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Chapter 10: Problem 8

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

Short Answer

Expert verified
(a) \( \omega(t) = \frac{6.0}{5}t^5 - \frac{4.0}{3}t^3 + 2.0 \) (b) \( \theta(t) = \frac{1.0}{5}t^6 - \frac{1.0}{3}t^4 + 2.0t + 1.0 \)

Step by step solution

01

Integration to find Angular Velocity

To find the angular velocity \( \omega(t) \), we integrate the given angular acceleration \( \alpha(t) = 6.0t^4 - 4.0t^2 \). The angular velocity is given by \( \omega(t) = \int \alpha(t) \, dt \). Integrating,\[ \omega(t) = \int (6.0t^4 - 4.0t^2) \, dt = \left( \frac{6.0}{5}t^5 - \frac{4.0}{3}t^3 \right) + C_1 \]where \( C_1 \) is the constant of integration.
02

Determine Constant for Angular Velocity

We use the initial condition that at \( t=0 \), \( \omega = +2.0 \) rad/s. Substitute this into the expression for \( \omega(t) \) to solve for \( C_1 \):\[ 2.0 = \left( \frac{6.0}{5}(0)^5 - \frac{4.0}{3}(0)^3 \right) + C_1 \]Thus, \( C_1 = 2.0 \). Therefore, the angular velocity is \[ \omega(t) = \frac{6.0}{5}t^5 - \frac{4.0}{3}t^3 + 2.0 \]
03

Integration to find Angular Position

To find the angular position \( \theta(t) \), integrate the angular velocity \( \omega(t) \).\[ \theta(t) = \int \omega(t) \, dt = \int \left( \frac{6.0}{5}t^5 - \frac{4.0}{3}t^3 + 2.0 \right) \, dt \]Performing the integration, \[ \theta(t) = \left( \frac{6.0}{30}t^6 - \frac{4.0}{12}t^4 + 2.0t \right) + C_2 \]which simplifies to \[ \theta(t) = \frac{1.0}{5}t^6 - \frac{1.0}{3}t^4 + 2.0t + C_2 \]
04

Determine Constant for Angular Position

We use the initial condition that at \( t=0 \), \( \theta = +1.0 \) rad. Substitute into the expression for \( \theta(t) \) to solve for \( C_2 \):\[ 1.0 = \left( \frac{1.0}{5}(0)^6 - \frac{1.0}{3}(0)^4 + 2.0(0) \right) + C_2 \]Thus, \( C_2 = 1.0 \). Therefore, the angular position is \[ \theta(t) = \frac{1.0}{5}t^6 - \frac{1.0}{3}t^4 + 2.0t + 1.0 \]

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Key Concepts

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

Angular Acceleration
Angular acceleration is a measure of how fast the angular velocity is changing with time. You can think of it as the equivalent of linear acceleration in rotational motion. It tells us how quickly something is speeding up or slowing down as it rotates.
  • Given by \(\alpha(t)\), it is the derivative of angular velocity with respect to time.
  • In our exercise, it is expressed as \(\alpha(t) = 6.0t^4 - 4.0t^2\), showing how the angular acceleration can change dramatically over time.
  • Understanding \(\alpha\) helps in predicting how rotational systems will behave over time.
In mechanical systems, different pieces might experience varied angular accelerations depending on factors such as applied torque and rotational inertia. Knowing \(\alpha(t)\) helps us understand not just the system's performance but also the stress it might undergo.
Angular Velocity
Angular velocity is the rate of change of an object's angular position with respect to time. It tells us how quickly something is rotating.
  • The angular velocity \(\omega(t)\) can be found by integrating the expression for angular acceleration.
  • For the problem at hand, we find \(\omega(t) = \frac{6.0}{5}t^5 - \frac{4.0}{3}t^3 + 2.0\) rad/s after integrating the given acceleration.
Important things to note about angular velocity:
  • It can have both positive and negative values depending on the direction of rotation.
  • Steady angular velocity implies constant speed of rotation, while a change indicates acceleration or deceleration.
In many scenarios, understanding angular velocity is crucial as it can affect how a system behaves dynamically during its operation.
Angular Position
Angular position is a vital concept that tells us the orientation or the angle of an object in rotational motion at any given time.
  • Represented as \(\theta(t)\), it can be found by integrating angular velocity \(\omega(t)\).
  • By doing so for our exercise, we obtain \(\theta(t) = \frac{1.0}{5}t^6 - \frac{1.0}{3}t^4 + 2.0t + 1.0\) rad.
  • The initial angular position \(\theta_0\) gives us the reference point from which all other positions are measured.
Angular position is crucial when calculating the trajectory or predicting the future positions of rotating objects. Effective understanding ensures accurate modeling of systems where precise rotational landing or orientation is critical, such as in robotics or satellite motion.

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Most popular questions from this chapter

A uniform helicopter rotor blade is \(7.80 \mathrm{~m}\) long, has a mass of \(110 \mathrm{~kg},\) and is attached to the rotor axle by a single bolt. (a) What is the magnitude of the force on the bolt from the axle when the rotor is turning at 320 rev/min? (Hint: For this calculation the blade can be considered to be a point mass at its center of mass. Why?) (b) Calculate the torque that must be applied to the rotor to bring it to full speed from rest in \(6.70 \mathrm{~s}\). Ignore air resistance. (The blade cannot be considered to be a point mass for this calculation. Why not? Assume the mass distribution of a uniform thin rod.) (c) How much work does the torque do on the blade in order for the blade to reach a speed of 320 rev/min?

A drum rotates around its central axis at an angular velocity of \(12.60 \mathrm{rad} / \mathrm{s} .\) If the drum then slows at a constant rate of \(4.20 \mathrm{rad} / \mathrm{s}^{2}\) (a) how much time does it take and (b) through what angle does it rotate in coming to rest?

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In \(5.0 \mathrm{~s},\) it rotates 25 rad. During that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) What is the instantaneous angular velocity of the disk at the end of the \(5.0 \mathrm{~s} ?\) (d) With the angular acceleration unchanged, through what additional angle will the disk turn during the next \(5.0 \mathrm{~s} ?\)

A vinyl record on a turntable rotates at \(33 \frac{1}{3}\) rev/min. (a) What is its angular speed in radians per second? What is the linear speed of a point on the record (b) \(15 \mathrm{~cm}\) and (c) \(7.4 \mathrm{~cm}\) from the turntable axis?

A pulley, with a rotational inertia of \(1.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its axle and a radius of \(10 \mathrm{~cm},\) is acted on by a force applied tangentially at its rim. The force magnitude varies in time as \(F=0.50 t+0.30 t^{2},\) with \(F\) in newtons and \(t\) in seconds. The pulley is initially at rest. At \(t=3.0 \mathrm{~s}\) what are its (a) angular acceleration and (b) angular speed?
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