Question

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

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

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: \(\omega = \frac{2\pi}{12\,s} = \frac{\pi}{6}\,rad/s\)

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: \(0^2 = \left(\frac{\pi}{6}\,rad/s\right)^2 + 2 \alpha \left(\frac{\pi}{2}\right)\)

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 \(\alpha\). \(\alpha = \frac{-\pi^2}{72} \cdot \frac{1}{\pi} = - \frac{\pi}{72} \, rad/s^2\)

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.

Key concepts

Angular Speed

Angular speed refers to the rate at which an object moves through an angle. For objects moving in a circular path, like a Ferris wheel, it is essential in understanding how quickly they complete a rotation. Typically represented by the symbol \(\omega\), angular speed is measured in radians per second (rad/s).
To calculate the angular speed, we use the formula:

• \(\omega = \frac{2\pi}{T}\)

where \(T\) is the time taken for one complete rotation.
In the context of the Ferris wheel problem, it takes 12 seconds to complete a single revolution. Hence, the angular speed of the Ferris wheel is \(\frac{\pi}{6}\,\text{rad/s}\). This ensures that the wheel covers the angle of \(2\pi\) radians (one full circle) every 12 seconds.

Angular Acceleration

Angular acceleration contrasts angular speed by identifying how quickly the angular speed itself changes over time. Denoted by \(\alpha\), it measures the rate of change of angular velocity and is expressed in radians per second squared (rad/s²).
When the Ferris wheel comes to a stop, its angular speed changes from an initial value to zero. This change over a specified angle (one quarter of a revolution, or \(\frac{\pi}{2}\,\text{radians}\)) allows us to calculate the angular acceleration using the following relationship:

• \(\omega_f^2 = \omega_i^2 + 2\alpha\theta\)

Here, \(\omega_i = \frac{\pi}{6}\,\text{rad/s}\) is the initial angular speed, \(\omega_f = 0\), and \(\theta = \frac{\pi}{2}\).
Through calculation, the wheel's angular acceleration is found to be \(- \frac{\pi}{72} \, \text{rad/s}^2\). The negative sign reflects deceleration, as the wheel slows to a halt.

Tangential Acceleration

Tangential acceleration is closely related to angular motion but describes the linear aspect. It represents how quickly the linear speed of a point on a rotating object changes as it moves along its circular path. This acceleration affects the object's tangential velocity.
Tangential acceleration, \(a_t\), can be determined from angular acceleration \(\alpha\) and the radius \(r\) of the circular path using the formula:

• \(a_t = \alpha \times r\)

With the Ferris wheel's radius of 9 meters, and angular acceleration previously calculated as \(-\frac{\pi}{72}\,\text{rad/s}^2\), we find the tangential acceleration:

• \(a_t = -\frac{\pi}{72} \times 9\,m = -\frac{9\pi}{72} \text{ m/s²}\)

The negative result indicates that the acceleration is opposite to the motion's initial direction, aligning with the wheel's slowing down. Understanding this helps in comprehending the physical experience of slowing motion on a Ferris wheel and similar rotating systems.