/*! 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 69 A vacuum cleaner belt is looped ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A vacuum cleaner belt is looped over a shaft of radius 0.45 \(\mathrm{cm}\) and a wheel of radius 1.80 \(\mathrm{cm} .\) The arrangement of the belt, shaft, and wheel is similar to that of the chain and sprockets in Fig. \(Q 9.4 .\) The motor turns the shaft at 60.0 \(\mathrm{rev} / \mathrm{s}\) and the moving belt turns the wheel, which in turn is connected by another shaft to the roller that beats the dirt out of the rug being vacuumed. Assume that the belt doesn't slip on either the shaft or the wheel. (a) What is the speed of a point on the belt? (b) What is the angular velocity of the wheel, in rad/s?

Short Answer

Expert verified
(a) 1.695 m/s, (b) 94.17 rad/s

Step by step solution

01

Understanding the Problem

We need to determine the linear speed of a point on a belt that loops over a shaft and a wheel, as well as the angular velocity of the wheel. We know the radii of both the shaft and the wheel and the rotational speed of the shaft.
02

Calculate the Linear Speed of the Belt (Part a)

The speed of the belt is equal to the linear speed of the point on the circumference of the shaft. Using the formula:\[ v_s = r_s \omega_s \]where: - \( v_s \) is the linear speed of the shaft (and the belt),- \( r_s = 0.45 \) cm is the radius of the shaft, converted to meters = 0.0045 m,- \( \omega_s = 60.0 \) rev/s is the angular speed of the shaft, converted to rad/s = \( 60.0 \times 2 \pi \) rad/s.Calculating:\[ v_s = 0.0045 \, \text{m} \times 60.0 \times 2 \pi \, \text{rad/s} = 1.695 \, \text{m/s} \]
03

Relate Belt Speed to Wheel Angular Velocity (Part b)

Since the belt doesn't slip, its speed when it contacts the wheel is the same as the shaft:\[ v_s = v_w = 1.695 \, \text{m/s} \] where \( v_w \) is the linear speed for the wheel.
04

Calculate the Angular Velocity of the Wheel

Use the relationship between linear speed and angular velocity for the wheel:\[ v_w = r_w \omega_w \]where:- \( r_w = 1.80 \) cm is the radius of the wheel, converted to meters = 0.018 m,- \( \omega_w \) is the angular velocity of the wheel.Rearranging the equation to solve for \( \omega_w \):\[ \omega_w = \frac{v_w}{r_w} = \frac{1.695}{0.018} \, \text{m/s} \approx 94.17 \, \text{rad/s} \]

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.

Linear Speed
Linear speed, or linear velocity, refers to how fast a point on a moving object travels along a straight path. In this context, we are analyzing a point on a vacuum cleaner belt.
The key formula connecting linear speed with rotational motion is \[ v = r \omega \] where:
  • \( v \) is the linear speed of the point,
  • \( r \) represents the radius of the rotating object, and
  • \( \omega \) is the angular velocity.
This formula shows how the rotation of a shaft or wheel gets translated into linear motion.
When the shaft turns, it drags the belt along, resulting in a linear speed equal to the speed at the shaft edge. This speed remains constant around the wheel since there's no slip. Calculating it involves converting angular speed from revolutions per second to radians per second, then multiplying by the shaft’s radius. This gives the speed at which the belt moves linearly over time.
Angular Velocity
Angular velocity quantifies how quickly an object rotates around an axis, expressed in radians per second (rad/s).
For rotating objects like wheels, angular velocity is crucial in linking the object's rotation to real-world effects, like driving a belt.
In the given problem, we find the wheel's angular velocity using the constant speed of the belt (since it doesn’t slip) and the formula \[ v_w = r_w \omega_w \].Given the wheel's radius, this formula can be rearranged to solve for \( \omega_w \), which gives us how fast the wheel spins in terms of its rotation angle per second. By knowing the wheel's angular velocity, we see how quickly it contributes to the roller's rotation for specific functions like beating dirt from a rug.
Angular velocity helps bridge the gap between linear movement and rotational motion, illustrating how different parts of a machine interact.
Moment of Inertia
Moment of inertia is a key physical concept in understanding rotational motion, yet it wasn't directly calculated in this exercise. Consider it as the rotational counterpart to mass in linear motion.
It determines an object's resistance to changes in its angular motion. Each part of a rotating system contributes to the total moment of inertia based on its mass and distance from the rotation axis.
While not part of the specific calculations for belt speed or wheel angular velocity, understanding the moment of inertia can deepen insights into how machines like vacuum cleaners operate. It influences how energy is distributed through rotating parts, affecting the machine’s efficiency and performance.
Even though we did not calculate it here, recognizing its significance helps in grasping the broader context of rotational motion in mechanical systems. It ensures balance in the entire mechanism, contributing to its effective and smooth operation.

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 rotating blade of a blender turns with constant angular acceleration 1.50 rad/s\(^{2}\) (a) How much time does it take to reach an angular velocity of 36.0 rad/s, starting from rest? (b) Through how many revolutions does the blade turn in this time interval?

The flywheel of a gasoline engine is required to give up 500 \(\mathrm{J}\) of kinetic energy while its angular velocity decreases from 650 \(\mathrm{rev} / \mathrm{min}\) to 520 \(\mathrm{rev} / \mathrm{min}\) . What moment of inertia is required?

A thin, rectangular sheet of metal has mass \(M\) and sides of length \(a\) and \(b .\) Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one corner of the sheet.

The Kinetic Energy of Walking. If a person of mass \(M\) simply moved forward with speed \(V\), his kinetic energy would be \(\frac{1}{2} M V^{2}\).However, in addition to possessing a forward motion, various parts of his body (such as the arms and legs) undergo rotation. Therefore, his total kinetic energy is the sum of the energy from his forward motion plus the rotational kinetic energy of his arms and legs. The purpose of this problem is to see how much this rotational motion contributes to the person's kinetic energy. Biomedical measurements show that the arms and hands together typically make up 13\(\%\) of a person's mass, while the legs and feet together account for 37\(\%\) . For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. In a brisk walk, the arms and legs each move through an angle of about \(\pm 30^{\circ}\) (a total of \(60^{\circ} )\) from the vertical in approximately 1 second. We shall assume that they are held straight, rather than being bent, which is not quite true. Let us consider a 75 -kg person walking at 5.0 \(\mathrm{km} / \mathrm{h}\) , having arms 70 \(\mathrm{cm}\) long and legs 90 \(\mathrm{cm}\) long. (a) What is the average angular velocity of his arms and legs? (b) Using the average angular velocity from part (a), calculate the amount of rotational kinetic energy in this person's arms and legs as he walks. (c) What is the total kinetic energy due to both his forward motion and his rotation? (d) What percentage of his kinetic energy is due to the rotation of his legs and arms?

About what axis will a uniform, balsa-wood sphere have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius, with the axis along a diameter?

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.