The Wall of Death in an amusement park is comprised of a vertical cylinder that can spin around the vertical axis. Riders stand against the wall of the spinning cylinder and the floor falls away leaving the riders held up by friction. The radius of the cylinder is 4.2 m and the coefficent of static friction between the rider and the wall is 0.47. Find the minimum number of revolutions per minute necessary so that the riders do not slip down the wall (note: to specify revolutions per minute as the units in your answer, enter rev/min).

Centripetal acceleration = w^2 R

where w is angular velocity in radians per second
then the force holding the rider against the wall is m a = m w^2 R
then the maximum friction force preventing a slide to oblivion = 0.47 m w^2 R
when m g is bigger than that we slide
so
9.8 = 0.47 w^2 (4.2)
solve for w
then convert w
w radians/s * (1 rev/2 pi radians) *(60 s/min) = answer in revs/min

To find the minimum number of revolutions per minute necessary so that the riders do not slip down the wall, we need to consider the maximum centrifugal force acting on the riders and compare it to the maximum friction force provided by the static friction.

The maximum centrifugal force acting on the riders is given by:

Fc = m * (omega)^2 * r

where m is the mass of the rider, omega is the angular velocity, and r is the radius of the cylinder.

The maximum friction force (Ff) that can be provided by the static friction is given by:

Ff = mu * N

where mu is the coefficient of static friction and N is the normal force.

Since the riders are held up by friction, the normal force (N) is equal to the weight (mg) of the riders, where g is the acceleration due to gravity.

Equating Fc and Ff, we get:

m * (omega)^2 * r = mu * mg

Canceling common terms, we have:

(omega)^2 = mu * g / r

To find the angular velocity (omega), we need to convert revolutions per minute to radians per second. There are 2*pi radians in one revolution and 60 seconds in one minute.

Converting revolutions per minute to radians per second, we have:

omega = (2*pi * rev/min) * (1 min / 60 s)

Substituting this into the equation for (omega)^2, we get:

(2*pi * rev/min)^2 * (1 min / 60 s)^2 = mu * g / r

Simplifying, we have:

(4*pi^2 * rev^2 / (60^2 * min^2)) = mu * g / r

Solving for rev/min, we get:

rev/min = sqrt((mu * g * 60^2 * min^2)/(4*pi^2 * r))

Plugging in the given values for the coefficient of static friction (mu = 0.47), the acceleration due to gravity (g = 9.8 m/s^2), and the radius of the cylinder (r = 4.2 m), we can calculate the minimum number of revolutions per minute necessary for the riders to not slip down the wall:

rev/min = sqrt((0.47 * 9.8 * 60^2 * min^2) / (4*pi^2 * 4.2))

Calculating this value, we find:

rev/min ≈ 5.95 rev/min

Therefore, the minimum number of revolutions per minute necessary so that the riders do not slip down the wall is approximately 5.95 rev/min.

To find the minimum number of revolutions per minute necessary so that the riders do not slip down the wall, we need to consider the force balance acting on the riders.

Friction is the force that prevents the riders from slipping down the wall. It is given by the equation:

Friction = coefficient of static friction * Normal force

The normal force is the force exerted by the wall on the riders, perpendicular to the surface of the wall. In this case, since the riders are standing against the wall, the normal force is equal to the weight of the riders.

The riders are held up by the friction force:

Friction = weight

Let's break down the forces involved:

1. Weight of the riders: This is the force pulling the riders downward and is given by the equation:

Weight = mass * acceleration due to gravity

2. Friction force: This is the force exerted by the wall on the riders, parallel to the surface of the wall.

Since the riders are not sliding, the frictional force must be equal to the weight of the riders. Therefore:

Friction = coefficient of static friction * Normal force
Weight = coefficient of static friction * Weight

Simplifying the equation:

Weight - coefficient of static friction * Weight = 0
(1 - coefficient of static friction) * Weight = 0

Solving for Weight:

Weight = 0 / (1 - coefficient of static friction) = 0

From this equation, we see that the weight of the riders must be zero for there to be no slipping. This means that the riders are effectively weightless, which is not possible. Therefore, it is not possible for the riders to remain on the wall without slipping, regardless of the number of revolutions per minute.

In conclusion, there is no minimum number of revolutions per minute that would prevent the riders from slipping down the wall.