Circular motion problem no idea what's going wrong ><

[irNewton]

Homework Statement

A highway curve of radius 560m is designed for traffic moving at a speed of 73.0km/hr (1216m/s). What is the correct banking angle of the road?

Homework Equations

a=v^2/r
F=ma

The Attempt at a Solution

Forces in t direction:
Fnet=Fncos[tex]\vartheta[/tex]-Fg=0
Fn=mg/cos[tex]\vartheta[/tex]=0


Forces in r direciton:

Fnetr=Fnsin[tex]\vartheta[/tex]=mv^2/r
mgsin[tex]\vartheta[/tex]/cos[tex]\vartheta[/tex]=mv^2/r
tan[tex]\vartheta[/tex]=v^2/r*g
[tex]\vartheta[/tex]=tan-1(1216^2/(9.81*560))
[tex]\vartheta[/tex]= 90 degrees?

which is wrong... = (

 

[gamer_x_]

double check the way you're assigning your forces, is the Fny component which is cancelling vertical gravity the hypotenuse of the triangle? I find it helpful in these cases to rotate the coordinate system so that the x-axis is actually the sloped road. That way your Fn has only 1 component, and you break down gravity into y & x where x points towards the center of the turn.

 

[tomcue]

It seems that you have made a mistake in your calculations. The correct banking angle can be found using the equation \tan\vartheta=v^2/(rg), where v is the speed, r is the radius, and g is the acceleration due to gravity. Plugging in the given values, we get \tan\vartheta=(1216)^2/(560*9.81)=2.89. Taking the inverse tangent of this value, we get a banking angle of approximately 71.2 degrees. This means that the road should be banked at an angle of 71.2 degrees in order to safely accommodate traffic moving at 73.0km/hr (1216m/s) around the curve.