A roller coaster is designed such that riders experience "weightlessness" as they go round the top of a hill whose radius of curvature is \(20\text{ m}\). The speed of the car at the top of the hill is between:
(2008)
\(13\text{ m/s}\) and \(14\text{ m/s}\)
\(14\text{ m/s}\) and \(15\text{ m/s}\)
\(15\text{ m/s}) and \(16\text{ m/s}\)
\(16\text{ m/s}\) and \(17\text{ m/s}\)
Solution:
At the top of a hill, "weightlessness" implies the normal force is zero. The centripetal force is provided solely by gravity: (mg = \frac{mv^2}{R}). Thus, (v = \sqrt{gR}). Using (g = 9.8\text{ m/s}^2) and (R = 20\text{ m}), (v = \sqrt{9.8 \times 20} = \sqrt{196} = 14\text{ m/s}). This speed lies between \(14\text{ m/s}\) and \(15\text{ m/s}\).
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