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 rev/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

Answer: The total acceleration of the seat 8 seconds after the angular acceleration starts is approximately 38.63 m/s² at an angle of 1.93° from the radial line.

Step-by-step solution

Determine the angular acceleration

First, we need to convert the final angular speed from rev/s to rad/s: \(0.600\;\text{rev/s}\times \frac{2\pi\;\text{rad}}{\text{rev}}=1.20\pi\;\text{rad/s}\). Now, we can use the equation of motion to determine the angular acceleration, α: \(ω_f = ω_i + αt\) \(1.20\pi = 0 + α(8)\) α = \(\frac{1.20\pi}{8}\) = \(0.15\pi \text{rad/s}^2\)

Find the centripetal and angular accelerations of the seat

To find the centripetal acceleration, \(a_c\), we can use the formula \(a_c = rω^2\), where \(r\) is the distance from the rotation axis to the seat (2.75 m) and \(ω\) is the final angular speed (1.20π rad/s): \(a_c = (2.75 \text{ m})(1.20\pi \text{rad/s})^2 = 2.75(1.44\pi^2) \approx 38.61\; \text{m/s}^2\) To find the angular acceleration, \(a_{tan}\), we can use the formula \(a_{tan} = rα\), where \(α\) is the angular acceleration found earlier (0.15π rad/s²): \(a_{tan} = (2.75 \text{ m})(0.15\pi \text{rad/s}^2) = 0.4125\pi \approx 1.30\; \text{m/s}^2\)

Calculate the total acceleration

We will find the magnitude of the total acceleration using the Pythagorean theorem: \(|a_T| = \sqrt{a_c^2 + a_{tan}^2} = \sqrt{(38.61)^2 + (1.30)^2} \approx 38.63\; \text{m/s}^2\) Now, we will find the direction of the total acceleration using trigonometry. Let θ be the angle between the radial line and the total acceleration vector: \(\tan(\theta) = \frac{a_{tan}}{a_c} = \frac{1.30}{38.61}\) \(\theta = \arctan(\frac{1.30}{38.61}) \approx 1.93^{\circ}\) So, the total acceleration of the seat 8 seconds after the angular acceleration starts is approximately \(38.63\; \text{m/s}^2\) at an angle of \(1.93^{\circ}\) from the radial line.

Key concepts

Angular Acceleration

Angular acceleration is a measure of how quickly an object changes its angular velocity. It is analogous to linear acceleration but it occurs in circular motion. In this carousel exercise, the angular acceleration helps us determine how fast the ride is speeding up in terms of its spinning motion.
To find the angular acceleration, we need the change in angular speed and the time over which this change occurs. The formula used is:

• \(\alpha = \frac{\omega_f - \omega_i}{t}\)

Where \(\omega_f\) is the final angular speed, \(\omega_i\) is the initial angular speed (which is zero in this case), and \(t\) is the time. Converting the given final angular speed of 0.600 rev/s to radians per second (rad/s) is essential, as it provides a standard unit for calculating \(\alpha\).
In the exercise, this resulted in an angular acceleration of \(0.15\pi \; \text{rad/s}^2\). This tells us that every second, the angular velocity of the carousel increases by \(0.15\pi \; \text{rad/s}\).

Centripetal Acceleration

Centripetal acceleration occurs whenever an object moves in a circular path, and it acts towards the center of the circle. This is essential in keeping the object moving in that path.
The formula for centripetal acceleration \(a_c\) is:

• \(a_c = r\omega^2\)

Where \(r\) is the radius (the distance from the rotation axis to the object) and \(\omega\) is the angular speed in rad/s.
In the carousel example, the seat is 2.75 meters from the center, and the final angular speed is used to compute \(a_c\), resulting in approximately \(38.61 \; \text{m/s}^2\). This high value underscores how significant centripetal force needs to be to keep the seat moving in a circle at that speed.

Total Acceleration

Total acceleration in circular motion comprises two components: the angular (tangential) acceleration and the centripetal acceleration. These two types of acceleration combine to form the total acceleration.

• Angular acceleration (\(a_{tan}\)) changes the speed along the circle's path.
• Centripetal acceleration (\(a_c\)) keeps the object moving in the circle.

To find the total acceleration's magnitude, we use the Pythagorean theorem:

• \(|a_T| = \sqrt{a_c^2 + a_{tan}^2}\)

For direction, trigonometry comes into play. The angle \(\theta\) between the radial line and total acceleration vector is found using:

• \(\tan(\theta) = \frac{a_{tan}}{a_c}\)

In the exercise, the total acceleration was calculated to be \(38.63 \; \text{m/s}^2\) at an angle of roughly \(1.93^{\circ}\) from the radial, indicating the direction slightly away from the center. Understanding this helps in visualizing how the seat moves and accelerates, portraying a complete picture of the circular motion dynamics.