/*! 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 77 Two thin rods of length \(L\) ar... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 77

Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in rad/s) about axes that pass perpendicularly through one end. Rod A is massless but has a particle of mass \(0.66 \mathrm{~kg}\) attached to its free end. \(\operatorname{Rod} \mathrm{B}\) has a mass \(0.66 \mathrm{~kg},\) which is distributed uniformly along its length. (a) Which has the greater moment of inertia \(-\) rod \(\mathrm{A}\) with its attached particle or \(\operatorname{rod} \mathrm{B} ?(\mathrm{~b})\) Which has the greater rotational kinetic energy? Account for your answers.

Short Answer

Expert verified
(a) Rod A has a greater moment of inertia. (b) Rod A also has greater rotational kinetic energy.

Step by step solution

01

Understanding the Moment of Inertia for Rod A

The moment of inertia for Rod A with a particle of mass \( m = 0.66 \, \text{kg} \) attached at a distance \( L \) from the axis can be given by \( I_A = mL^2 \). Since the rod itself is massless, we only consider the particle's contribution.
02

Calculating Moment of Inertia for Rod B

Rod B has mass \( m = 0.66 \, \text{kg} \) distributed uniformly. The moment of inertia for a rod pivoting about an end is \( I_B = \frac{1}{3}mL^2 \). Here, this formula accounts for the distribution of mass along the length of the rod.
03

Comparing Moment of Inertia for Rod A and Rod B

Compare \( I_A = mL^2 \) with \( I_B = \frac{1}{3}mL^2 \). Clearly, \( I_A \) is greater than \( I_B \) since \( mL^2 > \frac{1}{3}mL^2 \). Thus, Rod A has a greater moment of inertia.
04

Understanding Rotational Kinetic Energy

The rotational kinetic energy of an object is given by \( K = \frac{1}{2}I\omega^2 \). Since \( I_A > I_B \), use this formula to compare the rotational kinetic energies.
05

Calculating Rotational Kinetic Energy for Both Rods

For Rod A, the kinetic energy is \( K_A = \frac{1}{2} (mL^2) \omega^2 \). For Rod B, it is \( K_B = \frac{1}{2} (\frac{1}{3}mL^2) \omega^2 = \frac{1}{6} mL^2 \omega^2 \). Clearly, \( K_A > K_B \) since \( \frac{1}{2} > \frac{1}{6} \). Thus, Rod A has greater rotational kinetic energy.

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Key Concepts

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

Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its rotation. Imagine spinning a toy top; the energy that keeps it twirling is its rotational kinetic energy. For mathematical understanding, it's expressed as \( K = \frac{1}{2}I\omega^2 \), where:
  • \( K \) represents the rotational kinetic energy.
  • \( I \) is the moment of inertia, indicating how mass is spread out from the axis of rotation.
  • \( \omega \) is the angular speed, representing how quickly the object rotates.
In scenarios like comparing Rod A and Rod B, the differences in kinetic energy arise directly from differences in the moment of inertia, despite having the same angular speed. Because Rod A has a greater moment of inertia, its rotational kinetic energy is also larger compared to Rod B.
Angular Speed
Angular speed is how fast something is spinning around an axis. It's often measured in radians per second (rad/s). Think about how quickly a fan blade spins; the speed of this rotation is its angular speed, denoted by \( \omega \). This is different from linear speed, which you might encounter when observing how fast a car moves along a road.

Angular speed is crucial in rotational dynamics since it helps define the energy and momentum in rotating systems. In our scenario with rods A and B, both rods rotate at the same angular speed \( \omega \). Despite this similarity, the difference in their kinetic energies results from how their mass is arranged relative to their respective axes.
Mass Distribution
The distribution of mass in a rotating object profoundly influences its moment of inertia, which in turn affects its rotational kinetic energy. For Rod B, the mass of 0.66 kg is spread out evenly along its length. Meanwhile, Rod A features a concentrated mass at its free end, while the rod itself is massless.
  • This concentration in Rod A results in a higher moment of inertia \( I \), calculated as \( mL^2 \), compared to \( \frac{1}{3}mL^2 \) for Rod B.
  • Given that the kinetic energy \( K \) is directly related to \( I \), this distribution means that even though Rods A and B share the same mass and length, their mass distribution leads to different outcomes in their rotational behaviors.
Therefore, understanding mass distribution helps explain why some objects (like Rod A) have higher kinetic energy鈥攁 concept crucial in engineering and physics.

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Most popular questions from this chapter

Concept Questions Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in \(\mathrm{rad} / \mathrm{s}\) ) about axes that pass perpendicularly through one end. \(\operatorname{Rod} \mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{~kg}\) attached to its free end. Rod \(\mathrm{B}\) has a mass \(0.66 \mathrm{~kg}\), which is distributed uniformly along its length. (a) Which has the greater moment of inertia-rod A with its attached particle or rod B? (b) Which has the greater rotational kinetic energy? Account for your answers. Problem The length of each rod is \(0.75 \mathrm{~m}\), and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod A with its attached particle and of rod B. Make sure your answers are consistent with your answers to the Concept Questions.

A flat uniform circular disk (radius \(=2.00 \mathrm{~m}\), mass \(=1.00 \times 10^{2} \mathrm{~kg}\) ) is initially stationary. The disk is free to rotate in the horizontal plane about a frictionless axis perpendicular to the center of the disk. A \(40.0-\mathrm{kg}\) person, standing \(1.25 \mathrm{~m}\) from the axis, begins to run on the disk in a circular path and has a tangential speed of \(2.00 \mathrm{~m} / \mathrm{s}\) relative to the ground. Find the resulting angular speed of the disk (in \(\mathrm{rad} / \mathrm{s}\) ) and describe the direction of the rotation.

Interactive Solution \(9.55\) at illustrates one way of solving a problem similar to this one. A thin rod has a length of \(0.25 \mathrm{~m}\) and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of \(0.32 \mathrm{rad} / \mathrm{s}\) and a moment of inertia of \(1.1 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\). A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (mass \(=4.2 \times 10^{-3} \mathrm{~kg}\) ) gets where it's going, what is the angular velocity of the rod?

A ceiling fan is turned on and a net torque of \(1.8 \mathrm{~N} \cdot \mathrm{m}\) is applied to the blades. The blades have a total moment of inertia of \(0.22 \mathrm{~kg} \cdot \mathrm{m}^{2}\). What is the angular acceleration of the blades?

Two thin rectangular sheets \((0.20 \mathrm{~m} \times 0.40 \mathrm{~m})\) are identical. In the first sheet the axis of rotation lies along the \(0.20-\mathrm{m}\) side, and in the second it lies along the \(0.40-\mathrm{m}\) side. The same torque is applied to each sheet. The first sheet, starting from rest, reaches its final angular velocity in \(8.0 \mathrm{~s}\). How long does it take for the second sheet, starting from rest, to reach the same angular velocity?
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