Airplane flying in a horizontal circle

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SUMMARY

An airplane flying in a horizontal circle at a speed of 410 km/h with wings tilted at an angle of 42° requires specific calculations to determine the radius of the circle. The relevant equations include centripetal acceleration (Ac = v² / r) and Newton's second law (F = ma). The solution involves balancing the vertical and horizontal components of aerodynamic lift, leading to the formula r = v² / (tan a * g). After substituting the values, the calculated radius is approximately 19,000 meters.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula.
  • Knowledge of forces acting on an airplane, specifically aerodynamic lift.
  • Familiarity with trigonometric functions, particularly tangent.
  • Basic physics concepts including Newton's laws of motion.
NEXT STEPS
  • Study the derivation of centripetal acceleration in circular motion.
  • Learn about the principles of aerodynamic lift and its components.
  • Explore the application of trigonometry in physics problems.
  • Review unit conversion techniques in physics calculations.
USEFUL FOR

Students preparing for physics exams, particularly those focusing on mechanics and circular motion, as well as educators teaching these concepts in a classroom setting.

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Homework Statement



An airplane is flying in a horizontal circle at a speed of 410 km/h (Fig. 6-41). If its wings are tilted at angle a = 42° to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an “aerodynamic lift” that is perpendicular to the wing surface.



Homework Equations



Ac = v2 / r
F=ma

The Attempt at a Solution


I did this prob with a friend today and then again on my own still didn't get it right... Please help, I have a test tomorrow lol
I am going to use L as the force that's providing the lift.
and a as the angle.
vertical component of L has to equal to mg or else the plane will fall down.
horizontal component of L is making the plane go in a circle
so L cosa - mg = 0
L sin a = m*v2 /r
Lcos a = mg
L = mg/cos a
plugging back into L sin a = m*v2 /r
i got (mg/cosa)*sina = m*v2/r
the m's cancel
so tan a*g = v2/r
r = v2/(tan a *g)
plugging in numbers
i got r = 4102/tan42*9.8
= 168100/8.8239 = 19050.5 round to 3 significant figures, 19000
please look over and correct mistakes
 
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Check your units. Your final expression currently is: r = \frac{(410\,\rm km/hr)^2}{\tan(42^{\circ}) \cdot 9.8 \, \rm m/s^2}. This gives your answer in \frac{\rm km^2 \cdot s^2}{\rm m \cdot hr^2}
 
ah! units are going to be the death of me! thankyou!
 
No prob. It doesn't hurt to put units in your calculations. In physics, quantities without units are meaningless. (Unless they're dimensionless!:smile:)
 

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