/*! 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 25 Consider a particle of a rigid o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a particle of a rigid object rotating about a fixed axis. Show that the tangential and radial vector components of the linear acceleration are: $$ \overrightarrow{\mathbf{a}}_{\tan }=\overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\mathbf{r}} \quad \text { and } \quad \overrightarrow{\mathbf{a}}_{\mathrm{R}}=\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\mathbf{v}} $$

Short Answer

Expert verified
The tangential and radial components of linear acceleration are given by \( \overrightarrow{\mathbf{a}}_{\tan} = \overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\mathbf{r}} \) and \( \overrightarrow{\mathbf{a}}_{R} = \overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\mathbf{v}} \) respectively.

Step by step solution

01

Understanding Angular and Linear Quantities

Recognize that \( \overrightarrow{\boldsymbol{\alpha}} \) is the angular acceleration, \( \overrightarrow{\mathbf{r}} \) is the position vector, \( \overrightarrow{\boldsymbol{\omega}} \) is the angular velocity, and \( \overrightarrow{\mathbf{v}} \) is the tangential velocity of the particle. We need to find the tangential and radial acceleration components.
02

Tangential Acceleration Component

The tangential acceleration vector \( \overrightarrow{\mathbf{a}}_{\tan} \) can be found by the cross product of the angular acceleration \( \overrightarrow{\boldsymbol{\alpha}} \) with the position vector \( \overrightarrow{\mathbf{r}} \), which is given by:\[\overrightarrow{\mathbf{a}}_{\tan} = \overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\mathbf{r}}\]
03

Radial Acceleration Component

For the radial acceleration component \( \overrightarrow{\mathbf{a}}_{R} \), we use the angular velocity \( \overrightarrow{\boldsymbol{\omega}} \) and the tangential velocity \( \overrightarrow{\mathbf{v}} = \overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\mathbf{r}} \). The radial acceleration is derived from the cross product of angular velocity and the tangential velocity:\[\overrightarrow{\mathbf{a}}_{R} = \overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\mathbf{v}}\]

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.

Tangential Acceleration
Tangential acceleration occurs when a particle moves along a curved path, such as when an object rotates around a fixed axis. It is the component of linear acceleration parallel to the path of motion. Using the cross product, we can calculate this component easily, considering the angular acceleration and the position of the particle relative to the axis of rotation.
The formula for tangential acceleration is given by \[\overrightarrow{\mathbf{a}}_{\tan} = \overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\mathbf{r}}\]where:
  • \( \overrightarrow{\boldsymbol{\alpha}} \) is the angular acceleration vector.
  • \( \overrightarrow{\mathbf{r}} \) is the position vector from the axis of rotation to the particle.
Tangential acceleration tells us how fast the velocity of a point on the rotating object is changing in the direction tangential to its circular path. If the angular acceleration is zero, the tangential acceleration will also be zero, indicating no change in speed, though the object may still be in rotational motion.
Radial Acceleration
Radial acceleration, often referred to as centripetal acceleration, keeps a rotating particle on its circular path. This type of acceleration points towards the center of the rotation, continuously changing the direction of the velocity of the object without altering its speed along the path. We often find radial acceleration in situations where an object is moving in a circular motion at a constant speed.
To determine radial acceleration, the importance of angular velocity and the tangential velocity becomes evident. The formula for radial acceleration is achieved through another cross product:\[\overrightarrow{\mathbf{a}}_{R} = \overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\mathbf{v}}\]where:
  • \( \overrightarrow{\boldsymbol{\omega}} \) is the angular velocity vector.
  • \( \overrightarrow{\mathbf{v}} \) is the tangential velocity, found through \( \overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\mathbf{r}} \).
This concept is crucial to understanding forces that act in rotating systems, such as the tension in the string of a swinging ball or the gravitational pull experienced by celestial bodies.
Angular Velocity
Angular velocity is a vector quantity that represents how fast a point or rigid object rotates around an axis. It plays a pivotal role in both tangential and radial components of acceleration. To get a grip on how objects behave under rotational motion, understanding angular velocity is essential.
Represented by:\[\overrightarrow{\boldsymbol{\omega}} = \frac{d\theta}{dt}\]where \( \theta \) is the angular displacement and \( t \) is time. Unlike linear velocity, which relates to an object's speed in m/s, angular velocity is measured in radians per second (rad/s). This measurement shows the rate at which the angular displacement changes and indicates the speed of rotation and its directional aspect.
  • Key in determining vector components like tangential and radial acceleration.
  • Integral in calculating rotation in mechanical systems and physical phenomena like cyclonic systems and even galaxy formations.
Understanding angular velocity helps bridge the gap between rotational and linear motion, offering insights into various natural and engineered systems.

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

(1) (a) What is the angular momentum of a 2.8 -kg uniform cylindrical grinding whecl of radius 18 \(\mathrm{cm}\) when rotating at 1300 rpm? \((b)\) How much torque is required to stop it in 6.0 s?

(III) Hurricanes can involve winds in excess of 120 \(\mathrm{km} / \mathrm{h}\) at the outer edge. Make a crude estimate of \((a)\) the cnergy, and (b) the angular momentum, of such a hurricane, approximating it as a rigidly rotating uniform cylinder of air (density 13 \(\mathrm{kg} / \mathrm{m}^{3}\) ) of radius 85 \(\mathrm{km}\) and height 4.5 \(\mathrm{km}\) .

Water drives a waterwheel (or turbine) of radius \(R=3.0 \mathrm{~m}\) as shown in Fig. \(11-47 .\) The water enters at a speed \(v_{1}=7.0 \mathrm{~m} / \mathrm{s}\) and exits from the waterwheel at a speed \(v_{2}=3.8 \mathrm{~m} / \mathrm{s} . \quad(a)\) If \(85 \mathrm{~kg}\) of water passes through per second, what is the rate at which the water delivers angular momentum to the waterwheel? \((b)\) What is the torque the water applies to the waterwheel? \((c)\) If the water causes the waterwheel to make one revolution every \(5.5 \mathrm{~s}\), how much power is delivered to the wheel?

(II) A diver (such as the one shown in Fig. 2\()\) can reduce her moment of inertia by a factor of about 35 when changing from the straight position to the tuck position. If she makes 20 rotations in 1.5 \(\mathrm{s}\) when in the tuck position, what is her angular spced (rev/s) when in the straight position?

A particle is located at \(\overrightarrow{\mathbf{r}}=(4.0 \hat{\mathbf{i}}+3.5 \hat{\mathbf{j}}+6.0 \hat{\mathbf{k}}) \mathrm{m} .\) A force \(\overrightarrow{\mathbf{F}}=(9.0 \hat{\mathbf{j}}-4.0 \hat{\mathbf{k}}) \mathbf{N}\) acts on it. What is the torque, calculated about the origin?

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.