In an amusement park ride called The Roundup, passengers stand inside a 17.0 -di

Question

In an amusement park ride called The Roundup, passengers stand inside a 17.0 -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in the figure (Figure 1) .

Part A Suppose the ring rotates once every 5.40 . If a rider's mass is 57.0 , with how much force does the ring push on her at the top of the ride?

Part B Suppose the ring rotates once every 5.40 . If a rider's mass is 57.0 , with how much force does the ring push on her at the bottom of the ride?

Part C

Explanation / Answer

At the highest position of the ring, two forces act on the rider. 1) her weight, mg, downwards and 2) reactive force from the ring, F, downwards. Together they provide the necessary centripetal force, m?²r, for the circular motion, where ? is the angular speed of the ring and r, its radius. => F + mg = m?²r => F = m?²r - mg = 52 [(2p/3.8)² * 9 - 9.8] N = 52 [24.6 - 9.8] N = 769.6 N. If F' = reactive force from the ring upwards, F' - mg = m?²r => F' = m?²r + mg = 52 [(2p/3.8)² * 9 + 9.8] N = 52 [24.6 + 9.8] N = 1788.8 N. For the longest rotation period, F = 0 => m?²r = mg => ? = v(g/r) => 2p/T = v(g/r) => T = 2pv(r/g) = 2pv(9/9.8) s. = 6.02 s.

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