/*! 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 72 Attached to each end of a thin s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Attached to each end of a thin steel rod of length \(1.20 \mathrm{~m}\) and mass \(6.40 \mathrm{~kg}\) is a small ball of mass \(1.06 \mathrm{~kg}\). The rod is constrained to rotate in a horizontal plane about a vertical axis through its midpoint. At a certain instant, it is rotating at \(39.0 \mathrm{rev} / \mathrm{s}\). Because of friction, it slows to a stop in \(32.0 \mathrm{~s}\). Assuming a constant retarding torque due to friction, compute (a) the angular acceleration, (b) the retarding torque, (c) the total energy transferred from mechanical energy to thermal energy by friction, and (d) the number of revolutions rotated during the \(32.0 \mathrm{~s}\). (e) Now suppose that the retarding torque is known not to be constant. If any of the quantities (a), (b), (c), and (d) can still be computed without additional information, give its value.

Short Answer

Expert verified
(a) \( \alpha = -7.654 \text{ rad/s}^2 \); only (\alpha) can be computed if torque is variable; see steps for (b-d).

Step by step solution

01

Calculate the angular velocity

First, we convert the initial angular velocity from revolutions per second to radians per second using the formula: \( \omega_i = 39.0 \text{ rev/s} \times 2\pi \text{ rad/rev} = 78\pi \text{ rad/s} \).
02

Find angular acceleration

The angular acceleration can be calculated using the equation: \( \alpha = \frac{\omega_f - \omega_i}{\Delta t} \), where \( \omega_f = 0 \ \text{ rad/s} \) and \( \omega_i = 78\pi \ \text{ rad/s} \). Thus, \( \alpha = \frac{0 - 78\pi}{32} = -\frac{78\pi}{32} \approx -7.654 \text{ rad/s}^2 \).
03

Calculating moment of inertia

The moment of inertia \( I \) of the system consists of the rod and two balls attached at each end. For a thin rod about its center, \( I_{rod} = \frac{1}{12}m_{rod}L^2 = \frac{1}{12} \times 6.40 \text{ kg} \times (1.20 \text{ m})^2 \). For the two small balls treated as point masses located at the ends of the rod, \( I_{balls} = 2 \times m_{ball} \left( \frac{L}{2} \right)^2 = 2 \times 1.06 \text{ kg} \times \left( \frac{1.2}{2} \right)^2 \).Compute each and add for total \( I \).
04

Calculate the retarding torque

The retarding torque is given by the equation \( \tau = I \cdot \alpha \). Using the moment of inertia from Step 3 and \( \alpha = -7.654 \text{ rad/s}^2 \) from Step 2, calculate \( \tau \).
05

Calculate total energy transferred to thermal energy

Mechanical energy lost \( \Delta E = \frac{1}{2}I\omega_i^2 \). Substitute \( \omega_i \) from Step 1 and \( I \) from Step 3 to find \( \Delta E \).
06

Calculate the number of revolutions

The number of revolutions during the time is found from the total angular distance \( \theta \). First, calculate \( \theta \) using \( \theta = \omega_i \cdot \Delta t + \frac{1}{2} \cdot \alpha \cdot (\Delta t)^2 \) and convert from radians to revolutions using \( \theta / (2\pi) \).
07

