/*! 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] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
The time to reach the angular velocity of \( 36.0 \, \mathrm{rad} / \mathrm{s} \) is \( 24 \, \mathrm{s} \) and the blade turns approximately \( 69 \) revolutions during this interval.

Step by step solution

01

Find the time using the formula for angular velocity

We rearrange the equation \( \omega = \omega_0 + \alpha t \) to solve for time \( t \), giving \( t = (\omega - \omega_0) / \alpha \). Plugging in the known values: \( t = (36.0 \, \mathrm{rad} / \mathrm{s} - 0) / 1.5 \, \mathrm{rad} / \mathrm{s}^2 \) yields \( t = 24 \, s \).
02

Calculate the total angle covered

Now that we have the time, we substitute our values into the equation \( \theta = 0.5 \alpha t^2 \) to compute the total angle the blade covers in that time: \( \theta = 0.5 * 1.5 \, \mathrm{rad} / \mathrm{s}^2 * (24 \, s)^2 = 432 \, \mathrm{rad}\).
03

Convert angle from radians to revolutions

The angle found is in radians. To convert this into number of revolutions, we use the fact that one complete revolution equals \( 2\pi \) radians. So the number of revolutions \( n \) is computed as \( n = \theta / 2\pi \). Substituting the known value \( n = 432 \, \mathrm{rad} / 2\pi \) results in \( n \approx 69 \) revolutions.

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 Velocity
Angular velocity is a measure of how fast an object rotates or revolves. It tells you how much angle an object sweeps out in a unit of time. For circular motion, it is essentially the rate of change of angular displacement. The standard unit for measuring angular velocity is radians per second \(\left(\mathrm{rad}/\mathrm{s}\right)\).

  • Angular velocity can be constant or change over time.
  • It is analogous to linear velocity, but for rotational motion.
To calculate angular velocity, you can use the formula:\[\omega = \omega_0 + \alpha t\]where:
  • \(\omega\) is the final angular velocity.
  • \(\omega_0\) is the initial angular velocity (which is zero if starting from rest).
  • \(\alpha\) is the angular acceleration.
  • \(t\) is the time elapsed.
In the problem given, the blade reaches an angular velocity of \(36\,\mathrm{rad}/\mathrm{s}\) starting from rest over a specific time period. Calculating how long it takes can be done by rearranging the angular velocity formula to solve for the time, thus giving us the first step in handling such problems.
Revolutions
Revolutions refer to the complete turns around a center point or axis. When discussing angular motion, it tells us how many times an object goes full circle.

  • A single revolution means moving a complete \(2\pi\) radians.
  • It's a standard way to talk about full cycles or loops.
In the second part of the exercise, we calculate how many revolutions the blender blade completes in the given time interval while accelerating. Once we know the total angle it has rotated, in radians, we can easily find how many full cycles or revolutions it completes. This conversion uses the relationship between radians and revolutions. Consequently, it's vital to develop a sense of how one translates to the other.

By knowing the total angle in radians (calculated in the previous steps) and using the relationship that \(1\) revolution equals \(2\pi\) radians, you can find the number of revolutions:\[n = \frac{\theta}{2\pi}\]where:
  • \(n\) is the number of revolutions.
  • \(\theta\) is the angle in radians.
Having cut through the calculations, the blade makes approximately \(69\) revolutions.
Radians
Radians are the standard unit of angular measure in physics and engineering. Unlike degrees, which divide a circle into \(360\) equal parts, radians are defined so that a complete circle is \(2\pi\) radians.

  • One radian represents the angle of a circle created when the arc length is equal to the radius.
  • Radians provide a direct relationship between the linear and angular quantities.
The choice of using radians is practical: it naturally ties into mathematical formulations for circles and curves. This makes calculations involving angular velocity, and acceleration, much simpler.

Furthermore, almost every formula in rotational motion works more efficiently and elegantly with radians. For example, when computing the total angle turned (as required in our blender blade problem), using radians allows for direct use in kinetic equations:\[\theta = \frac{1}{2} \alpha t^2\]Here:
  • \(\theta\) denotes the angular displacement in radians.
  • \(\alpha\) is the angular acceleration.
  • \(t\) is the time.
The problem shows the power of radians as they streamline the process of converting angular measurements to other forms, such as revolutions, strengthening the connection between linear and rotational dynamics.

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 disk of radius \(25.0 \mathrm{~cm}\) is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. \(\mathbf{P 9 . 6 1}\) ). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is \(1.80 \mathrm{~m} / \mathrm{s}^{2}\). (a) Find \(A\). (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of \(15.0 \mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches \(15.0 \mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .)\)

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?

Engineers are designing a system by which a falling mass \(m\) imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum (Fig. \(\mathbf{P 9 . 6 4}\) ). There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is \(3.71 \mathrm{~m} / \mathrm{s}^{2} .\) In the earth tests, when \(m\) is set to \(15.0 \mathrm{~kg}\) and allowed to fall through \(5.00 \mathrm{~m},\) it gives \(250.0 \mathrm{~J}\) of kinetic energy to the drum. (a) If the system is operated on Mars, through what distance would the \(15.0 \mathrm{~kg}\) mass have to fall to give the same amount of kinetic energy to the drum? (b) How fast would the \(15.0 \mathrm{~kg}\) mass be moving on Mars just as the drum gained \(250.0 \mathrm{~J}\) of kinetic energy?

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

A uniform wheel in the shape of a solid disk is mounted on a frictionless axle at its center. The wheel has mass \(5.00 \mathrm{~kg}\) and radius \(0.800 \mathrm{~m} .\) A thin rope is wrapped around the wheel, and a block is suspended from the free end of the rope. The system is released from rest and the block moves downward. What is the mass of the block if the wheel turns through 8.00 revolutions in the first \(5.00 \mathrm{~s}\) after the block is released?

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.