A ferris wheel, radius 5.0m, rotates once every 8.4 seconds.
a) What is the centripetal acceleration of a 60kg person at the circumference?
b) calculate the force of the seat on the person at the highest point.
c) calculate the force of the seat on the person at the lowest point.
d) Assume the seat is horizontal, that it does not have a back and that the seat surface is rough. If the person on the seat is midway between the highest and lowest point on the ferris wheel, calculate the minimum coefficient of static friction between the seat and the person in order that the person does not fall off the ferris wheel.
So.
a) 2.8m/s^2
b) 420N [upwards]
c) 756 N [upwards]
d) I'm not sure how to approach this question. The answer in the back of the book says 0.286, but can someone describe how to get to the solution please
I know this is an old question, but I was just working through it myself and found the answer. So I will post it here for future people trying to solve this question.
Nu_s = ma
mg*u_s = ma
Therefore, u_s = a/g = 2.80/9.8
u_s = 0.286
:)
The outer rim speed is
V = 2 pi*R/t = 3.74 m/s
a) the centripetal acceleration is
V^2/R = 2.80 m/s^2
(b) M (g-a) = 420 N
(c) M (g+a) = 756 N
d) M a*mu_s = M g
mu_s = g/a = 3.5
You'd need something like Velcro for that high a static friction coefficient
I got 3.5 as well!
But the solutions manual says 0.286?
To solve part d) and find the minimum coefficient of static friction between the seat and the person on the Ferris wheel, we can start by considering the forces acting on the person at the midway point.
At the midway point, the person experiences a downward gravitational force (mg) and an upward normal force (N) from the seat. Since the seat is rough, there is also a maximum static friction force (fs) that can act horizontally between the person and the seat surface.
To prevent the person from falling off, the centripetal force acting towards the center of the Ferris wheel must be greater than the gravitational force. Therefore, we can equate the centripetal force with the net force acting towards the center:
Fc = N - mg
The centripetal force can be calculated using the formula:
Fc = m * (v^2 / r)
where m is the mass of the person, v is the tangential velocity of the person, and r is the radius of the Ferris wheel.
Now, let's find the tangential velocity (v) at the midway point. Since the Ferris wheel completes one revolution in 8.4 seconds, the tangential velocity can be calculated as the circumference divided by the time period:
v = 2πr / T
where T is the time period of revolution.
Substituting the given values, we find:
v = 2 * π * 5.0 m / 8.4 s = 3.74 m/s
Now, we can substitute this value of v, along with the given values of m and r, into the equation for the centripetal force:
m * (v^2 / r) = N - mg
Substituting the values:
60 kg * (3.74 m/s)^2 / 5.0 m = N - 60 kg * 9.8 m/s^2
Simplifying, we have:
N = 60 kg * (3.74 m/s)^2 / 5.0 m + 60 kg * 9.8 m/s^2
Calculating this expression, we find:
N ≈ 444.768 N
Now, we can find the maximum static friction force (fs) using the equation:
fs ≤ μs * N
where μs is the coefficient of static friction.
Since we want to find the minimum coefficient of static friction, we assume that the person is on the verge of slipping. Therefore, the static friction force is equal to its maximum value:
fs = μs * N
Substituting the values, we have:
μs * N = μs * 444.768 N ≥ mg
Simplifying, we find:
μs ≥ mg / N
Substituting the values of m and N, we have:
μs ≥ (60 kg * 9.8 m/s^2) / 444.768 N
Calculating this expression, we find:
μs ≥ 0.286
Therefore, the minimum coefficient of static friction needed to prevent the person from falling off the Ferris wheel is approximately 0.286.