# Circular Motion - Banked Curve

1. Nov 27, 2013

### Iser

1. The problem statement, all variables and given/known data

Designing an on ramp for the 401 the engineer wants cars to be able to make the turn with a radius of 50 m while travelling 40km/hr in conditions with no friction. What angle must he bank the curve at to make this possible?

2. Relevant equations

FR = (m)(aR)
FR = (m)(v2)/(r)

3. The attempt at a solution

I first started by drawing a diagram and noting the values I have for the question which are the velocity (40 km/h) and the radius (50 m). The value I we're looking for in the question is the angle the bank makes.

I first know FN(Sin θ) = (m)(v2)/R

which I can then isolate FN through which becomes

FN = (m*g)/(cos θ)

Since in the actual equation has no movement in the y axis it is:
FNsinθ = m * v2/r

Then I sub in the FN I got from the previous equation and sub it into this one getting:

(m*g/cos θ)(sin θ) = m * v2/r

which then simplifies into:

tan θ = v2/r*g

I then put in the original values I had into the equation and got 3.26 then I put 3.26 into the tan-1 thing in the calculator and got 73o degrees.

Is my answer and thought process correct?

2. Nov 27, 2013

### Staff: Mentor

Your thought process is perfect but your answer is not. (Convert the speed to standard units. )

3. Nov 28, 2013

### Iser

Converted 40 km/h to 11.111 m/s.

Now I input the following into the formula:

tan θ = (11.1 m/s)2/(50)(9.8)

Which equals 0.2514489...

I then input that into the tan-1 function into the calculator and get:

14.11o degrees

Is this correct?

4. Nov 28, 2013

Looks good!