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The angular position of a point on the rim of a rotating wheel is given by \(\theta=4.0 t-3.0 t^{2}+t^{3}\), where \(\theta\) is in radians and \(t\) is in seconds. What are the angular velocities at (a) \(t=2.0 \mathrm{~s}\) and \((\mathrm{b}) t=4.0 \mathrm{~s} ?\) (c) What is the average angular acceleration for the time interval that begins at \(t=2.0 \mathrm{~s}\) and ends at \(t=4.0 \mathrm{~s}\) ? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?

Short Answer

Expert verified
(a) 4.0 rad/s, (b) 28.0 rad/s, (c) 12.0 rad/s², (d) 6.0 rad/s², (e) 18.0 rad/s².

Step by step solution

01

Find the expression for angular velocity

The angular velocity is the first derivative of angular position \(\theta\) with respect to time \(t\). Given \(\theta = 4.0t - 3.0t^2 + t^3\), find \(\omega(t) = \frac{d\theta}{dt}\). Differentiate: \[ \omega(t) = \frac{d}{dt}(4.0t - 3.0t^2 + t^3) = 4.0 - 6.0t + 3t^2 \]
02

Calculate angular velocity at t = 2 s

Use the expression for angular velocity from Step 1: \[ \omega(2) = 4.0 - 6.0 \times 2 + 3 \times 2^2 \]\[ \omega(2) = 4.0 - 12.0 + 12.0 = 4.0 \] So, the angular velocity at \(t = 2.0\) s is \(4.0\) radians per second.
03

Calculate angular velocity at t = 4 s

Similarly, use the expression for \(\omega(t)\) to find: \[ \omega(4) = 4.0 - 6.0 \times 4 + 3 \times 4^2 \] \[ \omega(4) = 4.0 - 24.0 + 48.0 = 28.0 \] Thus, the angular velocity at \(t = 4.0\) s is \(28.0\) radians per second.
04

Determine average angular acceleration

Angular acceleration is the change in angular velocity over a time period. Calculate it from \(t = 2.0\) s to \(t = 4.0\) s:\[ \text{Average angular acceleration} = \frac{\omega(4) - \omega(2)}{4 - 2} \] \[ = \frac{28.0 - 4.0}{2} = 12.0 \] So, the average angular acceleration is \(12.0\) radians per second squared.
05

Find the expression for angular acceleration

Angular acceleration is the first derivative of angular velocity with respect to time. Start from \(\omega(t) = 4.0 - 6.0t + 3t^2\) and differentiate:\[ \alpha(t) = \frac{d\omega}{dt} = -6.0 + 6t \]
06

Calculate instantaneous angular acceleration at t = 2 s

Using \(\alpha(t)\) from Step 5:\[ \alpha(2) = -6.0 + 6 \times 2 = 6.0 \] Thus, the instantaneous angular acceleration at \(t = 2.0\) s is \(6.0\) radians per second squared.
07

Calculate instantaneous angular acceleration at t = 4 s

Using \(\alpha(t)\) from Step 5:\[ \alpha(4) = -6.0 + 6 \times 4 = 18.0 \] Therefore, the instantaneous angular acceleration at \(t = 4.0\) s is \(18.0\) radians per second squared.

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Key Concepts

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

Angular Velocity
Angular velocity, often symbolized by \( \omega \), is a measure of how fast an object rotates. It tells us the number of radians an object covers per second. In simpler terms, it describes the speed of rotation. To understand angular velocity, think of a spinning wheel on a bike. How fast that wheel spins is its angular velocity.

To find angular velocity from angular position, we need to perform differentiation. In mathematical terms, if a point's angular position \( \theta \) is described by the equation \( \theta = 4.0t - 3.0t^2 + t^3 \), then the angular velocity \( \omega(t) \) is determined by differentiating \( \theta \) with respect to time \( t \):
  • \( \omega(t) = \frac{d\theta}{dt} \)
  • This gives us \( \omega(t) = 4.0 - 6.0t + 3t^2 \).
By substituting specific time values into this expression, you can calculate the angular velocities at different points in time, helping you to understand how the wheel's speed changes.
Angular Acceleration
Angular acceleration, denoted \( \alpha \), represents how quickly an object's angular velocity changes. It's similar to acceleration in linear motion but applies to rotating systems. Angular acceleration can be thought of as how quickly a wheel speeds up or slows down its spinning.

