In an amusement park ride called The Roundup, passengers stand inside a 16.0 m -

Question

In an amusement park ride called The Roundup, passengers stand inside a 16.0 m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in the figure (Figure 1) .

1. Suppose the ring rotates once every 4.20 s . If a rider's mass is 59.0 kg , with how much force does the ring push on her at the top of the ride?

2. Suppose the ring rotates once every 4.20 s . If a rider's mass is 59.0 kg , with how much force does the ring push on her at the bottom of the ride?

3. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

Explanation / Answer

Angular velocity = 2 / 4.20s = 1.49 rad/s

Centripetal force required by passenger, Fc = mr²
Fc = 59kg x 8.0m x (1.49)² = 1047.88N

1) At top of ring passenger's weight provides (mg)N of the required centripetal force ..
. The track provides 1047.88N - (59.0kg X 9.80)N ..

. = 469.68 (downward force on passenger)

2) At bottom or ring, the track provides all the centripetal force and supports the passenger's weight..
Total force from track = 1047.88N + (59.0kg x 9.80)N .. = 1626N (upward)

3) Track reaction at the top becomes zero when the centripetal acceleration (r²) equals the free-fall rate of 9.80 m/s².. .. then both track and passenger have the same downward acceleration.

r² = 9.80
² = 9.80 / 8.0m = 1.225 (rad/s)²
= 1.10 rad/s
T = 2 / .. .. T = 2 / 1.10 .. .. T = 5.71 s

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