Banked Circular Motion without friction

AI Thread Summary
The discussion revolves around calculating the banking angle for a roadway designed for vehicles traveling at 28 m/s on a circular arc with a 190 m radius, ensuring no lateral friction is needed. The key equations involve centripetal force and the relationship between normal force and gravitational force. A user initially made an error by incorrectly assuming the normal force equaled mg cos(β) but later corrected this by using a different coordinate system. Another participant suggested that using mg tan(β) could yield better results for finding the angle. The conversation highlights the importance of correctly applying physics principles in solving problems related to banked curves.
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Homework Statement



A roadway is designed for traffic moving at a speed of 28 m s . A curved section of the
roadway is a circular arc of 190 m radius. The roadway is banked so that a vehicle can go
around the curve with the lateral friction forces equal to zero

Homework Equations



F_C = \frac{mv^2}{r}

The Attempt at a Solution



FBD.jpg


<br /> N\sin\beta = \frac{mv^2}{r}<br />
<br /> mg\cos\beta\sin\beta = \frac{mv^2}{r}<br />
<br /> 2\sin\beta\cos\beta = \frac{2v^2}{rg}<br />
<br /> \sin(2\beta) = \frac{2(28)(28)}{(190)(9.8)} <br />

I got the right answer if I didnt assume N = mg\cos\beta

Edit : Nevermind, careless mistake I was using different coordinate systems.
 
Last edited:
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hqjb said:

Homework Statement



A roadway is designed for traffic moving at a speed of 28 m s . A curved section of the
roadway is a circular arc of 190 m radius. The roadway is banked so that a vehicle can go
around the curve with the lateral friction forces equal to zero

Homework Equations



F_C = \frac{mv^2}{r}

The Attempt at a Solution



FBD.jpg


<br /> N\sin\beta = \frac{mv^2}{r}<br />
<br /> mg\cos\beta\sin\beta = \frac{mv^2}{r}<br />
<br /> 2\sin\beta\cos\beta = \frac{2v^2}{rg}<br />
<br /> \sin(2\beta) = \frac{2(28)(28)}{(190)(9.8)} <br />

I got the right answer if I didnt assume N = mg\cos\beta

Edit : Nevermind, careless mistake I was using different coordinate systems.

I assume you found that using <br /> mg\tan\beta = \frac{mv^2}{r}<br /> was more fruitfull?
 
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