A former amusement park attraction consisted of a huge vertical cylinder rotating rapidly around its axis. The rotation speed should be sufficient for the people found inside to be stuck to the wall when the floor was removed. what should be the minimum speed of rotation for this to happen? The coefficient of static friction between the people and the wall of the cylinder is 0.300 and the cylinders radius is 4 meters.
To determine the minimum speed of rotation required for people inside the amusement park attraction to be stuck to the wall when the floor is removed, we need to consider the forces involved.
When the floor is removed, the only force keeping people inside the cylinder is the static friction force between the people and the wall. So, the static friction force must be equal to or greater than the gravitational force acting on the people.
The static friction force can be calculated using the following formula:
Frictional force (Ff) = coefficient of static friction (μ) * Normal force (N)
The normal force (N) acting on a person inside the cylinder is equal to the person's weight, which is given by the formula:
Normal force (N) = Mass (m) * acceleration due to gravity (g)
In this case, we assume the mass (m) of a person to be the average mass, which is around 70 kg, and the acceleration due to gravity (g) is approximately 9.8 m/s^2.
First, let's calculate the normal force:
N = m * g = 70 kg * 9.8 m/s^2 = 686 N
Now, we can calculate the minimum static friction force required:
Ff = μ * N = 0.300 * 686 N = 205.8 N
The static friction force is also equal to the centripetal force required to keep the people stuck to the wall of the rotating cylinder. The centripetal force is given by the formula:
Centripetal force (Fc) = Mass (m) * Velocity^2 / Radius (r)
Since we want to find the minimum speed of rotation, we can assume a small value for mass (m) to simplify the calculation. Let's assume it to be 1 kg for simplicity.
Fc = m * v^2 / r = 1 kg * v^2 / 4 m
Setting the centripetal force equal to the minimum static friction force, we get:
1 kg * v^2 / 4 m = 205.8 N
Simplifying the equation, we have:
v^2 = 205.8 N * 4 m / 1 kg = 823.2 m^2/s^2
Finally, we can take the square root of both sides to find the minimum speed of rotation:
v = √(823.2 m^2/s^2) = 28.70 m/s
Therefore, the minimum speed of rotation for people inside the amusement park attraction to be stuck to the wall when the floor is removed should be approximately 28.70 m/s.