91Ó°ÊÓ

A Flywheel Has a Rotational Velocity At \(t_{1}=0\), a flywheel has a rotational velocity of \(4.7 \mathrm{rad} / \mathrm{s}\), a rotational acceleration of \(-0.25\) \(\mathrm{rad} / \mathrm{s}^{2}\), and a reference line at \(\theta_{1}=0 .\) (a) Through what maximum angle \(\theta^{\max }\) will the reference line turn in the positive direction? For what length of time will the reference line turn in the positive direction? At what times will the reference line be at (b) \(\theta=\frac{1}{2} \theta^{\max }\) and (c) \(\theta=-10.5\) rad (consider both positive and negative values of \(t\) )? (d) Graph \(\theta\) versus \(t\), and indicate the answers to \((\mathrm{a}),(\mathrm{b})\), and \((\mathrm{c})\) on the graph.

Step by step solution

01

- Identify known variables

Given variables: Initial angular velocity (\(\theta_{1}) = 0 \) rad, initial angular velocity \(\theta_{1}=0 \), angular acceleration \(\alpha = -0.25 \) rad/s\textsuperscript{2}. Initial rotation is \(\omega_{0}= 4.7 \) rad/s.

02

- Find Maximum Angle

To find the maximum angle, use the equation \[ \omega^2 = \omega_{0}^2 + 2\alpha\theta \] Since the max rotational speed will be zero, \(\omega = 0\): \[0 = (4.7)^2 + 2(-0.25)\theta^{max} \] Simplify to find \(\theta^{max}\): \(\theta^{max} = \frac{(4.7)^2}{2 \times 0.25} = 44.18 \) rad.

03

- Find Time for Maximum Angle

Use the equation \(\omega = \omega_{0} + \alpha\tau\) Solve for \(\tau\):\(0 = 4.7 + (-0.25)\tau \rightarrow \tau = \frac{4.7}{0.25} = 18.8 \) s.

04

- Find Time for Half Maximum Angle

For time at \(\theta= \frac{1}{2}\theta^{max}\): Use angle equation\(\theta = \omega_{0}t + 0.5\alpha t^2\): \ \(22.09 = 4.7\tau + (-0.125)\tau^2\) Solving this quadratic equation yields two values:\t \ \(\tau=4.439\) sor \(\tau = 99\) s.

05

- Find Time for Angle \(\theta = -10.5 \text{ rad} \)

Use angle equation again for negative angle: \(\theta = -10.5 \text{ rad}\)\(\theta = \omega_{0}t + \frac{1}{2}\alpha t^2\)\t Solving \(-10.5 = 4.7* \tau + 0.5(-0.25)(\tau^2)\).Solve quadratic yields \(\tau ~ 8.539\) s also fits \(\tau ~ -8.964\) s requires.

06

- Graph \( \theta \text{ Versus }\t )

Plotting results:For \( \theta^{max} \rightarrow 44.18\) rad at 18.8s.\ \theta=\ \frac{1}{2}\theta^{max} \rightarrow 22.09\rad at 4.439s\, 99 s\ \theta=-10.5 rad\ -8.539s\,8.964s at which.

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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 measure of how quickly an object rotates around a specific axis. It is often denoted by the Greek letter \(\theta\), and its standard unit is radians per second (\text{rad/s}). In this exercise, we start with an initial angular velocity of \(\theta_{1} = 0\text{ rad}\) and initial rotational velocity \(\theta_{1} = 4.7\, \text{rad/s}\). Angular velocity helps us determine how quickly a flywheel or other rotating object is spinning at any given moment. Understanding angular velocity is essential when trying to determine the rotational motion and behavior over time.

rotational acceleration

Rotational acceleration, denoted as \(\beta\), describes how an object's angular velocity changes over time. The units for rotational acceleration are usually \(\text{rad/s}^2\). In this problem, the rotational acceleration of the flywheel is given as \(\theta = -0.25\text{ rad/s}^2\). The negative sign indicates that the object is decelerating or rotating slower over time. To find the maximum angle the reference line will turn, we use the equation \[\theta^2 = \theta_{0}^2 + 2\alpha\theta \] By setting \(\theta = 0\) (since we want to find the point where the rotational speed is zero), we simplify and solve giving \(\theta^{max} = \frac{(4.7)^2}{2 \cdot 0.25} = 44.18 \text{rad}\).

solving quadratic equations

Quadratic equations are essential for solving problems involving rotational kinematics. In this exercise, we need to solve the equations: \[ 22.09 = 4.7 \omega + (-0.125)\omega^2 \] to determine when our angle is at half the maximum angle. Similarly, we have another quadratic equation to solve for \(\theta = -10.5 \text{ rad}\). These solutions provide the time values that indicate when the object reaches certain rotational states. Remember, solutions to quadratic equations can be obtained using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) This formula provides us with the necessary time values (\( \tau = 4.439\text{ s}\) or \( \tau = 99 \text{ s}\)) to determine these specific angular positions.

graphing rotational motion

Graphing rotational motion helps us visualize how the angle \(\theta\) changes over time (\(t\)). In this problem, we plot the given data points to see how the flywheel rotates. Important points to mark on the graph are:

• Maximum angle \(\theta^{max} = 44.18 \text{ rad}\) at \(18.8 \text{ s}\)
• Half maximum angle at \(4.439 \text{ s}\) and \(99 \text{ s}\)
• Angle \(\theta = -10.5 \text{ rad}\) at times around \(8.539 \text{ s}\) and \(-8.964 \text{ s}\)

By plotting these data points and understanding the equations behind them, students can better understand the dynamics of rotational motion and interpret such graphs effectively.