(a) [tex]4.14 rad/s^2[/tex]
The relationship beween centripetal acceleration and angular speed is
[tex]a=\omega^2 r[/tex]
where
[tex]\omega[/tex] is the angular speed
r is the radius of the circular path
Here we gave
[tex]a = 9g = 88.2 m/s^2[/tex] is the centripetal acceleration
r = 5.15 m is the radius
Solving for [tex]\omega[/tex], we find:
[tex]\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2[/tex]
(b) 21.3 m/s
The relationship between the linear speed and the angular speed is
[tex]v=\omega r[/tex]
where
v is the linear speed
[tex]\omega[/tex] is the angular speed
r is the radius of the circular path
In this problem we have
[tex]\omega=4.14 rad/s[/tex]
r = 5.15 m
Solving the equation for v, we find
[tex]v=(4.14 rad/s)(5.15 m)=21.3 m/s[/tex]
Explanation:
(a) Centripetal acceleration, [tex]a=9g=88.2\ m/s^2[/tex]
Radius, r = 5.15 m
Let [tex]\omega[/tex] is the angular speed. The relation between the angular speed and angular acceleration is given by :
[tex]a=\omega^2 r[/tex]
[tex]\omega=\sqrt{\dfrac{a}{r}}[/tex]
[tex]\omega=\sqrt{\dfrac{88.2}{5.15}}[/tex]
[tex]\omega=4.13\ rad/s[/tex]
(b) Let v is the linear speed of the person in the centrifuge have at this acceleration. It is given by :
[tex]v=r\times \omega[/tex]
[tex]v=5.15\times 4.13[/tex]
v = 21.26 m/s
Hence, this is the required solution.
