega^2\) expresses this relationship mathematically. It shows that if either the moment of inertia or the angular velocity increases, the rotational kinetic energy jumps up even more because of the squared term for \(\omega\). In practice, this means that when our oxygen molecule speeds up, its energy increases rapidly, highlighting the significance of controlling rotational speeds in machinery and technology to manage energy usage and safety.

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Most popular questions from this chapter

Most stars maintain an equilibrium size by balancing two forces - an inward gravitational force and an outward force resulting from the star's nuclear reactions. When the star's fuel is spent, there is no counterbalance to the gravitational force. Whatever material is remaining collapses in on itself. Stars about the same size as the Sun become white dwarfs, which glow from leftover heat. Stars that have about three times the mass of the Sun compact into neutron stars. And a star with a mass greater than three times the Sun's mass collapses into a single point, called a black hole. In most cases, protons and electrons are fused together to form neutrons-this is the reason for the name neutron star. Neutron stars rotate very fast because of the conservation of angular momentum. Imagine a star of mass \(5.00 \cdot 10^{30} \mathrm{~kg}\) and radius \(9.50 \cdot 10^{8} \mathrm{~m}\) that rotates once in 30.0 days. Suppose this star undergoes gravitational collapse to form a neutron star of radius \(10.0 \mathrm{~km} .\) Determine its rotation period.

The Crab pulsar \(\left(m \approx 2 \cdot 10^{30} \mathrm{~kg}, R=5 \mathrm{~km}\right)\) is a neutron star located in the Crab Nebula. The rotation rate of the Crab pulsar is currently about 30 rotations per second, or \(60 \pi \mathrm{rad} / \mathrm{s} .\) The rotation rate of the pulsar, however, is decreasing; each year, the rotation period increases by \(10^{-5}\) s. Justify the following statement: The loss in rotational energy of the pulsar is equivalent to 100,000 times the power output of the Sun. (The total power radiated by the Sun is about \(\left.4 \cdot 10^{26} \mathrm{~W} .\right)\)

A thin uniform rod (length \(=1.00 \mathrm{~m},\) mass \(=2.00 \mathrm{~kg})\) is pivoted about a horizontal frictionless pin through one of its ends. The moment of inertia of the rod through this axis is \(\frac{1}{3} m L^{2} .\) The rod is released when it is \(60.0^{\circ}\) below the horizontal. What is the angular acceleration of the rod at the instant it is released?

A cylinder with mass \(M\) and radius \(R\) is rolling without slipping through a distance \(s\) along an inclined plane that makes an angle \(\theta\) with respect to the horizontal. Calculate the work done by (a) gravity, (b) the normal force, and (c) the frictional force.

The flywheel of an old steam engine is a solid homogeneous metal disk of mass \(M=120 . \mathrm{kg}\) and radius \(R=80.0 \mathrm{~cm} .\) The engine rotates the wheel at \(500 .\) rpm. In an emergency, to bring the engine to a stop, the flywheel is disengaged from the engine and a brake pad is applied at the edge to provide a radially inward force \(F=100 .\) N. If the coefficient of kinetic friction between the pad and the flywheel is \(\mu_{\mathrm{k}}=0.200,\) how many revolutions does the flywheel make before coming to rest? How long does it take for the flywheel to come to rest? Calculate the work done by the torque during this time.
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