A Ferris wheel,rotating initially at an angular speed of 0.500 rad/s,accelerates over a 10.4-s interval at a rate of 0.040 0 rad/s².What angular displacement does the Ferris wheel undergo in this 10.4-s interval? A)7.4 rad B)0.21 rad C)2.2 rad D)0.0019 rad

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Quizplus · Accepted Answer

The angular displacement of the Ferris wheel can be calculated using the formula for angular displacement under constant angular acceleration: $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$, where $\omega_0$ is the initial angular speed, $\alpha$ is the angular acceleration, and $t$ is the time interval. Substituting the given values: $\theta = (0.500\, \text{rad/s}) \times 10.4\, \text{s} + \frac{1}{2} \times (0.040\, \text{rad/s}^2) \times (10.4\, \text{s})^2 = 5.2 + 2.2 = 7.4\, \text{rad}$.

A Ferris Wheel,rotating Initially at an Angular Speed of 0