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Alika [10]
3 years ago
15

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?
Physics
1 answer:
Viefleur [7K]3 years ago
4 0

Answer:

24.07415 rpm

Explanation:

\mu = Coefficient of friction = 0.63

v = Velocity

d = Diameter = 4.9 m

r = Radius = \frac{d}{2}=\frac{4.9}{2}=2.45\ m

m = Mass

g = Acceleration due to gravity = 9.81 m/s²

Here the frictional force balances the rider's weight

f=\mu F_n

The centripetal force balances the weight of the person

\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

Velocity is given by

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

Converting to rpm

2.52104\times \frac{60}{2\pi}=24.07415\ rpm

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

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Answer:

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Now from the above equation given that nR and V  are constant  we have that

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8 0
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Answer:

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