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.

Part A

Suppose the ring rotates once every 4.20 s . If a rider's mass is 58.0 kg , with how much force does the ring push on her at the top of the ride?

_____________________ N

Part B

Suppose the ring rotates once every 4.20 s . If a rider's mass is 58.0 kg , with how much force does the ring push on her at the bottom of the ride?

_____________________N

Part C

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

_____________________s

Answer 1

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 = 4.2s, v = pi*17m / 4.2s = 12.71 m/s
The apparent centrifugal force is F = m*v^2/r = 58kg * (12.7m/s)^2 / 8.5m = 1100.56 N
The force of gravity is f = mg = 58kg * 9.81m/s^2 = 569N

So at the top of the ride, net force = 1100.56N - 569N = 531.56 N
At the bottom of the ride, net force = 1100.56N + 569N = 1669.56 N

Minimum centrifugal force required = force of gravity = 569N
So m*v^2/r = 569 N
v = 9.13m/s = minimum tangential velocity to prevent 58kg riders falling off at the top
v = pi*d / T
So the maximum rotation period T = pi*17m / 9.13m/s = 5.84 seconds