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

Apparent weight is the normal force N that the seat exerts on you. For uniform circular motion the required centripetal acceleration is a_c = v^2/r directed toward the wheel’s center, so apply Newton’s second law in the radial (toward-center) direction.

Relevant formula and general result:
Sum of inward (toward center) forces = m a_c.
If θ is the angle measured from the top (so at the top θ = 0, at the bottom θ = π), the inward component of gravity is mg cosθ (positive when it points inward). Thus
N + mg cosθ = m v^2 / r,
so N(θ) = m v^2 / r − mg cosθ.

Evaluate at key positions (showing intermediate steps):
- At the top (θ = 0): cos0 = 1, so
  N_top = m v^2 / r − mg = m( v^2 / r − g ).
- At the bottom (θ = π): cosπ = −1, so
  N_bottom = m v^2 / r − mg(−1) = m( v^2 / r + g ).
- Halfway between top and bottom (θ = π/2 or 3π/2): cos(π/2)=0, so
  N_side = m v^2 / r.

Compare values:
N_bottom − N_top = m( v^2 / r + g ) − m( v^2 / r − g ) = 2 m g, so the normal force at the top is smaller than at the bottom by 2mg regardless of v or m. More generally N(θ) is minimized when cosθ is maximal (cosθ = 1), i.e. at the top.

Why the correct answer is a:
Because N(θ) = m v^2/r − mg cosθ is smallest when cosθ = 1 (the top), the apparent weight is smallest at the top. (If N_top ≤ 0 you would even lift off the seat.)

Why other options are incorrect:
b: The location of the minimum does not depend on v (only the magnitude of N does). c: Halfway (cosθ = 0) gives N = m v^2/r, which is larger than N_top. d: The bottom gives the maximum N, not the minimum. e: Changing mass scales all N values equally and does not change the position of the minimum.

Final confirmation: the correct answer is 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?单项选择题

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