Question

A carousel at a carnival has a diameter of \(6.00 \mathrm{~m}\). The ride starts from rest and accelerates at a constant angular acceleration to an angular speed of \(0.600 \mathrm{rev} / \mathrm{s}\) in \(8.00 \mathrm{~s}\) a) What is the value of the angular acceleration? b) What are the centripetal and angular accelerations of a seat on the carousel that is \(2.75 \mathrm{~m}\) from the rotation axis? c) What is the total acceleration, magnitude and direction, \(8.00 \mathrm{~s}\) after the angular acceleration starts?

Short answer

The angular acceleration is approximately \(0.075\pi\ \mathrm{rad/s^2}\). b) What are the centripetal and angular accelerations of a seat on the carousel that is \(2.75 \mathrm{~m}\) from the rotation axis? The centripetal acceleration is about \(25.9\ \mathrm{m/s^2}\), and the angular acceleration of the seat is approximately \(0.647\pi\ \mathrm{m/s^2}\). c) What is the total acceleration, magnitude, and direction, \(8.00 \mathrm{~s}\) after the angular acceleration starts? The total acceleration's magnitude is approximately \(26.2\ \mathrm{m/s^2}\), and its direction is \(4.35^{\circ}\) from the radial direction (counterclockwise).

Step-by-step solution

Convert angular speed from rev/s to rad/s

First, we need to convert the given angular speed from rev/s to rad/s. To do this, we can use the conversion factor: 1 rev = \(2\pi\) rad. \(0.600 \frac{\mathrm{rev}}{\mathrm{s}}\times \frac{2\pi \ \mathrm{rad}}{1\ \mathrm{rev}} = 1.2\pi \frac{\mathrm{rad}}{\mathrm{s}}\)

Use the constant angular acceleration equations

We can use the equation for angular displacement with constant acceleration: \(\omega^2 = \omega_0^2 + 2\alpha\theta\) Here, \(\omega\) is the final angular speed, \(\omega_0\) is the initial angular speed, \(\alpha\) is the angular acceleration, and \(\theta\) is the angular displacement.

Determine angular displacement

The carousel starts from rest, so \(\omega_0 = 0\). Also, we are given the time after which the final angular speed is attained, which is 8.00 s. Given angular speed \(\omega = 1.2\pi\ \mathrm{rad/s}\), the angular displacement for the time interval can be determined by: \(\theta = \omega t = (1.2\pi\ \mathrm{rad/s})\times (8 \mathrm{~s}) = 9.6\pi \mathrm{~rad}\)

Calculate the angular acceleration

Now we can substitute the values of \(\omega\), \(\omega_0\), and \(\theta\) in the equation to find the angular acceleration: \((1.2\pi\ \mathrm{rad/s})^2 = 0^2 + 2\alpha(9.6\pi\ \mathrm{~rad})\) Solve for \(\alpha\): \(\alpha = \frac{(1.2\pi\ \mathrm{rad/s})^2}{2\times (9.6\pi\ \mathrm{~rad})} = 0.075\pi\ \mathrm{rad/s^2}\) Therefore, the angular acceleration is approximately \(0.075\pi\ \mathrm{rad/s^2}\). #b) What are the centripetal and angular accelerations of a seat on the carousel that is \(2.75 \mathrm{~m}\) from the rotation axis?#

Calculate centripetal acceleration

We can determine the centripetal acceleration using the formula \(a_c = r\omega^2\), where \(r\) is the distance from the rotation axis. For the seat on the carousel 2.75 m from the rotation axis: \(a_c = (2.75\ \mathrm{m})(1.2\pi\ \mathrm{rad/s})^2 = 2.75 \times (1.2\pi)^2 \approx 25.9\ \mathrm{m/s^2}\)

Calculate the seat's angular acceleration

We can determine the seat's angular acceleration using the formula \(a_\alpha = r\alpha\), where \(r\) is the distance from the rotation axis. \(a_\alpha = (2.75\ \mathrm{m})(0.075\pi\ \mathrm{rad/s^2}) \approx 0.647\pi\ \mathrm{m/s^2}\) So, the centripetal acceleration is about \(25.9\ \mathrm{m/s^2}\), and the angular acceleration of the seat is approximately \(0.647\pi\ \mathrm{m/s^2}\). #c) What is the total acceleration, magnitude and direction, \(8.00 \mathrm{~s}\) after the angular acceleration starts?#

Compute total acceleration

We can determine the total acceleration using the vector sum of the centripetal and angular accelerations. Let's consider the centripetal acceleration in the radial direction and the angular acceleration in the tangential direction. Therefore, we will find the magnitude of the total acceleration using Pythagorean theorem: \(a_T = \sqrt{a_c^2 + a_\alpha^2} = \sqrt{(25.9\ \mathrm{m/s^2})^2 + (0.647\pi\ \mathrm{m/s^2})^2} \approx 26.2\ \mathrm{m/s^2}\)

Compute the direction

We will find the direction (angle) of the total acceleration with respect to the radial direction using the inverse tangent: \(\theta = \tan^{-1}\frac{a_\alpha}{a_c}=\tan^{-1}\frac{(0.647\pi\ \mathrm{m/s^2})}{(25.9\ \mathrm{m/s^2})} \approx 4.35^{\circ}\) The total acceleration's magnitude is approximately \(26.2\ \mathrm{m/s^2}\), and its direction is \(4.35^{\circ}\) from the radial direction (counterclockwise).