Centripetal Force of park ride

In summary, the conversation is about a question regarding a cylindrical ride at an amusement park and the calculation of the radius. The person has provided their attempted solution and is seeking help in finding the error. The correct formula is mentioned and the person realizes their mistake and thanks the other person for their help.
  • #1
rockmorg
22
0
I have a question on what it is I might be doing wrong here... I've tried this answer in the grader and it is wrong...

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 3.2 m/s and an 83 kg person feels a 495 N force pressing against his back. Radius of the chamber?

Here is what I have done..

v = 3.2 m/s
m = 83 kg
F = 495 N
r = ?

Fc = mv2/r
r = Fc(1/mv2)

r = 495 N (1/(83kg * 3.2 m/s2))
r = .582 m

Even just looking at it it looks wrong, heh...

Any help would be great!

Thanks,
-
Morgan
 
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  • #2
Notice this:

[tex] F = m \frac{v^2}{r} [/tex]

[tex] \frac{r}{F} F = \frac{r}{F} m \frac{v^2}{r} [/tex]

[tex] r = m \frac{v^2}{F} [/tex]
 
  • #3
I was just going to say that it looks like you didn't square the velocity.
 
  • #4
Ahhh thanks much - so I guess it was just a matter of arranging my equation? The mass needs to multiply out to those other variables for it to work...
 

1. What is centripetal force?

Centripetal force is the force that acts on an object moving in a circular path, towards the center of the circle. It keeps the object moving in a curved path instead of moving in a straight line.

2. How is centripetal force related to park rides?

Park rides, such as roller coasters and carousels, use centripetal force to keep riders moving in a circular motion. The design of the ride and the speed at which it moves creates the necessary centripetal force to keep riders safe and in place.

3. How is centripetal force calculated?

The formula for centripetal force is Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.

4. What factors affect the centripetal force of a park ride?

The centripetal force of a park ride is affected by the mass of the riders, the speed of the ride, and the radius of the circular path. A larger mass or higher speed will require a greater centripetal force, while a larger radius will require a smaller centripetal force.

5. How does centripetal force impact the safety of park rides?

Centripetal force is essential for the safety of park rides. If the force is not strong enough, riders may slide or fall off the ride. On the other hand, if the force is too strong, riders may experience discomfort or injury from the excessive force. Therefore, it is crucial for park ride designers to carefully calculate and design for the appropriate amount of centripetal force.

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