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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 \(+1.0\) 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) \( \omega(t) = 1.2t^5 - \frac{4.0}{3}t^3 + 2.0 \) rad/s; (b) \( \theta(t) = 0.2t^6 - \frac{t^4}{3} + 2.0t + 1.0 \) rad.

Step by step solution

01

Integrate Angular Acceleration to Find Angular Velocity

The angular acceleration \[ \alpha = \frac{d\omega}{dt} = 6.0t^4 - 4.0t^2 \]can be integrated to find the angular velocity \( \omega \):\[ \omega(t) = \int (6.0t^4 - 4.0t^2) \, dt \]Performing the integration:\[ \omega(t) = (6.0) \cdot \frac{t^5}{5} - (4.0) \cdot \frac{t^3}{3} + C_1 \]\[ \omega(t) = 1.2t^5 - \frac{4.0}{3}t^3 + C_1 \]Given the initial condition at \( t=0 \), \( \omega(0) = 2.0 \, \text{rad/s} \):\[ 2.0 = 1.2 \cdot 0^5 - \frac{4.0}{3} \cdot 0^3 + C_1 \]\[ C_1 = 2.0 \]So we have:\[ \omega(t) = 1.2t^5 - \frac{4.0}{3}t^3 + 2.0 \]
02

Integrate Angular Velocity to Find Angular Position

The angular velocity \[ \omega = \frac{d\theta}{dt} = 1.2t^5 - \frac{4.0}{3}t^3 + 2.0 \]can be integrated to find the angular position \( \theta \):\[ \theta(t) = \int \left(1.2t^5 - \frac{4.0}{3}t^3 + 2.0\right) \, dt \]Performing the integration:\[ \theta(t) = (1.2) \cdot \frac{t^6}{6} - \frac{4.0}{3} \cdot \frac{t^4}{4} + 2.0t + C_2 \]\[ \theta(t) = 0.2t^6 - \frac{t^4}{3} + 2.0t + C_2 \]Given the initial condition at \( t=0 \), \( \theta(0) = 1.0 \, \text{rad} \):\[ 1.0 = 0.2 \cdot 0^6 - \frac{0^4}{3} + 2.0 \cdot 0 + C_2 \]\[ C_2 = 1.0 \]So we have:\[ \theta(t) = 0.2t^6 - \frac{t^4}{3} + 2.0t + 1.0 \]

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Key Concepts

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

Angular Acceleration
Angular acceleration describes how the angular velocity of an object changes with time. It's an important concept in rotational motion that helps us understand how quickly a rotating object speeds up or slows down. In this exercise, we were given an expression for angular acceleration as \[ \alpha(t) = 6.0t^4 - 4.0t^2 \]This function shows how acceleration varies over time:
  • The term \(6.0t^4\) indicates that the acceleration increases rapidly with time since \(t^4\) grows faster than lower powers of \(t\).
  • The term \(-4.0t^2\) counteracts some of that acceleration, especially at smaller values of \(t\).
  • The units are radians per second-squared, a common unit when dealing with rotational motion.
Knowing angular acceleration allows us to derive other useful properties of rotational motion, like angular velocity and position. Understanding how these elements interact provides deeper insight into the motion of spinning objects.
Angular Velocity
Angular velocity represents how fast an object rotates. It gives the rate at which the angle changes over time. The original problem states that the angular velocity at time \(t=0\) is \(2.0 \text{ rad/s}\). We derive angular velocity by integrating the angular acceleration:\[ \omega(t) = \int (6.0t^4 - 4.0t^2) \, dt = 1.2t^5 - \frac{4.0}{3}t^3 + C_1 \]Where \(C_1\) is the constant of integration. Using the initial condition \(\omega(0) = 2.0 \text{ rad/s}\), we find:
  • \(C_1 = 2.0\)
Thus, the complete angular velocity expression becomes:\[ \omega(t) = 1.2t^5 - \frac{4.0}{3}t^3 + 2.0 \]This equation tells us how the wheel's rotation speed evolves:
  • Positive terms indicate increasing angular velocity over time.
  • The integration reveals how initial conditions affect the entire motion trajectory.
Angular Position
Angular position indicates the 'angle' or how far a point on the object has rotated from a reference position. Think of it as the endpoint of a journey in rotational terms. Initially, the angular position was given as \(1.0 \text{ rad}\) at \(t=0\).We determine angular position by integrating angular velocity:\[ \theta(t) = \int \left(1.2t^5 - \frac{4.0}{3}t^3 + 2.0\right) \, dt = 0.2t^6 - \frac{t^4}{3} + 2.0t + C_2 \]Applying the initial condition \(\theta(0) = 1.0 \text{ rad}\) allows us to solve for \(C_2\):
  • \(C_2 = 1.0\)
The final expression for the angular position is:\[ \theta(t) = 0.2t^6 - \frac{t^4}{3} + 2.0t + 1.0 \]Key observations:
  • Each term contributes differently to the overall path the wheel takes.
  • Changes in angular position over time correlate directly with how angular velocity changes.
  • Understanding this can help predict future positions based purely on current conditions.

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Most popular questions from this chapter

At \(7: 14\) A.M. on June 30,1908 , a huge explosion occurred above remote central Siberia, at latitude \(61^{\circ} \mathrm{N}\) and longitude \(102^{\circ} \mathrm{E} ;\) the fireball thus created was the brightest flash seen by anyone before nuclear weapons. The Tunguska Event, which according to one chance witness "covered an enormous part of the sky," was probably the explosion of a stony asteroid about 140 \(\mathrm{m}\) wide. (a) Considering only Earth's rotation, determine how much later the asteroid would have had to arrive to put the explosion above Helsinki at longitude \(25^{\circ} \mathrm{E}\). This would have obliterated the city. (b) If the asteroid had, instead, been a metallic asteroid, it could have reached Earth's surface. How much later would such an asteroid have had to arrive to put the impact in the Atlantic Ocean at longitude \(20^{\circ} \mathrm{W} ?\) (The resulting tsunamis would have wiped out coastal civilization on both sides of the Atlantic.)

A flywheel turns through 40 rev as it slows from an angular speed of \(1.5 \mathrm{rad} / \mathrm{s}\) to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular acceleration? (c) How much time is required for it to complete the first 20 of the 40 revolutions?

What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch? Answer in radians per second.

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?

The angular speed of an automobile engine is increased at a constant rate from 1200 rev/min to 3000 rev/min in \(12 \mathrm{~s}\). (a) What is its angular acceleration in revolutions per minute-squared? (b) How many revolutions does the engine make during this 12 sinterval?
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