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

Period = 2 pi r /v

top m(v^2/r -g)
bottom m(v^2/r + g)

To answer these questions, we first need to understand the concepts of circular motion and the forces involved.

(a) The period of the motion refers to the time it takes for the addict to complete one full revolution around the Ferris wheel. The period can be calculated using the formula:

T = (2πr) / v

where T is the period, r is the radius of the circular path, and v is the constant speed of the addict.

In this case, the radius of the circular path is given as 10 m, and the speed is given as 6.5 m/s. Substituting these values into the formula, we get:

T = (2π * 10) / 6.5

Now we can calculate the period by evaluating this expression:

T ≈ 6.12 s

Therefore, the period of the motion is approximately 6.12 seconds.

(b) The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, when the addict is at the highest point of the circular path, the only two forces acting on the addict are gravity and the normal force.

At the highest point, the normal force and the gravitational force must add up to give the net force in the downward direction. This means:

N - mg = mv²/r

where N is the magnitude of the normal force, m is the mass of the addict, g is the acceleration due to gravity (approximately 9.8 m/s²), v is the speed of the addict, and r is the radius of the circular path.

Substituting the given values, we have:

N - (70 kg)(9.8 m/s²) = (70 kg)(6.5 m/s)² / 10 m

Now we can solve for the magnitude of the normal force:

N ≈ 741 N

Therefore, the magnitude of the normal force on the addict from the seat when they go through the highest point of the circular path is approximately 741 Newtons.

(c) At the lowest point of the circular path, the normal force and the gravitational force must add up to give the net force in the upward direction. This means:

N + mg = mv²/r

Using the same values as before, we can solve for the magnitude of the normal force at the lowest point:

N + (70 kg)(9.8 m/s²) = (70 kg)(6.5 m/s)² / 10 m

Simplifying the equation, we get:

N ≈ 989 N

Therefore, the magnitude of the normal force on the addict from the seat when they go through the lowest point of the circular path is approximately 989 Newtons.