Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

Answers without the blur.

Just sign up for free and you're in.

Short Answer

The angular acceleration of a wheel is $α = 6.0 t^{4} - 4.0 t^{2}$ , with $α$ is in radians per second-squared and $t$ in seconds. At time $t = 0$ , the wheel has an angular velocity of $+ 2.0 \frac{rad}{s}$ and an angular position of $+ 1.0 rad$ .
Write expressions for (a) the angular velocity (rad/s) and (b) the angular position (rad) as functions of time (s).

Angular velocity of the wheel as a function of time is $ω = 1.2 t^{5} - 1.33 t^{3} + 2.0$
Angular position of the wheel as a function of time is $θ = 0 .2 t^{6} - 0 .33 t^{4} + 2 .0 t + 1 .0$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Listing the given quantities

The angular acceleration of a wheel is, $α = (6.0 t^{4} - 4.0 t^{2}) \frac{rad}{s^{2}}$

The angular velocity of a wheel at t = 0 is, $ω_{0} = + 2.0 \frac{rad}{s}$

The angular position of a wheel at t = 0 is, $θ_{0} = + 1.0 rad .$

Step 2: Determine the concept of angular acceleration and displacement

Find the angular velocity by taking time integral of angular acceleration and angular position by taking the time integral of angular velocity.

$ω = \int^{} α dt$
$θ = \int^{} ω dt$

Step 3: (a) Calculate angular velocity of wheel

Angular velocity is a time integral of angular acceleration. Hence,

$\begin{matrix} ω = \int^{} α dt \\ = \int^{} (6 .0 t^{4} - 4 .0 t^{2}) dt \\ = 6 .0 (\frac{t^{5}}{5}) - (4 .0) (\frac{t^{3}}{3}) + ω_{0} \\ = 1 .2 t^{5} - 1 .33 t^{3} + 2 .0 \end{matrix}$

Angular velocity of the wheel is $ω = 1.2 t^{5} - 1.33 t^{3} + 2.0$ .

Step 4: (b) Calculate angular position of wheel.

Angular position is a time integral of angular velocity. Hence:

$\begin{matrix} θ = \int^{} ω dt \\ = \int^{} (1.2 t^{5} - 1.33 t^{3} + 2.0) dt \\ = 1.2 (\frac{t^{6}}{6}) - 1.33 (\frac{t^{4}}{4}) + 2.0 t + θ_{0} \\ = 0.2 t^{6} - 0.33 t^{4} + 2.0 t + 1.0 \end{matrix}$

Therefore, angular position of the wheel is $= 0.2 t^{6} - 0.33 t^{4} + 2.0 t + 1.0$

Angular speed and angular position of an object is calculated from its angular acceleration.

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Listing the given quantities

Step 2: Determine the concept of angular acceleration and displacement

Step 3: (a) Calculate angular velocity of wheel

Step 4: (b) Calculate angular position of wheel.

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Listing the given quantities

Step 2: Determine the concept of angular acceleration and displacement

Step 3: (a) Calculate angular velocity of wheel

Step 4: (b) Calculate angular position of wheel.

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Famous Physicists

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics