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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. At \(t=0,\) 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) 128.0 rad/s, (d) 32.0 rad/s虏, (e) No, angular acceleration is not constant.

Step by step solution

01

Find Angular Position at t = 0

Substitute \(t = 0\) into the equation for angular position \(\theta(t) = 2.0 + 4.0t^2 + 2.0t^3\).\[\theta(0) = 2.0 + 4.0(0)^2 + 2.0(0)^3 = 2.0\]So, at \(t = 0\), the angular position is \(2.0\) radians.
02

Determine Angular Velocity Expression

The angular velocity \(\omega(t)\) is the derivative of the angular position \(\theta(t)\). Differentiate \(\theta(t) = 2.0 + 4.0t^2 + 2.0t^3\) with respect to \(t\).\[\omega(t) = \frac{d\theta}{dt} = 2(4.0)t + 3(2.0)t^2 = 8.0t + 6.0t^2\]This is the expression for angular velocity as a function of \(t\).
03

Calculate Angular Velocity at t = 0

Substitute \(t = 0\) into the angular velocity expression \(\omega(t) = 8.0t + 6.0t^2\).\[\omega(0) = 8.0(0) + 6.0(0)^2 = 0\]At \(t = 0\), the angular velocity is 0 rad/s.
04

Find Angular Velocity at t = 4.0 s

Insert \(t = 4.0\) into the angular velocity expression \(\omega(t)\).\[\omega(4.0) = 8.0(4.0) + 6.0(4.0)^2 = 32.0 + 96.0 = 128.0\]So, at \(t = 4.0\) seconds, the angular velocity is \(128.0\) rad/s.
05

Determine Angular Acceleration Expression

Angular acceleration \(\alpha(t)\) is the derivative of the angular velocity \(\omega(t)\). Differentiate \(\omega(t) = 8.0t + 6.0t^2\) with respect to \(t\).\[\alpha(t) = \frac{d\omega}{dt} = 8.0 + 2(6.0)t = 8.0 + 12.0t\]This expression gives angular acceleration as a function of \(t\).
06

Calculate Angular Acceleration at t = 2.0 s

Substitute \(t = 2.0\) into the angular acceleration formula \(\alpha(t) = 8.0 + 12.0t\).\[\alpha(2.0) = 8.0 + 12.0(2.0) = 8.0 + 24.0 = 32.0\]Thus, at \(t = 2.0\) seconds, the angular acceleration is \(32.0\) rad/s虏.
07

Examine if Angular Acceleration is Constant

The angular acceleration expression is \(\alpha(t) = 8.0 + 12.0t\), which depends on time \(t\). Since the acceleration changes with \(t\), it is not constant.

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Key Concepts

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

Angular Position
The angular position of an object describes its orientation relative to a reference direction as it moves in a circular path. In this context, it is given by the equation \[\theta(t) = 2.0 + 4.0t^2 + 2.0t^3\]. This expression shows how the angular position \(\theta\) changes over time \(t\). To find the position at a specific moment, such as \(t = 0\), we substitute this value into the equation:
  • \(\theta(0) = 2.0 + 4.0 \, (0)^2 + 2.0 \, (0)^3 = 2.0\) radians
This means that at the start, the object is located at 2.0 radians along its circular path. Understanding this concept helps us analyze rotational motion efficiently.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or spins around a central point. It is the rate of change of the angular position with respect to time. Mathematically, angular velocity \(\omega\) is the derivative of the angular position \(\theta\) with respect to time \(t\):
  • \(\omega(t) = \frac{d\theta}{dt} = 8.0t + 6.0t^2\)
This equation tells us how the angular velocity changes over time. For instance, at \(t = 0\), the calculation is as follows:
  • \(\omega(0) = 8.0 \, (0) + 6.0 \, (0)^2 = 0\) rad/s
And at \(t = 4.0\) seconds, it becomes:
  • \(\omega(4.0) = 8.0 \, (4.0) + 6.0 \, (4.0)^2 = 128.0\) rad/s
These calculations demonstrate the concept of angular velocity and how it is derived from the change in angular position over time.
Angular Acceleration
Angular acceleration quantifies the change in angular velocity over time, indicating how quickly a rotating object speeds up or slows down. It is determined by taking the derivative of the angular velocity with respect to time \(t\):
  • \(\alpha(t) = \frac{d\omega}{dt} = 8.0 + 12.0t\)
This formula shows that the angular acceleration is dependent on time, meaning it's not constant. To find the acceleration at specific times, substitute the time into the equation:
  • At \(t = 2.0\), \(\alpha(2.0) = 8.0 + 12.0 \, (2.0) = 32.0\) rad/s虏
This provides insight into how the rotational speed of the object is changing at different times. Understanding the nature of angular acceleration is crucial as it affects motion dynamics and stability in rotating systems.
Time Dependence
Time dependence in angular kinematics is essential as it explains how the motion characteristics of rotating objects evolve. The equations for angular position, velocity, and acceleration all include time \(t\) as a variable, which indicates change over time.
  • Angular position: \(\theta(t) = 2.0 + 4.0t^2 + 2.0t^3\)
  • Angular velocity: \(\omega(t) = 8.0t + 6.0t^2\)
  • Angular acceleration: \(\alpha(t) = 8.0 + 12.0t\)
These dependencies illustrate how each parameter varies, showing that both angular velocity and acceleration increase with time in this scenario. This understanding helps predict how the system will behave in future states, enabling effective modeling and control. The non-constant nature of these values emphasizes the need to consider time effects in analyzing rotational motion.

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Most popular questions from this chapter

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 10 rev/s; 60 revolutions later, its angular speed is \(15 \mathrm{rev} / \mathrm{s} .\) Calculate \((\mathrm{a})\) the angular acceleration, \((\mathrm{b})\) the time required to complete the 60 revolutions, (c) the time required to reach the 10 rev \(/\) s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 10 rev/s angular speed.

Two uniform solid spheres have the same mass of \(1.65 \mathrm{~kg}\), but one has a radius of \(0.226 \mathrm{~m}\) and the other has a radius of \(0.854 \mathrm{~m}\). Each can rotate about an axis through its center. (a) What is the magnitude \(\tau\) of the torque required to bring the smaller sphere from rest to an angular speed of \(317 \mathrm{rad} / \mathrm{s}\) in \(15.5 \mathrm{~s} ?\) (b) What is the magnitude \(F\) of the force that must be applied tangentially at the sphere's equator to give that torque? What are the corresponding values of (c) \(\tau\) and (d) \(F\) for the larger sphere?

Cheetahs running at top speed have been reported at an astounding \(114 \mathrm{~km} / \mathrm{h}\) (about \(71 \mathrm{mi} / \mathrm{h})\) by observers driving alongside the animals. Imagine trying to measure a cheetah's speed by keeping your vehicle abreast of the animal while also glancing at your speedometer, which is registering \(114 \mathrm{~km} / \mathrm{h}\). You keep the vehicle a constant \(8.0 \mathrm{~m}\) from the cheetah, but the noise of the vehicle causes the cheetah to continuously veer away from you along a circular path of radius \(92 \mathrm{~m}\). Thus, you travel along a circular path of radius \(100 \mathrm{~m} .\) (a) What is the angular speed of you and the cheetah around the circular paths? (b) What is the linear speed of the cheetah along its path? (If you did not account for the circular motion, you would conclude erroneously that the cheetah's speed is \(114 \mathrm{~km} / \mathrm{h},\) and that type of error was apparently made in the published reports.)

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 uniform cylinder of radius \(10 \mathrm{~cm}\) and mass \(20 \mathrm{~kg}\) is mounted so as to rotate freely about a horizontal axis that is parallel to and \(5.0 \mathrm{~cm}\) from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?
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