contestada

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 2.8 rad/s experiences a centripetal acceleration of 3.1 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 2.8 rad/s, the trainee experiences a total acceleration equal to 4.3 times the acceleration due to gravity.

(a) How long is the arm of the centrifuge? m

(b) What is the angular acceleration (magnitude) of the centrifuge in the second training exercise? rad/s2

Respuesta :

boffeemadrid

Answer:

3.87 m

7.55 rad/s²

Explanation:

[tex]a_c[/tex] = 3.1g

g = Acceleration due to gravity = 9.81 m/s²

Centripetal acceleration is given by

[tex]a_c=\dfrac{v^2}{r}\\\Rightarrow a_c=r\omega^2\\\Rightarrow r=\dfrac{a_c}{\omega^2}\\\Rightarrow r=\dfrac{3.1\times 9.81}{2.8^2}\\\Rightarrow r=3.87\ m[/tex]

The length of the arm of the centrifuge is 3.87 m

Now

a = 4.3g

Total acceleration is given by

[tex]a=\sqrt{a_t^2+a_c^2}\\\Rightarrow a_t=\sqrt{a^2-a_c^2}\\\Rightarrow a_t=\sqrt{(4.3\times 9.81)^2-(3.1\times 9.81)^2}\\\Rightarrow a_t=29.2331416033\ m/s^2[/tex]

Angular acceleration is given by

[tex]\alpha=\dfrac{a_t}{r}\\\Rightarrow \alpha=\dfrac{29.2331416033}{3.87}\\\Rightarrow \alpha=7.55\ rad/s^2[/tex]

The angular acceleration is 7.55 rad/s²

Otras preguntas