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Centripetal Acceleration for non uniform speed

  • Thread starter henry3369
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  • #1
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Homework Statement


An airplane flies in a loop (a circular path in a vertical plane) of radius 120m . The pilot's head always points toward the center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastest at the bottom.

At the top of the loop, the pilot feels weightless. What is the speed of the airplane at this point?

Homework Equations


F = mv2/r

The Attempt at a Solution


n + w = mv2/r
v = sqrt(gr) = 34.3 m/s

I was able to solve this problem using F = mv2/r, but I'm confused to why it worked. Doesn't this formula only apply to situations in which the speed is constant?
 

Answers and Replies

  • #2
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The formula applies to when you're at a single point on a circular path. For all the equation knows, you could be moving in a circle at that second and a second later you move off in a straight line. It deals with instantaneous conditions. However, under the assumption that the movement is uniform, we can apply it to the whole circular motion (hence the term uniform circular motion).

At the point on the circle, it considers your instantaneous velocity and radius to determine the centripital pull, so it should still work. You will just get different values for v when you are at different points on your circle
 
  • #3
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Henry: You could turn it around. Look at it from the pilot's seat. Newton says no force no acceleration. Also true in the direction perpendicular to the plane floor. In that direction F = ma pins down the second derivative of the coordinate. And that is the curvature of the trajectory. At any point, for non zero speed.
 

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