Circular Motion - Banked Curve

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Homework Help Overview

The problem involves determining the angle at which a banked curve must be designed for vehicles to navigate a turn without friction, given a specific radius and speed. The subject area is circular motion and dynamics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the process of converting speed to standard units and applying the relevant equations for circular motion. There is an exploration of the relationship between the angle of the bank, speed, and radius.

Discussion Status

Some participants have provided guidance on unit conversion and the application of trigonometric functions. There are differing results regarding the final angle, indicating ongoing exploration of the calculations involved.

Contextual Notes

Participants are working under the constraint of no friction and are questioning the correctness of their calculations and assumptions regarding the setup of the problem.

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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