Centripetal Force in an amusement park ride

In summary, in the amusement park ride The Roundup, passengers stand on a rotating ring with a diameter of 18.0 m. The ring rotates once every 4.90 s and the rider's mass is 58.0 kg. At the top of the ride, the two forces acting on the rider are the normal force and gravity, with the resultant equal to the centripetal force. The periodic time tells us that the angular velocity is 2pi / 4.9, and the formula for centripetal force is mw^2r. With this information, we can solve for the forces at the top and bottom of the ride (556177.07 N). However, the normal force calculation may be incorrect
  • #1
GoSS190
20
0
In an amusement park ride called The Roundup, passengers stand inside a 18.0 m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane. Suppose the ring rotates once every 4.90 s. and the rider's mass is 58.0 kg.

A.) With how much force does the ring push on her at the top of the ride?

B.) 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?

If anyone could help me with these questions that would be great. Thanks


I tried this using the equations to find v = (2pi(r)) / T

but then i realized that the velocity is different at the top than at the bottom. I am stumped as to how to find the velocity then find the force.

I think the equation for force at the top is (m(vtop)2) / R

Can anyone help me out with this though please
 
Last edited:
Physics news on Phys.org
  • #2
At the top of the ring, what are the forces acting on the body? The resultant of those two forces is equal to the centripetal force.
 
  • #3
The two forces are the normal force and gravity but i don't know how to find the normal force
 
  • #4
GoSS190 said:
The two forces are the normal force and gravity but i don't know how to find the normal force

Yes, the are the two forces, the normal force (R) and the weight(W) <-I assumed that is what you meant by gravity.

The resultant of those two R+W=Fc where Fc is the centripetal force.
 
  • #5
How do you find the R or the normal force cause the velocity is different at the top than at the bottom
 
  • #6
You don't need the velocity,the told you that "the ring rotates once every 4.90 s"

What does this tell you about the periodic time and hence the angular velocity?
 
  • #7
it would be 1 / 4.9 right
 
  • #8
yes and how to relate that to angular velocity?


What formulas do you know to calculate centripetal force?
 
  • #9
2pi / 4.9 is the angular velocity
 
  • #10
and centripetal force is therefore?
 
  • #11
That is the part that I don't understand really
 
  • #12
is it mw^2r
 
  • #13
yes it is, you can now solve part a and part b noting that Weight acts downwards and normal reaction acts upwards
 
  • #14
i got an answer of 556177.07 but that doesn't seem correct to me and I don't know what i did wrong
 

Related to Centripetal Force in an amusement park ride

What is centripetal force?

Centripetal force is the force that acts on an object moving in a circular path, directed towards the center of the circle.

How is centripetal force applied in an amusement park ride?

In an amusement park ride, centripetal force is applied through the rotation of the ride, causing passengers to move in a circular path.

What is the relationship between centripetal force and speed in an amusement park ride?

The faster the ride moves, the greater the centripetal force will be. This is because the force required to keep an object moving in a circular path increases with speed.

How is the centripetal force calculated in an amusement park ride?

The centripetal force in an amusement park ride can be calculated using the formula Fc = (mv^2)/r, where m is the mass of the object, v is the velocity, and r is the radius of the circular path.

What are some safety considerations when it comes to centripetal force in amusement park rides?

Operators of amusement park rides must carefully consider the speed and radius of the ride to ensure that the centripetal force is not too strong and does not pose a danger to passengers. Additionally, proper maintenance and inspection of the ride is crucial to ensure that it is functioning safely and effectively.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
7K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
13
Views
9K
Back
Top