A popular carnival ride consists of a large, round room that rotates fast enough that any person
standing against the wall will be stuck there, even as the floor is dropped away from their feet. The
coefficient of static friction between the person and the wall is μs
, and the radius of the cylinder is R.
a) Find an equation for the minimum period of rotation, T, needed to keep the people in place on the
wall. This equation will include μs
, R, g and π.
b) Solve for T if R = 4.00 m and μs = 0.60. What is the room's angular speed in rpm?
Centripetal acceleration is
a= v² /R= ω²R=(2π/T)²R
ma=F(fr)
m(2π/T)²R =μmg.
4π²R/T= μg.
T=4π²R/ μg
To find the equation for the minimum period of rotation, T, needed to keep the people in place on the wall, we need to consider the forces acting on the person.
a) Let's start by drawing a free body diagram of the person against the wall.
The force keeping the person against the wall is the static friction force, fs. The weight of the person is mg, where m is the mass of the person and g is the acceleration due to gravity (approximately 9.8 m/s^2). The net force acting towards the center of the circular room is mv^2/R, where v is the linear speed of the person. Since the person is in equilibrium, these forces are equal:
fs = mv^2/R = μs * mg
Simplifying this equation, we can solve for v:
v = μs * g * R
The period of rotation, T, is the time it takes for the person to complete one full rotation. Since the distance traveled in one rotation is the circumference of the circular room (2πR), we can relate the linear speed, v, to the angular speed, ω:
v = ω * R
Substituting the expression for v, we get:
ω * R = μs * g * R
Dividing both sides by R, we find:
ω = μs * g
The angular speed, ω, is given by 2π divided by the period of rotation, T. So we can write:
2π/T = μs * g
Solving for T, we get:
T = 2π/(μs * g)
Therefore, the equation for the minimum period of rotation, T, is T = 2π/(μs * g), where μs is the coefficient of static friction between the person and the wall, R is the radius of the cylinder, and g is the acceleration due to gravity.
b) To solve for T when R = 4.00 m and μs = 0.60, we can substitute these values into the equation:
T = 2π/(0.60 * 9.8)
Calculating the value, we find:
T ≈ 1.068 seconds
To find the room's angular speed in RPM (revolutions per minute), we can use the relationship between angular speed and period:
ω = 2π/T
Substituting the value of T we just found, we get:
ω ≈ 5.913 rad/s
To convert from radians per second to revolutions per minute, we can use the conversion factor:
1 revolution = 2π radians
1 minute = 60 seconds
So, multiplying ω by (1 revolution/2π radians) and (60 seconds/1 minute), we find:
ω ≈ (5.913 * 1 revolution/2π radians) * (60 seconds/1 minute) ≈ 56.44 RPM
Therefore, the room's angular speed is approximately 56.44 RPM.