In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates in a horizonta circle. A trainee riding in the chamber of a centrifuge rotating with a constant angular speed of 2.6 rad/s experiences a centripetal acceleration of 2.9 times the acceleration due to gravity. In a second training exercise, the centrifuge speeds up from rest with a constant angular acceleration. When the centrifuge reaches an angular speed of 2.6 rad/s, the trainee experiences a total acceleration equal to 5.1 times the acceleration due to gravity. (a) How long is the arm of the centrifuge? (b) What is the angular acceleration in the second training exercise?

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### Centrifuge Training for Pilots and Astronauts

**Problem Statement:**

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates in a horizontal circle. A trainee riding in the chamber of a centrifuge rotating with a constant angular speed of 2.6 rad/s experiences a centripetal acceleration of 2.9 times the acceleration due to gravity. In a second training exercise, the centrifuge speeds up from rest with a constant angular acceleration. When the centrifuge reaches an angular speed of 2.6 rad/s, the trainee experiences a total acceleration equal to 5.1 times the acceleration due to gravity. 

#### Questions:
**(a)** How long is the arm of the centrifuge?  
**(b)** What is the angular acceleration in the second training exercise?

### Detailed Explanation:

**(a) Length of the Arm of the Centrifuge:**

Given data:
- Angular speed, ω = 2.6 rad/s
- Centripetal acceleration, \( a_c \) = 2.9g (where g is the acceleration due to gravity)

The centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = r \cdot \omega^2 \]
where:
- \( r \) is the radius (or length of the arm)
- \( \omega \) is the angular speed

Rearranging the formula to solve for \( r \):
\[ r = \frac{a_c}{\omega^2} \]

Substituting the given values:
\[ r = \frac{2.9g}{(2.6)^2} \]

We know that the acceleration due to gravity, \( g \approx 9.81 \, m/s^2 \):

\[ r = \frac{2.9 \times 9.81 \, m/s^2}{(2.6)^2} \]
\[ r \approx \frac{28.449 \, m/s^2}{6.76} \]
\[ r \approx 4.21 meters \]

So, the length of the arm of the centrifuge is approximately 4.21 meters.

**(b) Angular Acceleration in the Second Training Exercise:**

Given data:
- Total acceleration, \( a_t = 5.1g \)
- Components of total acceleration:
Transcribed Image Text:### Centrifuge Training for Pilots and Astronauts **Problem Statement:** In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates in a horizontal circle. A trainee riding in the chamber of a centrifuge rotating with a constant angular speed of 2.6 rad/s experiences a centripetal acceleration of 2.9 times the acceleration due to gravity. In a second training exercise, the centrifuge speeds up from rest with a constant angular acceleration. When the centrifuge reaches an angular speed of 2.6 rad/s, the trainee experiences a total acceleration equal to 5.1 times the acceleration due to gravity. #### Questions: **(a)** How long is the arm of the centrifuge? **(b)** What is the angular acceleration in the second training exercise? ### Detailed Explanation: **(a) Length of the Arm of the Centrifuge:** Given data: - Angular speed, ω = 2.6 rad/s - Centripetal acceleration, \( a_c \) = 2.9g (where g is the acceleration due to gravity) The centripetal acceleration \( a_c \) is given by the formula: \[ a_c = r \cdot \omega^2 \] where: - \( r \) is the radius (or length of the arm) - \( \omega \) is the angular speed Rearranging the formula to solve for \( r \): \[ r = \frac{a_c}{\omega^2} \] Substituting the given values: \[ r = \frac{2.9g}{(2.6)^2} \] We know that the acceleration due to gravity, \( g \approx 9.81 \, m/s^2 \): \[ r = \frac{2.9 \times 9.81 \, m/s^2}{(2.6)^2} \] \[ r \approx \frac{28.449 \, m/s^2}{6.76} \] \[ r \approx 4.21 meters \] So, the length of the arm of the centrifuge is approximately 4.21 meters. **(b) Angular Acceleration in the Second Training Exercise:** Given data: - Total acceleration, \( a_t = 5.1g \) - Components of total acceleration:
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