A student of weight 661 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force FN on the student from the seat is 549 N.

(b)What is the magnitude of FN at the lowest point?
_________N

(c) If the wheel's speed is doubled, what is the magnitude of FN at the highest point?
________N

(d) What is the magnitude of FN at the lowest point under the same conditions as in (c)?
______ N

AT the top, force is m(g-a). YOu know mg, which is given, so calculate ma.

at the bottom, force is m(g+a).
If one doubles speed, then a quadruples.(a=v^2/r)

1000

To find the magnitude of the normal force (FN) at different points on the Ferris wheel, we need to consider the forces acting on the student at those points. The normal force is the force exerted by a surface perpendicular to the contact point.

(b) At the lowest point, the student experiences two forces: their weight (661 N) and the normal force (FN). Assuming the student remains upright, both forces act downwards. Therefore, the magnitude of the normal force at the lowest point is equal to the sum of the weight and the normal force:

FN = Weight + 549 N
FN = 661 N + 549 N
FN = 1210 N

Therefore, the magnitude of FN at the lowest point is 1210 N.

(c) If the wheel's speed is doubled, the acceleration of the student will also double. Since the student sits upright, the normal force will decrease at this higher speed. However, the weight of the student remains the same.

The magnitude of the normal force at the highest point can be found by balancing the forces acting on the student. At the highest point, the student experiences two forces: their weight (661 N) and the normal force (FN). Assuming the student remains upright, both forces act downwards.

At the highest point, the net force acting on the student is the centripetal force required for circular motion. This force is provided by the difference between the weight and the normal force:

Centripetal Force = Weight - FN
Centripetal Force = 661 N - 549 N
Centripetal Force = 112 N

The magnitude of the centripetal force is given by:

Centripetal Force = mass * acceleration

Assuming a constant mass, we can express the centripetal force as:

Centripetal Force = mass * velocity^2 / radius

Doubling the speed of the wheel means that the centripetal force also doubles. Therefore:

2 * Centripetal Force = mass * (2 * velocity)^2 / radius

Simplifying the equation gives:

2 * Centripetal Force = mass * 4 * velocity^2 / radius

Multiplying both sides by 2, we obtain:

4 * Centripetal Force = mass * 4 * velocity^2 / radius

Now, we equate this result to the difference between the weight and the normal force:

4 * Centripetal Force = Weight - FN

Substituting the known values:

4 * 112 N = 661 N - FN

Simplifying the equation gives:

448 N = 661 N - FN

Finally, solving for FN gives:

FN = 661 N - 448 N
FN = 213 N

Therefore, the magnitude of FN at the highest point when the wheel's speed is doubled is 213 N.

(d) In part (c), we found that the magnitude of FN at the highest point is 213 N when the wheel's speed is doubled.

Using the same reasoning as in part (b), the magnitude of the normal force at the lowest point can be found by balancing the forces acting on the student. At the lowest point, the net force acting on the student is the centripetal force required for circular motion. This force is provided by the difference between the weight and the normal force:

Centripetal Force = Weight - FN

Substituting the known values:

Centripetal Force = 661 N - FN

At this increased speed, the centripetal force is still doubled, so:

2 * Centripetal Force = 2 * (661 N - FN)
2 * Centripetal Force = 1322 N - 2FN

But we know from part (c) that the centripetal force is 448 N. Substituting this value, we can find FN:

2 * 448 N = 1322 N - 2FN

Simplifying the equation gives:

896 N = 1322 N - 2FN

2FN = 1322 N - 896 N
2FN = 426 N

FN = 426 N / 2
FN = 213 N

Therefore, the magnitude of FN at the lowest point under the same conditions as in part (c) is also 213 N.