Centrifugal force

  • Thread starter Telemachus
  • Start date
  • #1
832
30

Homework Statement


I have this exercise which says: An airplane moves along a circular path of 150 meters radius, contained in a vertical plane with constant speed module 200km / h.

a)Determine the force resulting from the interaction between the seat and the pilot, whose mass is 80kg, at the lowest point of the trajectory.

Determine the force that results from the interaction between the seat and the pilot at the highest point, that is, when the pilot is in inverted position.

Well, for the first part I've used the normal acceleration:

a)[tex]a_n=\displaystyle\frac{v^2}{\rho}=\displaystyle\frac{55.5^2}{150}=20.57m/s^2[/tex]

Then: [tex]F=ma_n=80kg20.57m/s^2=1645.6N[/tex]

b)Now this is what I did, I'm not sure about this, and it makes me doubt about the steps I've made before.

[tex]mg+F=ma_n \therefore{F=m(g-a_n)=861,6N}[/tex]

Is this right?

Bye there, thanks for posting.
 
Last edited:

Answers and Replies

  • #2
709
0
I think the word you're looking for is "centripetal", there is no such thing as "centrifugal" force.
 
  • #3
832
30
actually there is: http://en.wikipedia.org/wiki/Centrifugal_force its the force that holds the pilot over the seat when he turns upside down. Its a fictitional force, its a consequence of inertia.

By the way, I think that b is ok, and a) is wrong. the correction for a) would be:

[tex]F=m(a_n+g)[/tex]

If someone can tell me if this is okey I'd be really grateful.
 
  • #4
709
0
Correct, a reaction to the centripetal force. The use of the term is discouraged by every lecturer I've had because it confuses people.
"Centrifugal force is often confused with centripetal force. Centrifugal force is most commonly introduced as a force associated with describing motion in a non-inertial reference frame, and referred to as a fictitious or inertial force (a description that must be understood as a technical usage of these words that means only that the force is not present in a stationary or inertial frame).[1][2] There are three contexts in which the concept of the fictitious force arises when describing motion using classical mechanics.[3] In the first context, the motion is described relative to a rotating reference frame about a fixed axis at the origin of the coordinate system. For observations made in the rotating frame, all objects appear to be under the influence of a radially outward force that is proportional to the distance from the axis of rotation and to the square of the rate of rotation (angular velocity) of the frame. The second context is similar, and describes the motion using an accelerated local reference frame attached to a moving body, for example, the frame of passengers in a car as it rounds a corner.[3] In this case, rotation is again involved, this time about the center of curvature of the path of the moving body. In both these contexts, the centrifugal force is zero when the rate of rotation of the reference frame is zero, independent of the motions of objects in the frame.[4] The third context is the most general, and subsumes the first two, as well as stationary curved coordinates (e.g., polar coordinates)[5][6][7][8], and more generally any system of abstract coordinates as in the Lagrangian formulation of mechanics."
 
  • #5
709
0
Ps
regarding the "actual" question.
a_c=v^2/R
F=ma and at the instant the plane is at the top of the loop a_c=a
so F/m=v^2/R
F=mv^2/R Then you have the weight of the pilot to take into account as well.

anyways that all works out to your answer and F=mv^2/R will always work for these questions instead of essentially deriving it yourself.
 
  • #6
832
30
What about the gravity?
 
  • #7
709
0
What about it?
F=80*55.55^2/150=1646.1
w=ma =80*9.81=784.8
So F_net=1646.1-784.8=861.29N
and if you wanted to know the force when it was at the bottom of the loop all you would have to do is
F_net=1646.1+784.8=2430.89N

It all works out.
 
  • #8
709
0
and I think your original for part a was correct and your correction is wrong
 
  • #9
832
30
I think you've used the correction actually :P
 
  • #10
709
0
how do you mean? not unless I misinterpreted what your "correction" was.
 
  • #11
832
30
When you calculated the force you've made exactly this: [tex]F=m(a_n+g)[/tex]
 
  • #12
709
0
Sorry, yeah i did but i was thinking about b).
 
  • #13
832
30
This is b)

pat666 said:
What about it?
F=80*55.55^2/150=1646.1
w=ma =80*9.81=784.8

Its my mistake anyway, cause I've forgotten to say which one was b) :P

I'll correct the original post.

Oh, I can't edit now. Anyway, this was b): Determine the force that results from the interaction between the seat and the pilot at the highest point, that is, when the pilot is in inverted position.
 

Related Threads on Centrifugal force

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
605
  • Last Post
Replies
1
Views
6K
Top