a) 4.36 rad/s
b) 20.2 m/s
Explanation:
a)
For an object in uniform circular motion, the centripetal acceleration is related to the angular speed by the equation
[tex]a=\omega^2 r[/tex]
where
a is the centripetal acceleration
[tex]\omega[/tex] is the angular speed
r is the radius of the circle
For the person in this problem we have:
[tex]a=9g=88.2 m/s^2[/tex] is the centripetal acceleration
r = 4.64 m is the radius of the trajectory
Solving for [tex]\omega[/tex], we find the required angular speed to produce this centripetal acceleration:
[tex]\omega=\sqrt{\frac{a}{r}}=\sqrt{\frac{88.2}{4.64}}=4.36 rad/s[/tex]
b)
For a uniform circular motion, the relationship between angular speed and linear 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 circle
For the person in this problem we have:
[tex]\omega=4.36 rad/s[/tex] is the angular speed
r = 4.64 m is the radius of the circle
Therefore, the linear speed is:
[tex]v=(4.36)(4.64)=20.2 m/s[/tex]
