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pentagon [3]
3 years ago
14

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

n a horizontal circle. A trainee riding in the chamber of a centrifuge rotating with a constant angular speed of 1.7 rad/s experiences a centripetal acceleration of 3.2 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 1.7 rad/s, the trainee experiences a total acceleration equal to 4.4 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?
Physics
1 answer:
RSB [31]3 years ago
8 0

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is 2.7 rad/s^2

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

a_c=\omega^2 r

where:

\omega is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

\omega=1.7 rad/s is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity (g=9.8 m/s^2), so:

a_c=3.2 g = 3.2(9.8)=31.4 m/s^2

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

a=4.4 g

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration (a_c) and the tangential acceleration (a_t):

a=\sqrt{a_c^2+a_t^2}

We know that:

a = 4.4g

a_c = 3.2 g

So, we can find the tangential acceleration:

a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2

The angular acceleration is related to the tangential acceleration by

\alpha = \frac{a_t}{r}

where r = 10.9 m is the length of the centrifuge. Substituting,

\alpha = \frac{29.6}{10.9}=2.7 rad/s^2

Learn more about centripetal and angular acceleration here:

brainly.com/question/2562955

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

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