Question

A common carnival ride, called a gravitron, is a large r = 11 m cylinder in which people stand against the outer wall. the cylinder spins so that the riders move with a linear speed v. At a certain point the floor of the cylinder lowers and the people are surprised they don't slide down. The friction between the wall and their clothes is μs = 0.62. What is the minimum speed, in meters per second, that the cylinder must make a person move at to ensure they will "stick" to the wall?

Answer 1

Answer:

 v = 13.19 m / s

Explanation:

This problem must be solved using Newton's second law, we create a reference system where the x-axis is perpendicular to the cylinder and the Y-axis is vertical

 

X axis

       N = m a

Centripetal acceleration is

       a = v² / r

Y Axis

      fr -W = 0

      fr = W

The force of friction is

     fr = μ N

Let's calculate

    μ (m v² / r) = mg

   μ v² / r = g

   v² = g r / μ

   v = √ (g r /μ)

   v = √ (9.8 11 / 0.62)

   v = 13.19 m / s