91Ó°ÊÓ

\(9.88^{\circ}\) DATA You are analyzing the motion of a large flywheel that has radius \(0.800 \mathrm{~m}\). In one test run, the wheel starts from rest and turns in a horizontal plane with constant angular acceleration. An accelerometer on the rim of the flywheel measures the magnitude of the resultant acceleration \(a\) of a point on the rim of the flywheel as a function of the angle \(\theta-\theta_{0}\) through which the wheel has turned. You collect these results: $$ \begin{array}{l|cccccccc} \boldsymbol{\theta}-\boldsymbol{\theta}_{\mathbf{0}}(\mathbf{r a d}) & 0.50 & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 3.50 & 4.00 \\ \hline \boldsymbol{a}\left(\mathbf{m} / \mathbf{s}^{\mathbf{2}}\right) & 0.678 & 1.07 & 1.52 & 1.98 & 2.45 & 2.92 & 3.39 & 3.87 \end{array} $$ Construct a graph of \(a^{2}\left(\right.\) in \(\left.\mathrm{m}^{2} / \mathrm{s}^{4}\right)\) versus \(\left(\theta-\theta_{0}\right)^{2}\) (in rad \(^{2}\) ). (a) What are the slope and \(y\) -intercept of the straight line that gives the best fit to the data? (b) Use the slope from part (a) to find the angular acceleration of the flywheel. (c) What is the linear speed of a point on the rim of the flywheel when the wheel has turned through an angle of \(135^{\circ} ?\) (d) When the flywheel has turned through an angle of \(90.0^{\circ},\) what is the angle between the linear velocity of a point on its rim and the resultant acceleration of that point?

Step by step solution

01

Determining the Slope and Intercept

To find the slope and \(y\)-intercept a graph of \(a^{2}\) in \(m^{2} / s^{4}\) versus \((\theta-\theta_{0})^{2}\) in \(rad^{2}\) needs to be sketched. From the linear fit, one can identify the slope and the \(y\)-intercept of this line.

02

Calculating Angular Acceleration

Once the slope is known, one can calculate angular acceleration \(\alpha\) using the formula \( \alpha = \frac{ \text{slope} }{r^2} \), where \(r\) is the radius of the flywheel given as \( 0.800 m \).

03

Computing Linear Speed

For part (c), to find the linear speed of a point on the flywheel at a given angle, one needs to use the formula \( v = r * \omega \). The angular velocity \( \omega \) can be calculated using the formula \( \omega = \sqrt{2 * \alpha * \theta } \), where \( \theta \) is the angle given in radians.

04

Determining the Angle Between Velocity and Acceleration

In the last part, the angle between the linear velocity of a point on the flywheel and the resultant acceleration of that point when the flywheel turned through an angle of \(90.0^{\circ}\) needs to be calculated. Since at this point, we have the centripetal and tangential acceleration making a right angle, we can use the Pythagorean theorem to find the resultant acceleration and then find the angle

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Key Concepts

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

Flywheel Motion

A flywheel is a heavy wheel attached to a rotating shaft, designed to store rotational energy. Understanding flywheel motion involves analyzing how rotational dynamics influence both the angular and linear aspects of motion.
Flywheels operate based on principles of angular acceleration, where they begin from rest and gradually pick up speed due to a torque applied. The motion in a horizontal plane, as described in the exercise, implies that gravitational forces are not influencing the perpendicular direction, focusing solely on rotational forces.
In the exercise, the flywheel starts from rest, implying the initial angular velocity is zero. The term "constant angular acceleration" signals that the rate of change of angular velocity is uniform. This also means that the acceleration is related to the angle through which the flywheel turns, as captured by the measured data.

Linear Speed

Linear speed on the rim of a flywheel is directly connected to the angular velocity. As the flywheel rotates, a point on its edge moves with a speed dependent on both the radius of the wheel and its angular velocity.
Linear speed (\( v \)) can be determined using the relationship between angular velocity (\( \omega \)) and radius (\( r \)): \( v = r \cdot \omega \).
For the flywheel, the angular velocity is calculated from the angular acceleration once the angle has been specified in radians. The exercise presents a task to compute linear speed at an angle of \( 135^{\circ} \). Converting degrees to radians is crucial here since angular measurements in physics typically rely on radians for accuracy and standardization.
Using the given angular acceleration, students calculate angular velocity with the formula \( \omega = \sqrt{2 \cdot \alpha \cdot \theta} \), crucial for determining the linear speed on the rim.

Resultant Acceleration

Resultant acceleration on the rim of a flywheel encompasses both tangential and centripetal components. These components originate from the motion specifics and are vital in understanding full kinematic behaviors.
- **Tangential Acceleration**: This arises from the angular acceleration and results in a change in the magnitude of the velocity of a point on the edge of the flywheel.- **Centripetal Acceleration**: Exists due to the rotation's inherent circular path, directing towards the center of the circle and keeping the point moving in this path.
The exercise asks about the angle between linear velocity and resultant acceleration when the flywheel turns through \( 90^{\circ} \). At this juncture, tangential and centripetal accelerations form right-angle components. By employing vector addition principles and the Pythagorean theorem, one can deduce the resultant acceleration and determine the significant relative angle to linear velocity. This exact understanding helps grasp the mechanics dictating rotational dynamics efficiently.