In an amusement park ride called The Roundup, passengers stand inside a 19.0 m -

Question

In an amusement park ride called The Roundup, passengers stand inside a 19.0m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in the figure (Figure 1) .

a)

Suppose the ring rotates once every 4.30s . If a rider's mass is 57.0kg , with how much force does the ring push on her at the top of the ride?

b)

Suppose the ring rotates once every 4.30s . If a rider's mass is 57.0kg , with how much force does the ring push on her at the bottom of the ride?

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

Explanation / Answer

T = 4.3

2pi/w = 4.3

w = 1.4612

m = 57

R = 19/2 = 9.5 m

at the top

gravitation acts downwards = mg = 558.6 N

centrifugal force acts upwards = mw^2 R = 1156.16 N

Net force has to be balanced by the forcde of normal reaction on her

Ring pushes her with a force = 1156.16 - 558.6 = 597.56 N

b)

at the bottm gravitation and centrifugal force both act downward

Ring pushes her with a force = 1156.16 + 558.6 = 1714.76 N

c)

as time period increases w decreases

so centrifugal force decreases

for the rider to not fall

mw^2R > mg

at critical condition

mw^2R = mg

w = sqrt (g/R) = sqrt (9.8/9.5)

T = 2pi/w = 6.09 seconds

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