Banked Circular Motion without friction

  • Thread starter hqjb
  • Start date
  • #1
hqjb
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0

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



[itex]F_C = \frac{mv^2}{r}[/itex]

The Attempt at a Solution



FBD.jpg


[itex]
N\sin\beta = \frac{mv^2}{r}
[/itex]
[itex]
mg\cos\beta\sin\beta = \frac{mv^2}{r}
[/itex]
[itex]
2\sin\beta\cos\beta = \frac{2v^2}{rg}
[/itex]
[itex]
\sin(2\beta) = \frac{2(28)(28)}{(190)(9.8)}
[/itex]

I got the right answer if I didnt assume [itex]N = mg\cos\beta[/itex]

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

Answers and Replies

  • #2
PeterO
Homework Helper
2,436
62

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



[itex]F_C = \frac{mv^2}{r}[/itex]

The Attempt at a Solution



FBD.jpg


[itex]
N\sin\beta = \frac{mv^2}{r}
[/itex]
[itex]
mg\cos\beta\sin\beta = \frac{mv^2}{r}
[/itex]
[itex]
2\sin\beta\cos\beta = \frac{2v^2}{rg}
[/itex]
[itex]
\sin(2\beta) = \frac{2(28)(28)}{(190)(9.8)}
[/itex]

I got the right answer if I didnt assume [itex]N = mg\cos\beta[/itex]

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

I assume you found that using [itex]
mg\tan\beta = \frac{mv^2}{r}
[/itex] was more fruitfull?
 

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