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At \(t=0\), a flywheel has an angular velocity of \(4.7 \mathrm{rad} / \mathrm{s}, \mathrm{a}\) constant angular acceleration of \(-0.25 \mathrm{rad} / \mathrm{s}^{2}\), and a reference line at \(\theta_{0}=0 .\) (a) Through what maximum angle \(\theta_{\max }\) will the reference line turn in the positive direction? What are the (b) first and (c) second times the reference line will be at \(\theta=\frac{1}{2} \theta_{\max } ?\) At what (d) negative time and (e) positive time will the reference line be at \(\theta=10.5\) rad? (f) Graph \(\theta\) versus \(t\), and indicate the answers to (a) through (e) on the graph.

Short Answer

Expert verified
\( \theta_{\max} = 44.18 \text{ rad} \); first and second times are \( t = 6.0 \text{ s} \) and \( t = 29.48 \text{ s} \); negative and positive times for \( \theta = 10.5 \text{ rad} \) are \( t = -1.1 \text{ s} \) and \( t = 37.7 \text{ s} \). Graph shows the full trajectory.

Step by step solution

01

Determine Maximum Angle (Part a)

To find the maximum angle \( \theta_{\max} \), we first determine when the angular velocity reaches zero. The formula for the final angular velocity is given by \( \omega_f = \omega_i + \alpha t \). Setting \( \omega_f = 0 \) and substituting \( \omega_i = 4.7 \, \text{rad/s} \) and \( \alpha = -0.25 \, \text{rad/s}^2 \), we get \( 0 = 4.7 - 0.25t \). Solving for \( t \) gives \( t = 18.8 \text{ s} \). The maximum angle \( \theta_{\max} \) is found using \( \theta = \theta_0 + \omega_i t + \frac{1}{2} \alpha t^2 \). Substituting the known values: \( \theta_{\max} = 0 + 4.7 \times 18.8 + \frac{1}{2} \times (-0.25) \times (18.8)^2 = 44.18 \, \text{rad} \).
02

Determine Earliest Time for Half Maximum Angle (Part b)

To find the first time \( \theta = \frac{1}{2} \theta_{\max} = 22.09 \, \text{rad} \), use the equation \( \theta = 4.7t + \frac{1}{2}(-0.25)t^2 \). Rewriting, we have \( 0.125t^2 - 4.7t + 22.09 = 0 \). Solving this quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 0.125, b = -4.7, \) and \( c = 22.09 \) results in \( t = 6.0 \text{ s} \) and \( t = 29.48 \text{ s} \). The first time is \( t = 6.0 \text{ s} \).
03

Determine Second Time for Half Maximum Angle (Part c)

From the quadratic solution of \( 0.125t^2 - 4.7t + 22.09 = 0 \), we already found one solution at \( t = 29.48 \text{ s} \). Therefore, the second time \( t = 29.48 \text{ s} \).
04

Determine Negative Time for \( \theta = 10.5 \) rad (Part d)

For \( \theta = 10.5 \, \text{rad} \), use \( 0.125t^2 - 4.7t + 10.5 = 0 \). Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 0.125, b = -4.7, c = 10.5 \), compute to find \( t \). Solving, we get the roots \( t = -1.1 \text{ s} \) and \( t = 37.7 \text{ s} \). The negative time is \( t = -1.1 \text{ s} \).
05

Determine Positive Time for \( \theta = 10.5 \) rad (Part e)

From Step 4, we found the positive time \( t = 37.7 \text{ s} \).
06

Graph \( \theta \) vs. \( t \) and Annotate Answers (Part f)

Plot the function \( \theta(t) = 4.7t + \frac{1}{2}(-0.25)t^2 \). Mark \( \theta_{\max} = 44.18 \text{ rad} \) at \( t = 18.8 \text{ s} \). Indicate \( \theta = 22.09 \text{ rad} \) at \( t = 6.0 \text{ s} \) and \( 29.48 \text{ s} \). Also, mark \( \theta = 10.5 \text{ rad} \) at \( t = -1.1 \text{ s} \) and \( 37.7 \text{ s} \). The parabola opens downward, reaching a maximum at \( 18.8 \text{ s} \).

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Key Concepts

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

Angular Velocity
Angular velocity is a critical concept in angular kinematics, representing how quickly an object rotates about an axis. It’s denoted by the symbol \( \omega \) and usually measured in radians per second (\( \text{rad/s} \)). In this exercise, we start with an initial angular velocity \( \omega_i = 4.7 \, \text{rad/s} \), indicating the speed of rotation at the beginning when \( t = 0 \).
When considering angular motion, just like linear velocity measures the rate of change of linear displacement, angular velocity measures the rate of change of angular displacement. Essentially, it provides insight into how fast or slow a rotating object is spinning.
To calculate angular velocity at any point, we apply the equation \( \omega = \omega_i + \alpha t \), where \( \alpha \) is the angular acceleration. Angular velocity becomes zero when the object stops rotating momentarily before changing direction or coming to a complete rest.
Angular Acceleration
Angular acceleration \( \alpha \) signifies the rate of change of angular velocity over time. It is measured in radians per second squared (\( \text{rad/s}^2 \)). In this exercise, we deal with a constant angular acceleration of \( -0.25 \, \text{rad/s}^2 \). The negative sign of \( \alpha \) implies that the rotation is slowing down over time, reducing the angular velocity.
Angular acceleration helps to determine how the angular velocity changes at different times, influencing both the speed and direction of rotation. A positive value would typically mean speeding up while a negative implies slowing down, as observed here.
This deceleration continues until the angular velocity is zero. At this point, because the acceleration is constant and negative, the flywheel will eventually stop and then start rotating in the opposite direction, emphasizing the importance of the acceleration sign in rotational dynamics.
Quadratic Equation
Quadratic equations frequently arise in angular kinematics due to the quadratic term present in angular displacement equations. In this case, the equation \( 0.125t^2 - 4.7t + 22.09 = 0 \) is derived from the general kinematic formula \( \theta = \theta_0 + \omega_i t + \frac{1}{2} \alpha t^2 \).
Solving this type of equation often requires the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).
In our solution, using the quadratic formula helps find the times when the flywheel's angle becomes half of its maximum value. Understanding how to use and solve quadratic equations is crucial for determining key times and angles in rotational motion.
Reference Angle
The reference angle in angular motion defines the initial position or starting angle of the rotating object. In this exercise, it is indicated by \( \theta_0 = 0 \), meaning the reference line starts at zero radians.
As the flywheel spins, this reference angle acts as a baseline to calculate how much total rotation has occurred over time.
When solving for various times and angles during motion, it's crucial to consider the starting position set by the reference angle. It helps in determining how far or in what direction the reference line has moved from its original location. In problems involving rotational motion, identifying and using the reference angle makes understanding and calculating angular displacements more straightforward.

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Most popular questions from this chapter

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