Circular Motion - Banked Curve

AI Thread Summary
To determine the banking angle for a curve with a radius of 50 m at a speed of 40 km/h without friction, the initial calculations were based on incorrect speed units. After converting the speed to 11.111 m/s, the formula tan θ = v²/(r*g) was applied correctly. This yielded a value of approximately 0.251, leading to a final angle of 14.11 degrees when using the arctangent function. The revised calculations confirm the correct approach and answer. The discussion highlights the importance of unit conversion in physics problems.
Iser
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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



FR = (m)(aR)
FR = (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 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?
 
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Iser said:
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?
Your thought process is perfect but your answer is not. (Convert the speed to standard units. :wink:)
 
Doc Al said:
Your thought process is perfect but your answer is not. (Convert the speed to standard units. :wink:)

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?
 
Iser said:
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?
Looks good!
 
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