Question

In a carnival ride, passengers stand with their backs against the wall of a cylinder. The cylinder is set into rotation and the floor is lowered away from the passengers, but they remain stuck against the wall of the cylinder. For a cylinder with a 2.0-m radius, what is the minimum speed that the passengers can have for this to happen if the coefficient of static friction between the passengers and the wall is 0.25?

Answers
2.3 m/s
3.0 m/s
4.9 m/s
8.9 m/s
It depends on the mass of the passengers.

Answer 1

To solve the problem, it is necessary to apply the related concepts to Newton's second law as well as the Normal and Centripetal Force experienced by passengers.

By Newton's second law we understand that

`F = mg`

Where,

m= mass

g = Gravitational Acceleration

Also we have that Frictional Force is given by

`F_r = \mu N`

In this particular case the Normal Force N is equivalent to the centripetal Force then,

`N = (mv^2)/(r)`

Applying this to the information given, and understanding that the Weight Force is statically equivalent to the Friction Force we have to

`F = F_r`

`mg = \mu N`

`mg = \mu (mv^2)/(r)`

Re-arrange to find v,

`v= \sqrt{(gr)/(\mu)}`

`v = \sqrt{((9.8)(2))/(0.25)}`

`v = 8.9m/s`

From the last expression we can realize that it does not depend on the mass of the passengers.