In an old-fashioned amusement park ride, passengers stand inside a 4.9-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.63 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed." What is the minimum angular speed, in rpm, for which the ride is safe?

[tex]\mu m\frac{v^2}{r}=mg\\\Rightarrow \mu \frac{v^2}{r}=g\\\Rightarrow v=\sqrt{\frac{gr}{\mu}}\\\Rightarrow v=\sqrt{\frac{9.81\times 2.45}{0.63}}\\\Rightarrow v=6.17656\ m/s[/tex]

Velocity is given by

[tex]v=\omega r\\\Rightarrow \omega=\frac{v}{r}\\\Rightarrow \omega=\frac{6.17656}{2.45}\\\Rightarrow \omega=2.52104\ rad/s[/tex]

Converting to rpm

[tex]2.52104\times \frac{60}{2\pi}=24.07415\ rpm[/tex]

The minimum angular speed for which the ride is safe is 24.07415 rpm

batolisis batolisis · Answer 2 · 2021-10-27T13:39:31+02:00

The minimum angular speed for which the ride will be safe is ≈ 24.07 rpm

Given data :

Diameter of hollow steel cylinder = 4.9 m. Radius ( r ) = 4.9 / 2 = 2.45 m

coefficient of friction of clothing ( [tex]\alpha[/tex] ) = 0.63

g = 9.81 m/s²

First step : Determine the velocity using the centripetal forces relation

v = [tex]\sqrt{\frac{g*r}{\alpha } }[/tex] ----- ( 1 )

where ; g = 9.81 m/s, r = 2.45 m , [tex]\alpha = 0.63[/tex]

Insert values into equation 1

V = [√( 9.81 * 2.45 )/0.63 ]

= 6.177 m/s

Next : convert velocity to rad/sec ( angular velocity )

V = ω*r

∴ ω = V / r

= 6.177 / 2.45 = 2.52 rad/sec

Final step: The minimum angular speed expressed in rpm

angular velocity ( ω ) * [tex]\frac{60}{2\pi }[/tex]

= 2.52 * [tex]\frac{60}{2\pi }[/tex] ≈ 24.07 rpm

Hence the minimum angular speed in rpm = 24.07 rpm.

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