Brain Buster: Solving the Minimum Angular Velocity for a Thrilling Carnival Ride

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The discussion centers on calculating the minimum angular velocity required for a rotating cylinder ride at a carnival, where the floor is removed and riders cling to the wall due to centrifugal force. The correct formula for angular velocity is derived as ω = √(g/(μr)), where g is the acceleration due to gravity, μ is the coefficient of static friction, and r is the radius of the cylinder. Substituting the values (g = 9.81 m/s², μ = 0.5, r = 5.6 m) yields an angular velocity of approximately 2 s⁻¹. However, a discrepancy arises regarding significant figures, leading to an accepted answer of 1.872 s⁻¹. The conversation highlights the importance of clarity in applying significant figure rules in calculations.
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One of the amusements at many carnivals is a rotating cylinder. The customers step inside and stand with their backs to the wall. The cylinder spins very rapidly, and at some angular velocity the floor is pulled away. The thrill-seekers now hang like flies on the wall. If the radius of the cylinder is 5.6 m, and the coefficient of static friction between the people and the wall is 0.5, what is the minimum angular velocity, w, at which the floor can be withdrawn?
 
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i used the equation W=√g/µr but it was incorrect
 
This problem is pretty straightforward if you draw an FBD from the rotating reference frame. The normal force exerted by the cylinder is going to provide the centripetal acceleration.

so
a_c=r\omega^2
F_{normal}=ma_c
F_{friction}=\mu m a_c
F_g=mg
|F_{friction}|=|F_g
\mu m a_c = m g
\mu r \omega^2 = g
\omega=\sqrt{\frac{g}{\mu r}}

Looks like you were right.
Now plugging in numbers:
\omega=\sqrt{\frac{9.81}{0.5 \times 5.6}} \approx 2 s^{-1}
(0.5 has only one sig fig).
 
thanks for the help i used that equation but for some reason the computer didnt like it the course tells me to use the significant figure rule then doesn't use it later so the accepted answer was 1.872s-1 but thanks for the help again i really appreciate it.
 
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