About -- Forum -- Articles -- Tutorials -- Books -- Apparel -- Contact

You are not logged in.

Announcement

#1 2007-11-29 00:59:49

nick227
New Member
Registered: 2007-11-29
Posts: 3

cup inside a plane

Hey guys, I have this question for homework. The chapter is on non inertial reference frames. I think I have the answer, but it doesn't seem right. Can someone tell me if i'm doing it right?

There is a plane moving in a circle with radius 8km. The plane is moving at a velocity of 350km/hr. Inside the plane, there is a cup on a table that is 3m off of the ground. If the cup falls down, what is its lateral displacement?

WORK:

350km/hr = 97m/s

v=rLaTeX Image

97=8000w

w=.01215rad/sec

Finding the time the cup falls:

y=v0+(1/2)at^2

v0=0 - initial velocity

t=.78

The time at which the cup falls, the plane travels a certain length. you can find the angle between where the plane was and there it will be.

LaTeX Image=LaTeX Imaget

=.00977rad = .54299(degrees)

If you make that into a triangle with one angle being .54299 degrees. Then split the triangle in half, the hypotenuse is 8000m. The side across the is:
sin(.54299)*(8000) = 75.81m

so that the linear displacement of the plane. I said that was the same thing as the cup's horizontal displacement. Then did the pythagorean theorem to get the lateral displacement.

(3^2)+(75.81^2)=x^2

x=75.87m
thats the answer. did i just completely do it wrong?

Offline

 

#2 2007-11-30 08:01:31

Martin
Moderator
From: Earth
Registered: 2004-10-04
Posts: 369

Re: cup inside a plane

The cup's lateral displacement is the length of the chord connecting the two points on the plane's circular trajectory.


The truth is out there.

Offline

 

#3 2007-12-10 04:05:15

nick227
New Member
Registered: 2007-11-29
Posts: 3

Re: cup inside a plane

great! thanks for looking at the problem.

Offline

 

Board footer

Powered by PunBB
© Copyright 2002–2005 Rickard Andersson



Copyright © J. Christopher Moore Publishing, All Rights Reserved