a) The length of the arm of the centrifuge is 10.9 m
b) The angular acceleration is [tex]2.7 rad/s^2[/tex]
Explanation:
a)
In a uniform circular motion, the centripetal acceleration is given by
[tex]a_c=\omega^2 r[/tex]
where:
[tex]\omega[/tex] is the angular speed of the circular motion
r is the radius of the circle
For the centrifuge in this problem, we have:
[tex]\omega=1.7 rad/s[/tex] is the angular speed
The centripetal acceleration is 3.2 times the acceleration due to gravity ([tex]g=9.8 m/s^2[/tex]), so:
[tex]a_c=3.2 g = 3.2(9.8)=31.4 m/s^2[/tex]
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):
[tex]r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m[/tex]
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
[tex]a=4.4 g[/tex]
So, 4.4 times the acceleration due to gravity.
The total acceleration is the resultant of the centripetal acceleration ([tex]a_c[/tex]) and the tangential acceleration ([tex]a_t[/tex]):
[tex]a=\sqrt{a_c^2+a_t^2}[/tex]
We know that:
a = 4.4g
[tex]a_c = 3.2 g[/tex]
So, we can find the tangential acceleration:
[tex]a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2[/tex]
The angular acceleration is related to the tangential acceleration by
[tex]\alpha = \frac{a_t}{r}[/tex]
where r = 10.9 m is the length of the centrifuge. Substituting,
[tex]\alpha = \frac{29.6}{10.9}=2.7 rad/s^2[/tex]
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