A circular-motion addict of mass 70 kg rides a Ferris wheel around in a vertical circle of radius 10 m at a constant speed of 6.5 m/s.
(a) What is the period of the motion?
s
(b) What is the magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path?
N
(c) What is it when both go through the lowest point?
N
I will be happy to critique your thinking.
To solve this problem, we can use the principles of circular motion and centripetal force.
(a) To find the period of the motion, we can use the formula for the period of a circular motion:
T = 2πr/v,
where T is the period, r is the radius of the circular path, and v is the speed of the addict.
Given:
r = 10 m (radius of the Ferris wheel),
v = 6.5 m/s (speed of the addict).
Substituting these values into the formula, we get:
T = 2π(10 m) / (6.5 m/s) = 20π / 6.5 s ≈ 3.08 s.
Therefore, the period of the motion is approximately 3.08 seconds.
(b) The magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path can be determined by analyzing the forces acting on the addict at that point.
At the highest point, the normal force provides the centripetal force to keep the addict moving in a circular path.
The formula for the centripetal force is:
Fc = m * v^2 / r,
where Fc is the centripetal force, m is the mass of the addict, v is the speed of the addict, and r is the radius of the circular path.
Given:
m = 70 kg (mass of the addict),
v = 6.5 m/s (speed of the addict),
r = 10 m (radius of the Ferris wheel).
Substituting these values into the formula, we get:
Fc = (70 kg) * (6.5 m/s)^2 / (10 m) ≈ 296.75 N.
Therefore, the magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path is approximately 296.75 Newtons.
(c) At the lowest point, the normal force still provides the centripetal force to keep the addict moving in a circular path.
Using the same formula as above:
Fc = m * v^2 / r,
we can calculate the magnitude of the normal force when the addict is at the lowest point.
Substituting the given values, we have:
Fc = (70 kg) * (6.5 m/s)^2 / (10 m) ≈ 296.75 N.
Therefore, the magnitude of the normal force on the addict from the seat when both go through the lowest point of the circular path is approximately 296.75 Newtons.