How Does Physics Explain Forces on The Roundup Amusement Park Ride?

[holmeskaei]

Homework Statement

In an amusement park ride called The Roundup, passengers stand inside a 16.0 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 4.10 s. If a rider's mass is 54.0 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 4.10 s. If a rider's mass is 54.0 kg, with how much force does the ring push on her at the bottom of the ride?

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

Homework Equations

Fr=nr+(Fg)r=n+mg=mv^2/r
w=v/r
1 rev=2pi radians
Might be more needed..

The Attempt at a Solution

4.10s/1 rev=1 rev/2pi radians=6.44 rad/s=w
w=v/r=6.44rad/s/8m=51.52m/s
Then do I plug it into the Fr equation? I don't have the acceleration..
Any help or how to think about it would be advantageous.
Thanks.

 

[Delphi51]

Those calcs don't look right to me.
You could do w = 1 rev/4.1s = 2*pi/4.1 = 1.53 radians/s
Or just work with v = 2*pi*r/T.
The n+mg=mv^2/r looks good.

 

[nlightner]

As a scientist, your first step would be to gather all the necessary information and equations related to the problem. The given information includes the rotation period of the ring (4.10 s), the mass of the rider (54.0 kg), and the diameter of the ring (16.0 m). The equations needed to solve this problem are the centripetal force equation (Fr=mv^2/r), the angular velocity equation (w=v/r), and the conversion of revolutions to radians (1 rev=2pi radians).

For Part A, you can use the centripetal force equation to calculate the force exerted on the rider at the top of the ride. Since the rider is at the top, the force will be directed towards the center of the ring. Thus, the centripetal force will be equal to the rider's weight (Fg=mg) plus the force needed to maintain the circular motion (Fr). We can then set this equal to the centripetal force equation and solve for Fr:

Fr=nr+(Fg)r=n+mg=mv^2/r
n=(54.0 kg)(51.52 m/s)^2/8.00 m=2223.24 N
Therefore, the force exerted on the rider at the top of the ride is 2223.24 N.

For Part B, the same process can be used to calculate the force exerted on the rider at the bottom of the ride. However, at the bottom, the force will be directed away from the center of the ring. Thus, the centripetal force will be equal to the rider's weight (Fg=mg) minus the force needed to maintain the circular motion (Fr). We can then set this equal to the centripetal force equation and solve for Fr:

Fr=nr-(Fg)r=n-mg=mv^2/r
n=(54.0 kg)(51.52 m/s)^2/8.00 m=2223.24 N
Therefore, the force exerted on the rider at the bottom of the ride is 2223.24 N.

For Part C, we need to find the longest rotation period that will prevent the riders from falling off at the top. This means that the force exerted on the rider at the top needs to be equal to or greater than the rider's weight (Fg=mg). We can set these two values equal to each other