In an amusement park ride called The Roundup, passengers stand inside a 18.0 m -

Question

In an amusement park ride called The Roundup, passengers stand inside a 18.0m -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 3.50

s . If a rider's mass is 55.0kg , with how much force does the ring push on her at the top of the ride?

2-Suppose the ring rotates once every 3.50

s . If a rider's mass is 55.0kg , 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?

In an amusement park ride called The Roundup, passengers stand inside a 18.0m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in the figure (Figure 1). Suppose the ring rotates once every 3.50 If a rider's mass is 55.0kg , with how much force does the ring push on her at the top of the ride? Suppose the ring rotates once every 3.50 If a rider's mass is 55.0kg , with how much force does the ring push on her at the bottom of the ride? What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

Explanation / Answer

In a verticle circle, tension (T) acting on passengers

T(top) = (m/r)* v^2(top) - mg

T(bot) = (m/r)* v^2(bot) +mg


v (top) = r w = 2 pi r / T = 2*3.14*18/2*3.50 = 16.148 m/s

T (top) = [55/9] * 260.7579 - 55 * 9.8 = 1593.52 - 539
= 1054.52 newtons

it is more than v (min) required to pass top point

conservation of energy gives v (bot) as

0.5 m v^2(bot) + 0 = 0.5 m v^2(top) + mg (2r)

v^2(bot) = v^2(top) + 4 g r = 260.7579 + 352.8 = 613.5579

T(bot) = (m/r)* [613.5579] +mg = 3749.52+539 = 4288.52 newtons

c)part
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v^2(top) = minimum to be in circle = r g and T(top) =0 that is just sufficient to provide acceleration = g at top so that it crosses top point.

v (top) = sqrt [(18/2) * 9.8] = 9.391 m/s

for T (top)=0 just pass without fall
V (top) = 9.391 m/s = r w = 2 pi r / T(period)
T = 2*3.14*18/2*9.391 = 6.018 secs

so it should revolve once in 6.018 secs

V is inversely proportional to period, v (min) T maximum

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