Question

7(0) = 4 + 3j 3. Hibbeler 13-32 4. Hibbeler 13-54 5. The human centrifuge in Germany shown on the back of the page is like several others throughout the world used to train and test astronauts and pilots. The distance from the hub to the middle of the hinge for the test compartment is 5m and the distance from the center of the hinge to the center of mass of the system (occupant and compartment) is 1m. If the angular velocity of the centrifuge is 4 rad/s, find the angle of the compartment. If the cm of the occupant is at the same spot as the system cm, find the foce on the occupant in units of "g's"

Answer 1

Let the angle that the hub makes with the vertical be x.

Then the horizontal and vertical distances of the cm from the point where the hub is connected to the arm are 1*sin(x) and 1*cos(x) respectively.

Then the horizontal and vertical distances of the cm from the hing are (5+1*sin(x)) and 1*cos(x) respectively.

With an angular speed of 4 rad/s, the centrifugal force on the hub is

$F_h = m*\omega^2*R$

where m is the mass of the hub, $\omega$ is the angular velocity and R is (5+1*sin(x))

The vertical force is graviational force given by

$F_v=m*g$

The torque balance about the point where the hub is connected to the arm gives:

$m*\omega^2*R*1*cos(x)-m*g*1*sin(x)=0$

$=>\omega^2*Rcos(x)=gsin(x)$

16 5 + sin(r))cos()-9.8sin(x)

$=>5cos(x)+sin(x)cos(x)=0.6125sin(x)$$=>5+sin(x)=0.6125tan(x)$

$=>x=1.46898 rad = 84.166^{\circ}$

The anglis is 84.166° with the vertical

$=>F_h = m*\omega^2*R=m*16*(5+0.995)=95.917*m N$

a_v = g = 9.8 m/s²

The acceleration values on the person would be same as the ones on the hub as the person is placed at the cm of the hub

So, the total acceleration experienced by the person is

$a= \sqrt{a_h^2+a_v^2} = 96.4165m/s^2$

acceleration is 96.4165 m/s²