How Do You Calculate the Flight Radius of a Plane Tilting at 40 Degrees?

[Oerg]

Homework Statement

A plane flies at 480km/h, with wings tilted at 40 degrees to the horizontal. What is the radius of circle in which plane is flying? (hc0108, ignore the contents in this rbacket)

The Attempt at a Solution

I ahve no idea how you can calculate the radius from that, and the solution was given rather shabbily astan\theta=v_p^2/rg

I seriously have no idea how you could calculate the radius by just knowing that its wings are tilted at that angle, what if the plane is a point mass? then the question makes no sense at all.

 

[mgb_phys]

Unless I'm missing something that isn't enough information.

 

[Astronuc]

Well assuming the plane in flying in the horizontal, i.e. neither gaining or losing altitude, the vertical force must equal the weight (mg) of the plane, so that it is still flying. The centripetal force is also related to the mass by F = mv²/r.

It's a bit like a banking car on a frictionless surface, but the differential pressure across the wing provides the centripetal force on plane just as the road provides the centripetal force on the car.

See the frictionless case in - http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html

I recommend drawing a free body diagram and look at the forces involved and determine the relationship between the weight of the plane (mg) and the centripetal force.

One knows g, v_p and theta, so solve for the radius, r.

 

[Oerg]

Ok, let me have a go at this:

If the plane is tilted 40 degrees to the horiziontal, then the upward force is 50 degrees to the horizontal. The centripetal force of acceleration is then given by

F_C=F_Ucos50

The centripetal force of acceleration can then be given by

F_Ucos50=\frac{mv_^2}{r}

which is then

F_Ucos50=\frac{F_Usin50gv^2}{r}

r=tan50gv^2

 

[Kushal]

w = mg
m = W/g = Fu sin 50 / g

so Fu cos 50 = [(Fu sin 50)(v^2)] / rg
r = (v^2 tan 50)/g

 

[Oerg]

w = mg
m = W/g = Fu sin 50 / g

so Fu cos 50 = [(Fu sin 50)(v^2)] / rg
r = (v^2 tan 50)/g

oh my yeh careless mistake, the g should be in the denominator

 

[Oerg]

looks like my suspicion that the question/answer was wrong was not unfounded afterall, the answer is 2160m which was calculated from theta=40 which is wrong. The correct answer should be 1520m.

 

[PhanthomJay]

looks like my suspicion that the question/answer was wrong was not unfounded afterall, the answer is 2160m which was calculated from theta=40 which is wrong. The correct answer should be 1520m.

No, please refer to Astronuc's
post and the site he referenced. This problem is similar to the car on a banked frinctionless roadway. The theta you should be using is 40 degrees, in accord with the book answer. Check your FBD and algebra and trig and equations.

 

[Oerg]

Hi, this is my FBD and i see nothing wrong in it

EDIT: Ok, nvm i see what's wrong here, my fbd and workings are correct, just that my final equation was the reciprocal of his since i used 50.