/*! 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 4 The angular position of a point ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 4

The angular position of a point on a rotating wheel is given by \(\theta=2.0+4.0 t^{2}+2.0 t^{3}\), where \(\theta\) is in radians and \(t\) is in seconds \(A t\) \(t=0\), what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at \(t=4.0 \mathrm{~s}\) ? (d) Calculate its angular acceleration at \(t=2.0 \mathrm{~s}\) (e) Is its angular acceleration constant?

Short Answer

Expert verified
(a) 2.0 rad, (b) 0 rad/s, (c) 128 rad/s, (d) 32 rad/s虏, (e) No, it changes with time.

Step by step solution

01

Determine Angular Position

To find the point's angular position at \( t = 0 \), substitute \( t = 0 \) into the given equation \( \theta = 2.0 + 4.0t^2 + 2.0t^3 \). This gives us \( \theta = 2.0 + 4.0(0)^2 + 2.0(0)^3 = 2.0 \) radians.
02

Calculate Angular Velocity

Angular velocity \( \omega \) is the derivative of angular position with respect to time. Differentiate \( \theta = 2.0 + 4.0t^2 + 2.0t^3 \) to get \( \omega = \frac{d\theta}{dt} = 0 + 8.0t + 6.0t^2 \). At \( t = 0 \), \( \omega = 8.0(0) + 6.0(0)^2 = 0 \) rad/s.
03

Angular Velocity at t=4.0 seconds

For \( t = 4.0 \) s, substitute \( t = 4.0 \) into \( \omega = 8.0t + 6.0t^2 \). This gives us \( \omega = 8.0(4.0) + 6.0(4.0)^2 = 32.0 + 96.0 = 128.0 \) rad/s.
04

Calculate Angular Acceleration at t=2.0 seconds

Angular acceleration \( \alpha \) is the derivative of angular velocity \( \omega \) with respect to time. Differentiating \( \omega = 8.0t + 6.0t^2 \), we get \( \alpha = \frac{d\omega}{dt} = 8.0 + 12.0t \). At \( t = 2.0 \) s, \( \alpha = 8.0 + 12.0(2.0) = 8.0 + 24.0 = 32.0 \) rad/s虏.
05

Check If Angular Acceleration is Constant

Since angular acceleration \( \alpha = 8.0 + 12.0t \) depends on time \( t \), it is not constant. The term \( 12.0t \) indicates that \( \alpha \) changes with 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 Position
Angular position refers to the orientation of a point or line with respect to a fixed direction in a rotational system. It is often denoted by \( \theta \) and measured in radians. In the context of our exercise, the angular position is given by the equation \( \theta = 2.0 + 4.0t^2 + 2.0t^3 \).
  • "\( \theta = 2.0 \)" indicates the starting angular position when \( t = 0 \), revealing that the system is not starting from an angle of zero.
  • "\( 4.0t^2 \)" and "\( 2.0t^3 \)" components tell us that the angular position changes with time and is influenced by both quadratic and cubic time terms.
This implies that as time progresses, the position moves further along its circular path, showcasing a more complex motion compared to a simple rotation.
Angular Velocity
Angular velocity, symbolized as \( \omega \), characterizes how fast an object rotates, typically measured in radians per second (rad/s). It is the rate of change of angular position.To determine \( \omega \) from the angular position \( \theta = 2.0 + 4.0t^2 + 2.0t^3 \), differentiate with respect to time \( t \). This yields:\[ \omega = \frac{d\theta}{dt} = 8.0t + 6.0t^2 \]- At \( t = 0 \), \( \omega = 0 \) indicating no initial rotational motion.- When \( t = 4.0 \), substituting into the equation gives a velocity of 128.0 rad/s.The angular velocity increases over time due to the presence of the quadratic term, showing the wheel is speeding up its rotation as time goes by.
Angular Acceleration
Angular acceleration, denoted by \( \alpha \), refers to the rate of change of angular velocity. It tells us how quickly the rotation speed is changing.To find \( \alpha \), differentiate the angular velocity function \( \omega = 8.0t + 6.0t^2 \) with respect to time:\[ \alpha = \frac{d\omega}{dt} = 8.0 + 12.0t \]
  • At \( t = 2.0 \) s, \( \alpha = 32.0 \) rad/s虏, indicating a significant increase in rotation speed.
  • The presence of the term \( 12.0t \) reveals that angular acceleration changes over time, it is not a fixed value.
It shows how acceleration increases the speed of rotation as time progresses. This non-constant acceleration implies the system undergoes a complex, dynamic change.
Time-Dependent Functions
In the context of angular motion, time-dependent functions describe how quantities like angular position, velocity, and acceleration change over time.The exercise showcases these elements fitting together in equations that depend on time \( t \):
  • Angular position \( \theta(t) = 2.0 + 4.0t^2 + 2.0t^3 \) evolves as time increases.
  • Angular velocity \( \omega(t) = 8.0t + 6.0t^2 \) accelerates with time, showing immediate changes in speed.
  • Angular acceleration \( \alpha(t) = 8.0 + 12.0t \) further highlights how rotation speed changes progressively.
These functions capture the rotational dynamics where each relies on t, illustrating a system that does not have uniform motion. Instead, each step presents a new, dependent result.

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) Show that the rotational inertia of a solid cylinder of mass \(M\) and radius \(R\) about its central axis is equal to the rotational inertia of a thin hoop of mass \(M\) and radius \(R / \sqrt{2}\) about its central axis. (b) Show that the rotational inertia \(I\) of any given body of mass \(M\) about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass \(M\) and a radius \(k\) given by $$ k=\sqrt{\frac{I}{M}} $$ The radius \(k\) of the equivalent hoop is called the radius of gyration of the given body.

A thin rod of length \(0.75 \mathrm{~m}\) and mass \(0.42 \mathrm{~kg}\) is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed \(4.0 \mathrm{rad} / \mathrm{s}\). Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a \(30 \mathrm{~kg}\) box as shown in Fig. \(10-59 .\) The outer radius \(R\) of the device is \(0.50 \mathrm{~m}\), and the radius \(r\) of the hub is \(0.20 \mathrm{~m}\). When a constant horizontal force \(\vec{F}_{\text {app }}\) of magnitude \(140 \mathrm{~N}\) is applied to a rope wrapped around the outside of the device, the box, which is suspended from a rope wrapped around the hub, has an upward acceleration of magnitude \(0.80\) \(\mathrm{m} / \mathrm{s}^{2}\). What is the rotational inertia of the device about its axis of rotation?

A pulley wheel that is \(8.0 \mathrm{~cm}\) in diameter has a \(5.6\) -m-long cord wrapped around its periphery. Starting from rest, the wheel is given a constant angular acceleration of \(1.5 \mathrm{rad} / \mathrm{s}^{2}\). (a) Through what angle must the wheel turn for the cord to unwind completely? (b) How long will this take?

A \(32.0 \mathrm{~kg}\) wheel, essentially a thin hoop with radius \(1.20 \mathrm{~m}\), is rotating at 280 rev/min. It must be brought to a stop in \(15.0 \mathrm{~s}\). (a) How much work must be done to stop it? (b) What is the required average power?
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Fields in Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Physical Quantities And Units

Read Explanation

Modern Physics

Read Explanation

Famous Physicists

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.