Acommon 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?

 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

think about the oscillation of the toy shown in the video. when the toy is at the highest point of its motion, its velocity is zero and its acceleration is negative.

lol

explanation: