Angle for a banked curve (friction-less surface)

[physgrl]

Homework Statement

At what angle should the roadway on a curve with a 50 m radius be banked to allow cars to negotiate the curve at 12 m/s even if the roadway is frictionless?

a. 0°
b. 12.2°
c. 17.1°
d. 35.0°
e. 73.2°

Homework Equations

β=arctan(v^2/r*g)

The Attempt at a Solution

im using β=arctan(v^2/r*g) but the key says the answer is b?
can anyone help me figure out why its not 16.4??

 

[grzz]

Why not try to derive the formula?

 

[physgrl]

i did...i though of the triangle as if the gravity was pointing downward the centripetal force outward and then the angle should be the same as the angle for the curve right?

 

[dacruick]

If your acceleration is towards the centre the force will be too. You need to find the angle of the road that gives a normal force with the horizontal component equal to the force needed to keep the circular motion described by v = 12 m/s and r = 50m

I'm aware that my response is a mouthful , feel free to ask me for clarification.

 

[physgrl]

do you mean like a triangle with the vertical 9.81*m and the horizontal (12^2/50)*m ??

 

[dacruick]

do you mean like a triangle with the vertical 9.81*m and the horizontal (12^2/50)*m ??

Yes. But the horizontal component will be a fraction of the normal force, which depends on the angle.

 

[I like Serena]

Welcome to PF, physgrl!

I derived the same formula as you have.
When I fill in the values I also get 16.4 degrees.
So I believe your answer is correct (assuming you did not make a typo).

 

[dacruick]

Welcome to PF, physgrl!

I derived the same formula as you have.
When I fill in the values I also get 16.4 degrees.
So I believe your answer is correct (assuming you did not make a typo).

I believe your answer is incorrect, as I got 17.1 as my answer. You should investigate why you are using arctan.

 

[I like Serena]

I believe your answer is incorrect, as I got 17.1 as my answer. You should investigate why you are using arctan.

Here's my interpretation of the forces.

We have:
$$F_{resultant} = {m v^2 \over r}$$
$$\tan \beta = {F_{resultant} \over m g }$$
The OP's formula follows from this.

 

[ehild]

16.4° is correct, the answer key is wrong. It happens quite often.

ehild

 

[dacruick]

Oooh I see where I went wrong. My apologies to all. I just always have trouble with the normal force being more than gravity. Its very unintuitive to me.

 

[physgrl]

Thanks to everyone!

 

[I like Serena]

You're welcome!