Analyze possible computations if torque is variable

If the retarding torque is not constant, only the angular acceleration \( \alpha \) from Step 2 can be computed because it was derived from initial and final angular velocities alone. The calculations for torque, energy, and revolutions rely on the torque being constant.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Acceleration
Angular acceleration is a measure of how quickly an object's rotational speed changes. In this exercise, the steel rod with two masses attached at its ends decelerates due to friction, meaning it experiences negative angular acceleration. To find angular acceleration, use the formula:
  • \( \alpha = \frac{\omega_f - \omega_i}{\Delta t} \)
It describes how the angular velocity \( \omega \) changes over a time interval \( \Delta t \). Initial velocity \( \omega_i \) is \( 78\pi \text{ rad/s} \) (converted from 39 rev/s), and the final \( \omega_f \) is 0 rad/s when the rod stops. The solution involves inserting these values into the formula, giving us \( \alpha = -7.654 \text{ rad/s}^2 \). Negative sign indicates deceleration, highlighting the rod's motion slowing down due to retarding forces. Angular acceleration plays a crucial role in understanding rotational dynamics and how forces can change rotational motion.
Understanding this concept is essential for analyzing how rotational systems behave under different forces and constraints.
Moment of Inertia
Moment of inertia (I) is a property indicating how much resistance an object has to being rotated. It depends heavily on the object's mass distribution relative to the axis of rotation. For this rod and balls system:
  • The rod's moment of inertia: \( I_{\text{rod}} = \frac{1}{12}m_{\text{rod}}L^2 \)
  • The balls as point masses at the rod's ends: \( I_{\text{balls}} = 2 \times m_{\text{ball}} \left( \frac{L}{2} \right)^2 \)
Calculating these individually and then combining them provides the total moment of inertia of the system. Moment of inertia is crucial to determining the angular acceleration for a given torque. A higher moment of inertia means more torque is needed to achieve the same angular acceleration. This concept is paramount in rotational dynamics, where understanding how mass distribution affects movement can influence design and function of rotational systems like wheels and turbines.
Retarding Torque
Retarding torque is the force that slows down the rotational motion. Torque in rotational dynamics is the rotational equivalent of force in linear motion. Here, friction is the source of retarding torque, calculated by \( \tau = I \cdot \alpha \), where \( I \) is the moment of inertia and \( \alpha \) is angular acceleration.
  • This exercise shows how frictional forces affect rotation, transforming some of the mechanical energy into thermal energy.
  • The retarding torque ensures that the system eventually comes to a halt by reducing the rotational speed to zero.
By understanding torque, we can better predict and control the rotational behavior of objects, which is important in systems like engines and brakes. Grasping how retarding torque works is vital for engineering applications, where managing the effects of energy dissipation and motion slowing is often a critical requirement.
Energy Transfer
Energy transfer in rotational dynamics often occurs from mechanical energy to thermal energy, primarily due to friction. Initially, the whole system has mechanical energy due to rotation. As the rod slows because of retarding torque, this energy is decreased.
  • The lost energy converts to thermal energy, signifying that friction does work on the system.
  • Here, the initial mechanical energy is given by \( \Delta E = \frac{1}{2}I\omega_i^2 \).
This energy change is crucial because it exemplifies a primary aspect of energy analysis in physics: energy conservation in a closed system. Even if the mechanical energy appears to vanish, it just transforms into another form, often less useful. This transformation highlights the importance of frictional forces in real-world applications, such as braking systems where controlling energy transfer is essential to prevent overheating and ensure safety.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The angular speed of an automobile engine is increased at a constant rate from 1200 rev/min to 3000 rev \(/ \min\) in 12 s. (a) What is its angular acceleration in revolutions per minute-squared? (b) How many revolutions does the engine make during this 12 s interval?

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period \(T\) of rotation is found by measuring the time between pulses. The pulsar in the Crab nebula has a period of rotation of \(T=0.033 \mathrm{~s}\) that is increasing at the rate of \(1.26 \times 10^{-5} \mathrm{~s} / \mathrm{y}\). (a) What is the pulsar's angular acceleration \(\alpha ?\) (b) If \(\alpha\) is constant, how many years from now will the pulsar stop rotating? (c) The pulsar originated in a supernova explosion seen in the year \(1054 .\) Assuming constant \(\alpha,\) find the initial \(T\)

Four identical particles of mass \(0.50 \mathrm{~kg}\) each are placed at the vertices of a \(2.0 \mathrm{~m} \times 2.0 \mathrm{~m}\) square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square, (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?

Starting from rest, a wheel has constant \(\alpha=3.0 \mathrm{rad} / \mathrm{s}^{2} .\) During a certain \(4.0 \mathrm{~s}\) interval, it turns through 120 rad. How much time did it take to reach that \(4.0 \mathrm{~s}\) interval?

Cheetahs running at top speed have been reported at an astounding \(114 \mathrm{~km} / \mathrm{h}\) (about \(71 \mathrm{mi} / \mathrm{h})\) by observers driving alongside the animals. Imagine trying to measure a cheetah's speed by keeping your vehicle abreast of the animal while also glancing at your speedometer, which is registering \(114 \mathrm{~km} / \mathrm{h}\). You keep the vehicle a constant \(8.0 \mathrm{~m}\) from the cheetah, but the noise of the vehicle causes the cheetah to continuously veer away from you along a circular path of radius \(92 \mathrm{~m}\). Thus, you travel along a circular path of radius \(100 \mathrm{~m} .\) (a) What is the angular speed of you and the cheetah around the circular paths? (b) What is the linear speed of the cheetah along its path? (If you did not account for the circular motion, you would conclude erroneously that the cheetah's speed is \(114 \mathrm{~km} / \mathrm{h},\) and that type of error was apparently made in the published reports.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.