Assume that I am on a Ferris wheel that is rotating. We know that my apparent weight will be smallest at the top of the wheel. For which of the following Ferris wheel speeds will that apparent weight be the smallest?单项选择题
A
a. 9.8 m/s
B
b. 15 m/s
C
c. 30 m/s
D
d. 0 m/s

Question

Assume that I am on a Ferris wheel that is rotating. We know that my apparent weight will be smallest at the top of the wheel. For which of the following Ferris wheel speeds will that apparent weight be the smallest?单项选择题

A

a. 9.8 m/s

B

b. 15 m/s

C

c. 30 m/s

D

d. 0 m/s

BlackTom AI · Accepted Answer

At the top of the Ferris wheel the centripetal acceleration is downward, so the normal force (apparent weight) is reduced. The relevant relation is from Newton’s 2nd law in the radial direction:

1) Relevant concepts and formula
- Centripetal acceleration a_c = v^2 / r toward the center.
- At the top: mg (down) minus N (upwards taken negative sign convention) provides the required centripetal force: mg - N = m v^2 / r.
- Solve for the normal force (apparent weight) N: N = mg - m v^2 / r.

2) Step-by-step reasoning
- N depends on v^2 with a minus sign: increasing speed increases the centripetal term m v^2 / r, therefore decreases N.
- Thus the larger the speed v, the smaller the apparent weight at the top.

3) Example calculation (illustrative; radius r and mass m are not given in the problem, so choose convenient values to show trend)
- Let m = 1 kg and r = 10 m (example only).
- For v = 0 m/s: N = mg - m v^2 / r = 9.8 - 0 = 9.8 N.
- For v = 9.8 m/s: N = 9.8 - (9.8^2)/10 = 9.8 - 96.04/10 = 9.8 - 9.604 = 0.196 N.
- For v = 15 m/s: N = 9.8 - (15^2)/10 = 9.8 - 225/10 = 9.8 - 22.5 = -12.7 N (negative indicates loss of contact without a restraint).
- For v = 30 m/s: N = 9.8 - (30^2)/10 = 9.8 - 900/10 = 9.8 - 90 = -80.2 N (much more negative).
These numbers show N decreases (and can reach zero or negative) as v increases.

4) Why the correct answer is correct
- Because N = mg - m v^2 / r decreases monotonically as v increases, the largest listed speed produces the smallest apparent weight.

5) Why the other options are incorrect
- d (0 m/s): No centripetal acceleration, so apparent weight equals true weight (largest).
- a (9.8 m/s) and b (15 m/s): intermediate speeds give intermediate (larger) normal forces than the highest speed 30 m/s.

Conclusion: The smallest apparent weight occurs at the largest speed, so choice c. 30 m/s is correct.

Assume that I am on a Ferris wheel that is rotating. We know that my apparent weight will be smallest at the top of the wheel. For which of the following Ferris wheel speeds will that apparent weight be the smallest?单项选择题

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