Circular motion- finding angle of the banking

[wxwolf]

Homework Statement

A banked circular highway curve is designed for traffic moving at 60km/h the radius of the curve is 200m traffic is moving along the highway at 40km/h on a rainy day. what is the minimum coefficient of friction between the tires and the road that will allow cars to take the turn without sliding off the road (assume the cars do not have negative lift)


Im trying to find the angle of the banking first. i used [theta=tangent^-1(v^2/rg)] where r is the radius, v is the velocity and g is gravity.

when i do the problem out though:

=tan^-1(60km/h)^2/(.2km)(9.8m/s^2)
=tan^-1(3600)/(1.96)
=tan^-1(1836.73)
=89.96

there is no way that banking is almost 90 degrees. Where am i going wrong?

once i do that, I can plug it in and solve the problem for the kinetic coefficient.

 

[Doc Al]

You're messing up with units. Use standard units for speed (m/s) and distance (m).

 

[baba89]

First of all, it is important to note that the formula you are using to find the angle of banking assumes that the curve is frictionless. In reality, there will always be some friction between the tires and the road, so the actual angle of banking will be less than 90 degrees. This could be the reason why your calculation is giving you an unrealistic result.

To find the correct angle of banking, you need to consider the forces acting on the car. In this case, the centripetal force (F=mv^2/r) is balanced by the component of the normal force (N) acting towards the center of the curve. This can be represented by the equation N=mgcos(theta), where theta is the angle of banking.

Using this equation, we can rearrange it to solve for theta: theta=cos^-1(v^2/rg)

Plugging in the values given in the problem, we get: theta=cos^-1((40km/h)^2/(200m)(9.8m/s^2))=cos^-1(0.0816)=83.5 degrees.

This is a more realistic angle of banking for a highway curve and can be used to solve for the minimum coefficient of friction.