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Angular accelleration(need verification)

  • Thread starter Susanne217
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Homework Statement



At the buttom of a loop in the vertical (r-theta) plane at the altitude of 400 meters an airplane has a horizontal velocity of 600 km/h (no horizontal accelleration). The radius of the curvature of the loop is 1200 meters.

Find the accelleration of plane in m/s^2 and the angular accelleration of the plane in rad/s^2


The Attempt at a Solution



I assume that the loop is circular and thus the velocity of the plane in the loop is

[tex]v = \frac{2\pi \cdot r}{t} = \frac{2\pi \cdot 1200 m}{3600 sec} = 2.0944 m/s[/tex]

the angular accelleration is the in this case

[tex]v^2/1200 m = 0.00365 rad/s^2[/tex]

In the book its that the angular accelleration is 0.0365 rad/s^2.

what am I doing wrong?

What accelleration of the plane how do I calculate that?

Sincerley
Susanne
 

Answers and Replies

  • #2
tiny-tim
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At the buttom of a loop in the vertical (r-theta) plane at the altitude of 400 meters an airplane has a horizontal velocity of 600 km/h (no horizontal accelleration). The radius of the curvature of the loop is 1200 meters.
Hi Susanne! :smile:

(buttom … is that a pirate spelling? :biggrin:)

Sorry, but you've gone completely loopy on this.

v is given, as 600 km/h,

(your formula for v is actually a formula for the speed if it took an hour to complete the circle :rolleyes:)

and your v2/r is the formula for (linear) centripetal acceleration not angular acceleration. :redface:

Get some sleep :zzz:, and try again! :smile:

(oh, and assuming the speed is constant, "the acceleration of the plane" is the centripetal acceleration)
 

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