To find angular acceleration, we again use the concept of differentiation. First, we obtain the expression for angular acceleration by differentiating the angular velocity \( \omega(t) = 4.0 - 6.0t + 3t^2 \). Hence:
  • \( \alpha(t) = \frac{d\omega}{dt} = -6.0 + 6t \).
This expression allows us to find the instantaneous acceleration at any point in time. For example, at \( t = 2 \, \text{s} \), substituting this value gives the acceleration at that moment. Similarly, for \( t = 4 \, \text{s} \), you plug it in to find the acceleration at the end of the observed time interval.

Average angular acceleration over a time period tells us the overall rate of change of angular velocity. Calculated by the formula \( \frac{\omega(4) - \omega(2)}{4 - 2} \), it provides a broad view of how much the spinning speed changed over that time frame.
Differentiation
Differentiation is a crucial mathematical tool that helps us understand how a function changes. In the context of rotational motion, it's used to find velocities and accelerations based on angular position functions.

Imagine the angular position of a point on a rotating wheel given by \( \theta = 4.0 t - 3.0 t^{2} + t^{3} \). The process of differentiation involves finding how this position changes over time, leading to new equations that describe changes more effectively:
  • Differentiating \( \theta \) with respect to \( t \) gives the angular velocity \( \omega = \frac{d\theta}{dt} \).
  • Further differentiation of \( \omega \) with respect to \( t \) gives the angular acceleration \( \alpha = \frac{d\omega}{dt} \).
Differentiation transforms problems of movement into more tangible, computable forms, allowing us to predict future dynamic behaviors based on current conditions.
Rotational Kinematics
Rotational kinematics is the study of motion in rotating objects without considering the forces causing the motion. It involves quantities like angular position, angular velocity, and angular acceleration, describing how a wheel or disk moves over time.

This field uses equations and concepts similar to linear kinematics, but they apply to angles, radians, and rotational aspects of an object. Consider the key rotational kinematic equations:
  • \( \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 \)
  • \( \omega = \omega_0 + \alpha t \)
  • \( \omega^2 = \omega_0^2 + 2\alpha (\theta - \theta_0) \)
These equations connect time, angles, velocity, and acceleration, offering a framework to predict how objects will rotate. Understanding these allows you to solve problems relating to how wheels, gears, and other rotational systems operate and change dynamically over time.

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Most popular questions from this chapter

A small ball of mass \(0.75 \mathrm{~kg}\) is attached to one end of a \(1.25-\mathrm{m}\) -long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is \(30^{\circ}\) from the vertical, what is the magnitude of the gravitational torque calculated about the pivot?

The flywheel of an engine is rotating at \(25.0 \mathrm{rad} / \mathrm{s}\). When the engine is turned off, the flywheel slows at a constant rate and stops in \(20.0 \mathrm{~s}\). Calculate (a) the angular acceleration of the flywheel, (b) the angle through which the flywheel rotates in stopping, and (c) the number of revolutions made by the flywheel in stopping.

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The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in \(2.2 \mathrm{~h}\). (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 75 revimin, what is the tangential component of the linear acceleration of a flywheel particle that is \(50 \mathrm{~cm}\) from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period \(T\) of rotation is found by measuring the time between pulses. The pulsar in the Crab nebula has a period of rotation of \(T=0.033 \mathrm{~s}\) that is increasing at the rate of \(1.26 \times 10^{-5} \mathrm{~s} / \mathrm{y} .\) (a) What is the pulsar's angular acceleration \(\alpha ?\) (b) If \(\alpha\) is constant, how many years from now will the pulsar stop rotating? (c) The pulsar originated in a supernova explosion seen in the year 1054 . Assuming constant \(\alpha\), find the initial \(T\).

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