/*! 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 11 The rotating blade of a blender ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 11

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?

Short Answer

Expert verified
(a) 24.0 s; (b) 68.7 revolutions.

Step by step solution

01

Understanding Angular Motion

To solve the problem, remember that angular acceleration (\(\alpha\)) is the rate of change of angular velocity (\(\omega\)) over time (\(t\)). The initial angular velocity (\(\omega_0\)) is zero since the blender starts from rest.
02

Applying the Angular Velocity Formula

The formula for final angular velocity in angular motion is \(\omega = \omega_0 + \alpha t\). Substitute \(\omega = 36.0\, \text{rad/s}\), \(\omega_0 = 0\, \text{rad/s}\), and \(\alpha = 1.50\, \text{rad/s}^2\) into the equation and solve for \(t\).
03

Solving for Time

Rearrange the formula to solve for time: \(t = \frac{\omega - \omega_0}{\alpha}\).Plug in the values: \(t = \frac{36.0\, \text{rad/s} - 0\, \text{rad/s}}{1.50\, \text{rad/s}^2}\), which simplifies to \(t = 24.0\, \text{s}\).
04

Determining Angular Displacement

To find angular displacement (\(\theta\)), use the formula \(\theta = \omega_0 t + \frac{1}{2}\alpha t^2\). Since \(\omega_0 = 0\, \text{rad/s}\), it simplifies to \(\theta = \frac{1}{2} \cdot 1.50\, \text{rad/s}^2 \cdot (24.0\, \text{s})^2\).
05

Calculating Angular Displacement

Compute the expression: \(\theta = \frac{1}{2} \cdot 1.50 \cdot 576.0\), which results in \(\theta = 432.0\, \text{rad}\).
06

Converting to Revolutions

Remember that one revolution equals \(2\pi\, \text{rad}\). Convert the angular displacement in radians to revolutions: \(\text{Revolutions} = \frac{432.0}{2\pi}\).
07

Final Calculation

Compute the number of revolutions: \(\text{Revolutions} = \frac{432.0}{6.2832} \approx 68.7\). Therefore, the blade turns approximately 68.7 revolutions during this time.

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 91影视!

Key Concepts

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

Angular Velocity
Angular velocity is a key concept in rotational motion. It indicates how fast an object rotates or spins around a particular axis.
Measured in radians per second (rad/s), it describes the rate of change of angular position relative to time.
In the context of a blender's blade, if you know its angular velocity, you can determine how quickly it reaches different speeds during operation.
  • The formula is given by \( \omega = \omega_0 + \alpha t \), where \( \omega \) is the final angular velocity and \( \omega_0 \) is the initial angular velocity.
  • In many practical scenarios, like our example, an object may start from rest. This means \( \omega_0 = 0 \).
  • This simplification makes calculations straightforward, as you only need the final velocity and the time or acceleration to analyze the motion.
Thus, understanding angular velocity helps predict and control the timing and speed in mechanisms like the blender.
Angular Acceleration
Angular acceleration is the change in angular velocity over time, much like how linear acceleration measures the change in speed.
It is represented by \( \alpha \) and measured in radians per second squared (rad/s虏).
In the scenario of the blender blade, it shows how quickly the blade is speeding up its rotation.
  • The angular acceleration tells us how fast the rotational speed of an object is increasing.
  • Positive values indicate an increase in rotational speed, while negative values (deceleration) would imply slowing down.
  • It's essential for calculating how long it takes the blender to reach a specific speed from rest. You use the relationship \( \omega = \omega_0 + \alpha t \).
This information is vital for designing and controlling any machinery that involves rotating components, ensuring efficiency and safety.
Angular Displacement
Angular displacement, \( \theta \), shows how much an object has rotated or moved in a circular path. Measured in radians, it represents the angle through which a point or line has been rotated in a specified body without regard to the total path traversed along the circumference.
  • For a rotating body starting from rest, the angular displacement can be calculated using \( \theta = \omega_0 t + \frac{1}{2}\alpha t^2 \). With \( \omega_0 = 0 \), this simplifies to \( \theta = \frac{1}{2}\alpha t^2 \).
  • In the problem of the blender's blade, finding angular displacement helps us understand how much the blade has moved within a fixed period.
  • The concept is crucial in determining the extent of rotation or movement in mechanisms where direction and magnitude of rotation matter.
This information is essential especially in systems that rely on precise angles, such as gears or rotary encoders.
Revolutions
Revolutions refer to the number of full 360-degree turns an object makes around an axis.
In physics, understanding how many times something spins completely around is important in rotational dynamics.
It's typically measured by converting the angular displacement from radians to revolutions.
  • The conversion is done by dividing the angular displacement \( \theta \) by \( 2\pi \), since one revolution equals \( 2\pi \) radians.
  • In our example, once the angular displacement is calculated, you can find out how many revolutions the blender's blade completes through \( \text{Revolutions} = \frac{\theta}{2\pi} \).
  • This count of spins or cycles helps in understanding how work is done over time by rotationally dynamic systems, like motors and drills.
Measuring in revolutions is practical, especially when considering machines and tools that operate continuously in cycles.
Angular Kinematics
Angular kinematics deals with the description of rotational motion without reference to the forces causing it.
This area parallels basic linear kinematics but focuses more on the aspects of circular motion, involving parameters like angular displacement, velocity, and acceleration.
  • The core equations, similar to those for linear motion, help to predict the behavior of rotating objects: \( \theta = \omega_0 t + \frac{1}{2}\alpha t^2 \) and \( \omega = \omega_0 + \alpha t \).
  • Angular kinematics provides a foundational framework to analyze scenarios such as the acceleration time of a blender blade or the total number of revolutions it undergoes.
  • Understanding these principles is key in the design and analysis of anything that rotates, ensuring components work reliably and efficiently over time.
By applying these concepts to real-world problems, engineers and students can develop insights into how rotational systems operate effectively.

