An amusement ride consists of a large horizontal circular platform turning at 12 rpm. People attempt to sit on this platform without slipping toward the outer edge. The coefficient of static friction is 0.15. Determine the distance from the center of the platform that a 150-pound person may sit before slipping occurs.
To determine the distance from the center of the platform that a person may sit before slipping occurs, we need to consider the centripetal force acting on the person and compare it to the maximum static friction force.
1. First, let's calculate the angular velocity of the platform in radians per second. We know that the platform turns at a rate of 12 rpm (revolutions per minute). To convert this to radians per second, we use the following conversion factor:
* 1 revolution = 2π radians
* 1 minute = 60 seconds
So, the angular velocity (ω) in radians per second is:
ω = (12 rev/minute) * (2π radians/1 rev) * (1 minute/60 seconds) = π/5 rad/s
2. Now, let's calculate the centripetal force (Fc) acting on the person. The centripetal force can be calculated using the formula:
Fc = m * ω^2 * r
Where:
Fc = Centripetal force
m = Mass of the person (150 pounds)
ω = Angular velocity (in rad/s)
r = Distance from the center of the platform
First, let's convert the mass from pounds to kilograms:
150 pounds * (0.4536 kg/1 pound) = 68.04 kg
Now, we can substitute the values into the formula:
Fc = (68.04 kg) * ((π/5 rad/s)^2) * r
Fc = 68.04 * (π^2/25) * r
3. To prevent slipping, the centripetal force must be less than or equal to the maximum static friction force between the person and the platform. The maximum static friction force (fs) can be calculated using the formula:
fs = μs * N
Where:
fs = Maximum static friction force
μs = Coefficient of static friction (given as 0.15)
N = Normal force
The normal force (N) acting on the person is equal to the person's weight (W):
N = W = 68.04 kg * 9.8 m/s^2 = 666.792 N
Substituting the values into the formula, we get:
fs = (0.15) * 666.792 N = 100.0188 N
4. Now, equating the centripetal force to the maximum static friction force, we have the inequality:
Fc ≤ fs
Substituting the values:
68.04 * (π^2/25) * r ≤ 100.0188
r ≤ (100.0188 * 25) / (68.04 * π^2)
Calculating this expression gives us the maximum allowed value of r.
5. Plugging the values into a calculator, we find that r ≤ 1.88 meters.
Therefore, a 150-pound person may sit at a maximum distance of approximately 1.88 meters from the center of the platform before slipping occurs.