In a certain amusement-park ride, riders stand with their backs against the wall of a spinning vertical cylinder. The floor falls away and the riders are held up by friction. If the radius of the cylinder is 4.2 m, find the minimum number of revolutions per minute to prevent riders from dropping when the coefficient of static friction between a rider and the wall is 0.41.
.41 omega^2 r = g
omega^2 = 9.81/(.41*4.2) = 5.7
omega = 2.39 radians/second
2.39 radians/second *30 seconds/min *1 rev/2 pi radians = 11.3 rpm
With that answer, it's right, but there's 60 seconds per min
To find the minimum number of revolutions per minute required to prevent riders from dropping, we need to determine the critical speed at which the friction force between the riders and the wall is at its maximum. At this speed, the net force on the riders will be equal to zero.
The maximum friction force can be found using the equation:
F_friction = μ_s * N
Where:
F_friction is the maximum static friction force,
μ_s is the coefficient of static friction,
N is the normal force, which is equal to the weight of the riders.
The normal force N can be calculated as:
N = m * g
Where:
m is the mass of an average rider,
g is the acceleration due to gravity.
Assuming an average rider has a mass of 70 kg and taking the acceleration due to gravity as 9.8 m/s^2, we can calculate the normal force:
N = 70 kg * 9.8 m/s^2 = 686 N
Next, we can calculate the maximum static friction force:
F_friction = 0.41 * 686 N = 281.26 N
The maximum static friction force is equal to the centripetal force acting on the riders when they are at the critical speed. The centripetal force can be calculated using the equation:
F_centripetal = m * (v^2 / r)
Where:
F_centripetal is the centripetal force,
m is the mass of an average rider,
v is the velocity of the riders,
r is the radius of the cylinder.
To solve for the velocity v, we rearrange the equation:
v = sqrt(F_centripetal * r / m)
At the critical speed, the net force on the riders is zero, so we can equate the centripetal force to the maximum static friction force:
F_centripetal = F_friction
Substituting the known values, we can solve for the velocity v:
281.26 N = 70 kg * (v^2 / 4.2 m)
v^2 = (281.26 N * 4.2 m) / 70 kg
v^2 = 16.8768 N*m / kg
v = sqrt(16.8768 N*m / kg) = 4.11 m/s
Finally, we can calculate the minimum number of revolutions per minute required:
The distance traveled in one revolution is equal to the circumference of the cylinder, which is 2π * r.
The time taken for one revolution is equal to the circumference divided by the velocity, which is (2π * r) / v.
To convert this time to minutes, we divide it by 60:
t_revolution = [(2π * r) / v] / 60
Substituting the known values:
t_revolution = [(2π * 4.2 m) / 4.11 m/s] / 60
t_revolution = 0.935 minutes
Therefore, the minimum number of revolutions per minute required to prevent riders from dropping is:
1 minute / 0.935 minutes = 1.07 revolutions per minute (approximately)
So, the minimum number of revolutions per minute is approximately 1.07.
To find the minimum number of revolutions per minute (RPM) to prevent the riders from dropping, we need to consider the forces acting on the riders while they are in the amusement park ride.
The only force preventing the riders from dropping is the frictional force between the riders and the wall of the spinning cylinder. This frictional force acts towards the center of the cylinder and provides the necessary centripetal force to keep the riders in place.
The centripetal force required can be calculated using the equation:
F = m * R * ω^2
Where:
- F is the centripetal force
- m is the mass of a rider
- R is the radius of the cylinder
- ω is the angular velocity of the cylinder
In this case, we know the radius of the cylinder (R = 4.2 m) and the coefficient of static friction (µ = 0.41). The frictional force can be calculated as:
F_friction = µ * m * g
Where:
- m is the mass of a rider
- g is the acceleration due to gravity
Since the frictional force provides the necessary centripetal force, we can equate the two equations:
µ * m * g = m * R * ω^2
The mass cancels out:
µ * g = R * ω^2
We need to solve for ω, the angular velocity. Rearranging the equation, we have:
ω = sqrt(µ * g / R)
By substituting the given values, we can calculate the angular velocity:
ω = sqrt(0.41 * 9.8 / 4.2)
ω ≈ 2.728 rad/s
Now, we need to convert the angular velocity to RPM. Since 1 revolution is equal to 2π radians, we have:
ω = 2π * RPM / 60
Rearranging the equation, we can solve for RPM:
RPM = ω * (60 / 2π)
Plugging in the value for ω, we can find the minimum number of revolutions per minute:
RPM = 2.728 * (60 / (2 * π))
RPM ≈ 26.08
Therefore, the minimum number of revolutions per minute to prevent riders from dropping is approximately 26.08 RPM.