A 92.72 kg passenger is in a cabin of a Ferris wheel rotating in a vertical circular path with a constant speed of 17.7 m/s. At the bottom of the circular path, the passenger stands on an electronic balance which is calibrated to a gravitational field strength of 9.81 Nkg-1 . If the Ferris wheel has a radius of 94 m, calculate the reading of the balance in kg.

Question

BlackTom AI · Accepted Answer

Normal force N at the bottom must supply both weight and the upward centripetal force, so N = m(g + v^2/r). The electronic balance reports an equivalent mass by dividing the normal force by the calibrated gravitational field g (9.81 N kg^-1), so the reading in kg is m_read = N/g = m(1 + v^2/(g r)).

Relevant values:
- m = 92.72 kg
- v = 17.7 m s^-1
- r = 94.0 m
- g = 9.81 N kg^-1

Step-by-step calculation:
1. Compute v^2: v^2 = (17.7)^2 = 313.29 m^2 s^-2.
2. Compute centripetal acceleration a_c = v^2 / r = 313.29 / 94 = 3.333723404... m s^-2.
3. Sum with gravity: g + a_c = 9.81 + 3.333723404 = 13.143723404 m s^-2.
4. Compute normal force: N = m(g + a_c) = 92.72 × 13.143723404 ≈ 1218.686 N.
5. Convert to balance mass reading: m_read = N / g = 1218.686 / 9.81 ≈ 124.22 kg.

Alternatively using the compact form:
m_read = m [1 + v^2/(g r)] = 92.72 × [1 + 313.29/(9.81×94)]
= 92.72 × (1 + 0.3397434) ≈ 92.72 × 1.3397434 ≈ 124.22 kg.

Why this is correct: at the bottom of a vertical circle the required centripetal acceleration is upward, so the normal force must exceed the weight by m v^2/r; the balance measures that normal force and displays it as an equivalent mass using g.

There are no multiple-choice alternatives here; any answer that ignores the additional centripetal term (giving 92.72 kg) or uses incorrect signs/values would be incorrect.

Conclusion: the balance reads approximately 124.22 kg, matching the provided answer.

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