Centripetal Acceleration of curved exit ramp

[rash219]

Centripetal Acceleration !

Homework Statement

An Engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. He does so by banking the road in such a way that the force causing the centripetal acceleration will be supplied by the component of the normal force toward the center of the path
a. Show that for a given speed v and radius r the curve must be banked at an angle \Theta such that tan\Theta = v^2/r * g

Homework Equations

a_c (centripetal acceleration) = V^2 / r
\SigmaF_y = m * a = 0

The Attempt at a Solution

i hope this diag. makes sense to you...

According to the question a_c = n * Sin\Theta ---- (1)

Then

\SigmaF_y = m * a = 0
(n * Cos \Theta) - (m * g) = 0
n = (m * g) / (Cos \Theta) -------- (2)

substitute 2 in 1 for n

a_c = (m * g) / (Cos \Theta) * Sin\Theta
= (m * g) Tan \Theta

now a_c (centripetal acceleration) = V^2 / r

therefore (V^2 / r) = (m * g) Tan \Theta

and Tan \Theta = (V^2) /(m * g * r)

what am i doing wrong ?.?

 

[PhanthomJay]

Check your equation (1) again, you have identified the centripetal force, not the centripetal acceleration.

 

[rash219]

Thanks! worked out right...