Circular Motion - Banked Curve

[Iser]

Homework Statement

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 traveling 40km/hr in conditions with no friction. What angle must he bank the curve at to make this possible?

Homework Equations

F_R = (m)(a_R)
F_R = (m)(v²)/(r)

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 F_N(Sin θ) = (m)(v²)/R

which I can then isolate F_N through which becomes

F_N = (m*g)/(cos θ)Since in the actual equation has no movement in the y-axis it is:
F_Nsinθ = m * v²/r

Then I sub in the F_N 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 73^o degrees.

Is my answer and thought process correct?

 

[Doc Al]

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 73^o degrees.

Is my answer and thought process correct?

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

 

[Iser]

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

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:

14.11^o degrees

Is this correct?

 

[Doc Al]

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:

14.11^o degrees

Is this correct?

Looks good!

 

[Nickie]

Your thought process is correct, but your answer is incorrect. The angle of banking should be 14.5 degrees, not 73 degrees. This can be verified by plugging in the values into the equation you derived: tan θ = v2/r*g.

tan θ = (40 km/h)^2 / (50 m * 9.8 m/s^2)

tan θ = 1600 km^2/h^2 / 490 m^2/s^2

tan θ = 3.2653

θ = tan^-1(3.2653)

θ = 74.5 degrees

Therefore, the correct angle of banking for the curve is approximately 14.5 degrees. This is important for ensuring that cars can safely make the turn without slipping or losing control due to the centrifugal force acting on them. It also minimizes the wear and tear on the tires.