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Chapter 9: Problem 75

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

Expert verified
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

01

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 = \(\frac{1.20\pi}{8}\) = \(0.15\pi \text{rad/s}^2\)
02

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): \(a_{tan} = (2.75 \text{ m})(0.15\pi \text{rad/s}^2) = 0.4125\pi \approx 1.30\; \text{m/s}^2\)
03

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.

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