Angular Motion: Calculating Minimum Radius of Curvature for Safe Plane Maneuvers

AI Thread Summary
The discussion revolves around calculating the minimum radius of curvature for a pilot to safely turn a plane upward after a dive, given a speed of 770 km/h and the maximum acceleration of 9g. Participants emphasize the importance of centripetal acceleration and the relevant formula, Fc = mv²/r, to derive the radius. It is noted that mass cancels out in the calculations, simplifying the problem. The final calculated radius is approximately 518.69 meters, and there is a light-hearted exchange about the difficulty of the question. The conversation highlights the value of guiding peers through problem-solving rather than just providing answers.
SS2006
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The human body can safely stand an acceleration 9 times that due to gravity. With what minimum radius of curvature may a pilot sfely turn the plane upward at the end of a dive if the plane's speed is 770km/h.


i did a quesotin before this, which had a roller coaster
i did mgh at a = 1/2 mv square at a + mgh b. my v square was derived
froma free body diagram i drew, and i set mvsquare/r = fnet
and rearranged until i had the equation equal to V.
and then solved for H (from mgh at a) and i got the answer.
For this i don't know wat to do, can i use the same formula, but i don't have both heights, just the first one
i just wanted a startup
 
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The problem has to do with centripetal acceleration.

Can you see how to set up eqns for centripetal acceln/force from the info given ?
 
energy isn't necessary for this question. It's all centripetal motion.
 
reallyy! so this is different from the roller coaster q
im surprised now
what formulas may help
 
somebody...
 
What is the formula for the centripetal force on a body traveling in a circle,of radius R, with an angular velocity of ω ?
 
v = w square tiems r?

or
mv square over r
 
Last edited:
SS2006 said:
v = mv square over r

Ok, this becomes,

Fc = mv²/r

and you want to find r.

What is the maximum value that Fc can be such that the pilot only experiences 9g ?

Then solve for r.
 
kkk i see
what bout mass
can i lave that 1?
 
  • #10
The mass will cancel out.
 
  • #11
ans...518.69 meters
how disgracful of u guys. can't even solve such easy question.
haha...got outbeaten by a 14 year old nutcase
 
  • #12
vijay123 said:
ans...518.69 meters
how disgracful of u guys. can't even solve such easy question.
haha...got outbeaten by a 14 year old nutcase
It is much more productive, if one guides a poster through the question rather than providing answers or solutions. What has SS2006 gained from your post?
 
  • #13
One would hope only confirmation, because if he's still waiting on an answer then his assignment might be a bit late. ;)
 
  • #14
Hawknc said:
One would hope only confirmation, because if he's still waiting on an answer then his assignment might be a bit late. ;)
Didn't spot the date:wink:
 
  • #15
lol...sry guys..i jus did dat for fun...
ok here it goes...
mv(squared)/radius=m9g
if you sub. in the values of the porblem by making r the subject of the formula, i got the ans.
radius=v(sqaured)/9g
were v is velcity, m is mass and g is 9.8ms(sqaured).
remeber to convert the km/h into m/sec.
sry for any hurt feelings
that wus jus for fun,
regards
vijay
 
  • #16
lol, i glanced through the pages and found this, the prob. wus easy yet know one gave a definate ans to the solution. that's were i budged in...sry..guys.
 
  • #17
how we should cancel m by this formule mv^2/r = ma ?
 
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