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

A circular saw blade with radius 0.120 \(\mathrm{m}\) starts from rest and turns in a vertical plane with a constant angular acceleration of 3.00 \(\mathrm{rev} / \mathrm{s}^{2}\) After the blade has turned through 155 rev, a small piece of the blade breaks loose from the top of the blade.After the piece breaks loose, it travels with a velocity that is ini tially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of 0.820 m to the floor. How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?

A wheel changes its angular velocity with a constant angular acceleration while rotating about a fixed axis through its center. (a) Show that the change in the magnitude of the radial acceleration during any time interval of a point on the wheel is twice the product of the angular acceleration, the angular displacement, and the perpendicular distance of the point from the axis. (b) The radial acceleration of a point on the wheel that is 0.250 \(\mathrm{m}\) from the axis changes from 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) to 85.0 \(\mathrm{m} / \mathrm{s}^{2}\) as the wheel rotates through 20.0 rad. Calculate the tangential acceleration of this point. (c) Show that the change in the wheel's kinetic energy during any time interval is the product of the moment of inertia about the axis, the angular acceleration, and the angular displacement. (d) During the 20.0 -rad angular displacement of part (b), the kinetic energy of the wheel increases from 20.0 \(\mathrm{J}\) to 45.0 \(\mathrm{J} .\) What is the moment of inertia of the wheel about the rotation axis?

A classic 1957 Chevrolet Corvette of mass 1240 kg starts from rest and speeds up with a constant tangential acceleration of 2.00 \(\mathrm{m} / \mathrm{s}^{2}\) on a circular test track of radius 60.0 \(\mathrm{m} .\) Treat the car as a particle. (a) What is its angular acceleration? (b) What is its angular speed 6.00 s after it starts? (c) What is its radial acceleration at this time? (d) Sketch a view from above showing the circular track, the car, the velocity vector, and the acceleration component vectors 6.00 s after the car starts. (e) What are the magnitudes of the total acceleration and net force for the car at this time? (f) What angle do the total acceleration and net force make with the car' velocity at this time?

A uniform 3.00 -kg rope 24.0 \(\mathrm{m}\) long lies on the ground at the top of a vertical cliff. A mountain climber at the top lets down half of it to help his partner climb up the cliff. What was the change in potential energy of the rope during this maneuver?

A metal sign for a car dealership is a thin, uniform right triangle with base length \(b\) and height \(h .\) The sign has mass \(M .\) (a) What is the moment of inertia of the sign for rotation about the side of length \(h ?\) (b) If \(M=5.40 \mathrm{kg}, b=1.60 \mathrm{m},\) and \(h=1.20 \mathrm{m},\) what is the kinetic energy of the sign when it is rotating about an axis along the \(1.20-\mathrm{m}\) side at 2.00 \(\mathrm{rev} / \mathrm{s} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Solid State Physics

Read Explanation

Physics of Motion

Read Explanation

Physical Quantities And Units

Read Explanation

Electrostatics

Read Explanation

Fundamentals of Physics

Read Explanation
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.