Finding the force on an amusement park ride

[jheld]

Homework Statement

In an amusement park ride called The Roundup, passengers stand inside a 16-m diameter ring. After the ring has acquired sufficient speed, it tilts into a vertical plane (figure shown in book).
A Suppose the ring rotates once every 4.5 s. If a rider's mass is 55 kg, with how much force does the ring push on her at the top of the ride? At the bottom?
B. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

Homework Equations

F = ma = mv^2/r
v = 2pi*r/t
d = 16, therefore r = 8
t = 4.5 s
m = 55 kg

The Attempt at a Solution

v = 2($$\Pi$$(8)/4.5 = 11.17 m/s
I think that in itself is correct.

F = 55(11.17^2)/8 = 857.8 N, but this is the wrong answer.

I'm just wondering where I am going wrong.

 

[LowlyPion]

Homework Statement

In an amusement park ride called The Roundup, passengers stand inside a 16-m diameter ring. After the ring has acquired sufficient speed, it tilts into a vertical plane (figure shown in book).
A Suppose the ring rotates once every 4.5 s. If a rider's mass is 55 kg, with how much force does the ring push on her at the top of the ride? At the bottom?
B. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

Homework Equations

F = ma = mv^2/r
v = 2π *r/t
d = 16, therefore r = 8
t = 4.5 s
m = 55 kg

The Attempt at a Solution

v = 2 π (8)/4.5 = 11.17 m/s
I think that in itself is correct.

F = 55(11.17^2)/8 = 857.8 N, but this is the wrong answer.

I'm just wondering where I am going wrong.

For a) you calculated the radial acceleration OK but didn't subtract the weight m*g to get your total force.

For b) if you subtract at the top, then you add at the bottom.

Finally, at what speed will the force at the top balance to 0? When is the speed such that m*g = mv²/r

 

[baba89]

I would approach this problem by first identifying the forces acting on the rider. At the top of the ride, the rider experiences two forces: the gravitational force pulling her down and the normal force from the ring pushing her up. At the bottom of the ride, the rider also experiences two forces: the gravitational force pulling her down and the centripetal force from the rotation of the ring pushing her towards the center.

To find the force at the top of the ride, we can use the equation F = ma, where F is the net force on the rider, m is the mass of the rider, and a is the acceleration. In this case, the net force is equal to the normal force, since the gravitational force is balanced by the normal force. So we can rewrite the equation as F = N = ma. We can also use the equation for centripetal force, F = mv^2/r, and substitute in the values we know. This gives us N = mv^2/r = (55 kg)(11.17 m/s)^2/8 m = 857.9 N.

At the bottom of the ride, the net force is equal to the centripetal force, so we can use the same equation as before, F = mv^2/r, to find the force. This gives us F = (55 kg)(11.17 m/s)^2/8 m = 857.9 N.

To find the longest rotation period that will prevent the riders from falling off at the top, we can use the equation for centripetal force again, but this time we rearrange it to solve for the velocity, v = √(Fr/m). The longest rotation period will occur when the velocity is at its minimum, so we can plug in the maximum force (the weight of the rider) and the minimum radius (the diameter of the ring) to get the minimum velocity. This gives us v = √(mg/r) = √(55 kg)(9.8 m/s^2)/8 m = 8.27 m/s. We can then use the equation v = 2πr/t to solve for the longest rotation period, t = 2πr/v = 2π(8 m)/8.27 m/s = 6.04 s. This means that if the rotation period is longer than 6.04 seconds, the riders will not have enough centrip