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Chapter 9: Problem 57

A boy is on a Ferris wheel, which takes him in a vertical circle of radius \(9.0 \mathrm{~m}\) once every \(12.0 \mathrm{~s}\). a) What is the angular speed of the Ferris wheel? b) Suppose the wheel comes to a stop at a uniform rate during one quarter of a revolution. What is the angular acceleration of the wheel during this time? c) Calculate the tangential acceleration of the boy during the time interval described in part (b).

Short Answer

Expert verified
Answer: The angular speed of the Ferris wheel is \(\frac{\pi}{6}\,rad/s\), the angular acceleration during a quarter of a revolution is \(- \frac{\pi}{72} \, rad/s^2\), and the boy's tangential acceleration is \(-\frac{9\pi}{72}\,m/s^2\).

Step by step solution

01

Find the angular speed

To find the angular speed (\(\omega\)) of the Ferris wheel, we can use the formula: \(\omega = \frac{2\pi}{T}\), where \(T\) is the time period for one complete revolution. We are given that the Ferris wheel takes 12 seconds for one complete revolution. Therefore, the angular speed is: \(\omega = \frac{2\pi}{12\,s} = \frac{\pi}{6}\,rad/s\)
02

Determine the angular acceleration

During the one quarter of a revolution, the Ferris wheel stops completely, which means the final angular speed (\(\omega_f\)) is zero. To find the angular acceleration (\(\alpha\)), we can use the formula: \(\omega_f^2 = \omega_i^2 + 2\alpha\theta\), where \(\omega_i\) is the initial angular speed, and \(\theta\) is the angle covered during the one quarter of a revolution (which is \(\frac{\pi}{2}\) radians). Let's substitute the values into the above equation: \(0^2 = \left(\frac{\pi}{6}\,rad/s\right)^2 + 2 \alpha \left(\frac{\pi}{2}\right)\)
03

Solve for angular acceleration

We can now solve for the angular acceleration (\(\alpha\)). Rearrange the equation in step 2 to isolate \(\alpha\): \(\alpha = \frac{-\left(\frac{\pi}{6}\right)^2}{2\left(\frac{\pi}{2}\right)}\) Now, simplify and solve for \(\alpha\). \(\alpha = \frac{-\pi^2}{72} \cdot \frac{1}{\pi} = - \frac{\pi}{72} \, rad/s^2\)
04

Calculate the tangential acceleration

The boy's tangential acceleration (\(a_t\)) can be found using the relation between angular and tangential acceleration: \(a_t = \alpha \, r\), where \(r\) is the radius of the Ferris wheel. We know the radius is 9 meters, and the angular acceleration was found in step 3. Plugging in these values, we get: \(a_t = -\frac{\pi}{72} \cdot 9\,m = -\frac{9\pi}{72}\,m/s^2\) So, the boy's tangential acceleration during the time interval described in part (b) is \(-\frac{9\pi}{72}\,m/s^2\). The negative sign indicates that the acceleration is in the opposite direction to the initial motion, as the Ferris wheel is coming to a stop.

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