/*! 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 82 A uniform rod rotates in a horiz... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is \(6.00 \mathrm{~m}\) long, weighs \(10.0 \mathrm{~N}\), and rotates at 240 rev/min. Calculate (a) its rotational inertia about the axis of rotation and (b) the magnitude of its angular momentum about that axis.

Short Answer

Expert verified
a) Rotational inertia is 12.24 kg m². b) Angular momentum is approximately 964.63 kg m²/s.

Step by step solution

01

Convert Rotation Speed to Radians per Second

The rod rotates at 240 revolutions per minute. First, convert this speed to radians per second. One revolution is equal to \(2\pi\) radians, and there are 60 seconds in a minute. Therefore, the angular speed \(\omega\) is calculated as follows:\[ \omega = 240 \text{ rev/min} \times \frac{2\pi \text{ rad/rev}}{60 \text{ s/min}} = 8\pi \text{ rad/s}. \]
02

Calculate Rotational Inertia using Rod's Formula

The rotational inertia \(I\) of a uniform rod about an axis through its end is calculated by the formula:\[ I = \frac{1}{3} m L^2 \]where \(m\) is the mass of the rod and \(L\) is its length. The weight of the rod is given as \(10.0 \text{ N}\), which corresponds to mass \(m = \frac{10.0 \text{ N}}{9.8 \text{ m/s}^2} = 1.02 \text{ kg}.\) Now substitute the values:\[ I = \frac{1}{3} \times 1.02 \text{ kg} \times (6.00 \text{ m})^2 = 12.24 \text{ kg m}^2. \]
03

Calculate Angular Momentum

The angular momentum \(L\) of a rotating object is given by the product of its rotational inertia and its angular velocity:\[ L = I \omega. \]Substitute the values of \(I\) and \(\omega\) calculated in the previous steps:\[ L = 12.24 \text{ kg m}^2 \times 8\pi \text{ rad/s} = 306.84\pi \text{ kg m}^2/s \approx 964.63 \text{ kg m}^2/s. \]

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Key Concepts

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

Rotational Inertia
Rotational inertia, also known as the moment of inertia, is a property of a rotating object that determines how much torque is needed for a desired angular acceleration about an axis. It's the rotational equivalent of mass in linear motion. Simply put, it measures how much an object resists changes to its rotation.

For a long rod like the one in this exercise, the rotational inertia depends on both its mass and its length. Specifically, for a rod rotating around an axis at one end, the formula is \( I = \frac{1}{3} m L^2 \), where \( m \) is the mass and \( L \) is the length. That's why in this exercise, the rotational inertia is calculated as \( 12.24 \text{ kg} \cdot \text{m}^2 \).

Understanding rotational inertia helps in analyzing systems involving rotational motion, such as engines, turbines, and even human bodies in sports to understand motion and balance. The greater the rotational inertia, the harder it is to change the object's rotational speed.
Angular Momentum
Angular momentum is the quantity of rotation an object has, taking into account both its mass distribution and its speed of rotation. It's a fundamental concept in rotational dynamics, similar to linear momentum in translational motion.

The angular momentum \( L \) of a rotating object can be calculated using the formula \( L = I \omega \), where \( I \) is the rotational inertia and \( \omega \) is the angular velocity. In this exercise, the resulting angular momentum of the rod is approximately \( 964.63 \text{ kg} \cdot \text{m}^2/s \). This amount quantifies the amount of "spin" or rotational motion the rod has about the axis.

Understanding angular momentum is essential in everything from designing mechanical devices to analyzing the orbits of celestial bodies. It is conserved in isolated systems, meaning that it remains constant if no external torque acts on the system.
Uni-axial Rotation
Uni-axial rotation refers to the rotation about a single, fixed axis. In contrast to more complex forms of rotation, such as those involving multiple axes, uni-axial rotation is a more straightforward subject to analyze.

In scenarios like the rotating rod in this exercise, the axis is vertical, while the rod itself rotates in a horizontal plane. This simple setup allows for the direct application of rotational dynamics concepts, such as calculating rotational inertia and angular momentum. Understanding uni-axial rotation is crucial for studying many real-world situations, such as wheels and rotors which rotate around a central axis.

This focus simplifies the analysis and is a foundational concept in understanding more complex rotational systems.
Conversion of Angular Speed
Angular speed refers to how fast an object rotates or spins around a particular axis. This measure is often initially provided in revolutions per minute (rev/min), especially in mechanical or engineering contexts. However, for analysis involving physics, it's frequently converted into radians per second (rad/s).

In this exercise, the rod's angular speed is given as 240 rev/min. To convert it to the standard unit of rad/s, remember that one revolution is equivalent to \( 2\pi \) radians. Using this conversion factor, the angular speed becomes \( 8\pi \) rad/s. This conversion is integral for using formulas in rotational dynamics, like those for angular momentum, which require angular speed in radians per second.
  • Converting units is important for ensuring consistency and accuracy in calculations.
  • Radians are the standard unit in angular measurements for scientific calculations.

Proper conversion allows the equations of motion to be applied accurately, leading to more precise and reliable results.

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Most popular questions from this chapter

During a jump to his partner, an aerialist is to make a quadruple somersault lasting a time \(t=1.87 \mathrm{~s}\). For the first and last quarter- revolution, he is in the extended orientation shown in Fig. \(11-55\), with rotational inertia \(I_{1}=19.9 \mathrm{~kg} \cdot \mathrm{m}^{2}\) around his center of mass (the dot). During the rest of the flight he is in a tight tuck, with rotational inertia \(I_{2}=3.93 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) What must be his angular speed \(\omega_{2}\) around his center of mass during the tuck?

A wheel of radius \(0.250 \mathrm{~m}\), moving initially at \(43.0 \mathrm{~m} / \mathrm{s}\), rolls to a stop in \(225 \mathrm{~m}\). Calculate the magnitudes of its (a) linear acceleration and (b) angular acceleration. (c) Its rotational inertia is \(0.155\) \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about its central axis. Find the magnitude of the torque about the central axis due to friction on the wheel.

Force \(\vec{F}=(-8.0 \mathrm{~N}) \hat{\mathrm{i}}+(6.0 \mathrm{~N}) \hat{\mathrm{j}}\) acts on a particle with position vector \(\vec{r}=(3.0 \mathrm{~m}) \hat{\mathrm{i}}+(4.0 \mathrm{~m}) \hat{\mathrm{j}} .\) What are (a) the torque on the particle about the origin, in unit-vector notation, and (b) the angle between the directions of \(\vec{r}\) and \(\vec{F}\) ?

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a mass of \(150 \mathrm{~kg}\), a radius of \(2.0 \mathrm{~m}\), and a rotational inertia of \(300 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axis of rotation. A \(60 \mathrm{~kg}\) student walks slowly from the rim of the platform toward the center. If the angular speed of the system is \(1.5 \mathrm{rad} / \mathrm{s}\) when the student starts at the rim, what is the angular speed when she is \(0.50 \mathrm{~m}\) from the center?

In a playground, there is a small merry-go-round of radius \(1.20 \mathrm{~m}\) and mass \(180 \mathrm{~kg} .\) Its radius of gyration (see Problem 79 of Chapter 10 ) is \(91.0 \mathrm{~cm}\). A child of mass \(44.0 \mathrm{~kg}\) runs at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) along a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate (a) the rotational inertia of the merry-go-round about its axis of rotation, (b) the magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round, and (c) the angular speed of the merry-go- round and child after the child has jumped onto the merry-go-round.

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