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Vertical circular motion in uniform circular motion

by seanpk92
Tags: circular, motion, uniform, vertical
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seanpk92
#1
Mar6-11, 10:00 AM
P: 6
1. The problem statement, all variables and given/known data
A pilot flies an airplane in a vertical circular loop with a radius of R = 1200 m. The plane is gaining speed as the pilot makes a dive, and its speed is measured to be 150 m/s when the plane reaches the bottom of the circle. If the total acceleration of the plane at the bottom is 2.3g, find (a) the tangential acceleration of the plane, (b) the direction of the plane’s acceleration at the bottom of the loop.


2. Relevant equations
What I can think of doing, is:
an=v2/R
and
a = an + at

3. The attempt at a solution
I tried making a free body diagram with the total force and an pointing toward the center, and the v and the at point to the right. At the bottom of the circle.
an = 1502/1200 = 18.75m/s2
a = 2.3g = 2.3/9.8m/s2 = 0.2346m/s2

but this doesn't give me my answer. The answer is (a) 12.5 m/s2 and (b) beta = 56.3o

(i dont know how to do a, so I cant start b)
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PhanthomJay
#2
Mar6-11, 10:27 AM
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P: 6,035
Quote Quote by seanpk92 View Post
1. The problem statement, all variables and given/known data
A pilot flies an airplane in a vertical circular loop with a radius of R = 1200 m. The plane is gaining speed as the pilot makes a dive, and its speed is measured to be 150 m/s when the plane reaches the bottom of the circle. If the total acceleration of the plane at the bottom is 2.3g, find (a) the tangential acceleration of the plane, (b) the direction of the plane’s acceleration at the bottom of the loop.


2. Relevant equations
What I can think of doing, is:
an=v2/R
yes, good
and
a = an + at
the tangential and centripetal acceleration vectors are at right angles to each other, so you need to add them vectorially to get the total acceleartaion.

3. The attempt at a solution
I tried making a free body diagram with the total force and an pointing toward the center,
yes
]and the v and the at point to the right.
yes, and the tangential force points in the same direction
At the bottom of the circle.
an = 1502/1200 = 18.75m/s2
a = 2.3g = 2.3/9.8m/s2 = 0.2346m/s2
2.3 g's means the acceleration is 2.3 times the gravitational acceleration
but this doesn't give me my answer. The answer is (a) 12.5 m/s2 and (b) beta = 56.3o

(i dont know how to do a, so I cant start b)
Correct you formula for total acceleration
seanpk92
#3
Mar6-11, 10:47 AM
P: 6
Thank you so much PhantomJay!

PhanthomJay
#4
Mar6-11, 11:01 AM
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Vertical circular motion in uniform circular motion

Quote Quote by seanpk92 View Post
Thank you so much PhantomJay!
(You're) Welcome to PF!


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