Assume that I am on a Ferris wheel rotating so that the seats move with a constant speed v. Where is my apparent weight the smallest?

Question

BlackTom AI · Accepted Answer

Centripetal acceleration for circular motion is a = v^2/r directed toward the center. The apparent weight is the normal force N the seat exerts on you; compare N at top and bottom by summing forces in the radial direction.

1) Knowledge points:
- Centripetal acceleration magnitude a = v^2/r, always toward the center.
- Newton’s second law in the radial direction: sum of radial components of forces = m a_radial.
- Apparent weight equals the normal force magnitude felt from the seat.

2) Step-by-step reasoning and algebra:
Take upward as positive at the bottom and downward as positive at the top (so “toward the center” is positive in each case).

At the bottom:
- Toward center is upward. Forces toward center: N (up) minus weight mg (down).
- Newton: N − mg = m(v^2/r).
- Solve: N_bottom = mg + m v^2/r.

At the top:
- Toward center is downward. Forces toward center: weight mg (down) minus N (up).
- Newton: mg − N = m(v^2/r).
- Solve: N_top = mg − m v^2/r.

3) Intermediate steps shown above; these give exact expressions for normal force at top and bottom.

4) Why top is smallest:
Compare N_top and N_bottom:
- N_top = mg − m v^2/r
- N_bottom = mg + m v^2/r
Clearly N_top < N_bottom for any positive v. Thus the apparent weight is smallest at the top of the ride. (If v is very large N_top could become zero or negative—meaning you would lift off the seat—but it is still the minimum normal force.)

5) Why other options are incorrect:
- b (At the bottom): incorrect because N_bottom is larger (mg + m v^2/r).
- c (depends on mass): incorrect—the position of the minimum does not depend on m; both N expressions scale with m so the comparison is independent of mass.
- d (depends on speed v): while the numerical value of N depends on v, the location of the minimum (the top) does not; for any v the top gives the smaller normal force.
- e (halfway): incorrect because intermediate positions have normal forces between the top and bottom values; they are not the minimum.

Conclusion: The apparent weight is smallest at the top, confirming answer a.

Assume that I am on a Ferris wheel rotating so that the seats move with a constant speed v. Where is my apparent weight the smallest?Single choice

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