Roller Coaster Physics: Force & Max Speed

Click For Summary
The discussion revolves around calculating the forces and speeds of a roller coaster at two points, A and B. At point A, with a speed of 19.6 m/s and a radius of 10.0 m, the initial force calculation using EF=m(v^2/r) is questioned, indicating a misunderstanding of the net forces involved. For point B, the maximum speed must account for gravitational forces and the curvature of the track, suggesting that point B is convex down. The participants emphasize the importance of correctly identifying the forces acting on the roller coaster at both points to solve the problems accurately. Clarification on the positions of points A and B is also requested to aid in the calculations.
tnhoots
Messages
36
Reaction score
0

Homework Statement


A roller coaster has a mass of 510kg when fully loaded with passengers. The radius at point A is 10.0 m and the radius at point B is 15.0 m.

-If the vehicle has a speed of 19.6 m/s at point A, what is the force exerted by the track on the car at this point? (N)

-What is the maximum speed the vehicle can have at B and still remain on the track? (m/s)



Homework Equations



EF=m (v^2 / r)

The Attempt at a Solution



*For the first problem I thought you would use the equation EF=m (v^2 / r). Which would be EF=510 (19.6^2 / 10.0). Which equals 510*(384.16/10.0). Equals 19592.16. However, this is wrong! Isn't that the right equation?

*For the second problem, I'm not sure where to start. But I'm assuming that I will need the correct force value before I can begin.

Any suggestion as to where I'm going wrong?
 
Physics news on Phys.org
It all depends on the location of points A and B, perhaps if you could post a diagram or describe the position of the two points we could help you further.
 
assuming a is on a curve that is convex up, (try it the other way and you'll find its off the track i think), you have two a accelerations to account for, g and the v^2/r which is 19.6^2/10=38.4. Compute the net a, then you have the force pushing on the track.

b must be convex down (or upside down, but then it would be a minimal speed): v^2/15=g as the two accelerations are opposing.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
6K
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 15 ·
Replies
15
Views
7K
  • · Replies 15 ·
Replies
15
Views
5K