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

Question

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

Part A

Suppose the ring rotates once every 4.70 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?

Part B

Suppose the ring rotates once every 4.70 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?

Part C

Explanation / Answer

The centripetal accelerating force is always towards the centre of the ring.
But the apparent centrifugal force experienced by the passenger is in the opposite direction.
So its a downward force assisting gravity at the bottom of the ride, and upward against gravity at the top.
Tangential velocity v = circumference / rotation period T
When T = 4.7s, v = pi*17m / 4.7s = 11.3 m/s
The apparent centrifugal force is F = m*v^2/r = 59kg * (11.3m/s)^2 / 8.5m = 886.31N
The force of gravity is f = mg = 59kg * 9.81m/s^2 = 579N

So at the top of the ride, net force = 886.31N - 579N = 307.31N
At the bottom of the ride, net force = 886.31N + 579N = 1465.31N

Minimum centrifugal force required = force of gravity = 579N
So m*v^2/r = 579N
v = 9.65m/s = minimum tangential velocity to prevent 59kg riders falling off at the top
v = pi*d / T
So the maximum rotation period T = pi*17m / 9.65m/s = 5.53 seconds

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