Car on banked turn - net force

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  • #26
Andrew Mason
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beginner16 said:
just one more time -

If banked curve (again no friction )has an angle theta , then only possible force pointing in direction to the center of circle must have a magnitude of m*g*tan(theta) . Right?
Wrong. The force pointing in the direction of the centre of the circle is [itex]Nsin\theta[/itex]. N depends on the speed as well as the angle.

AM
 
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Andrew Mason said:
Wrong. The force pointing in the direction of the centre of the circle is [itex]Nsin\theta[/itex]. N depends on the speed as well as the angle.

AM

I don't understand your reply . N*sin(theta) has same in magnitude as m*g*tan(theta) . If horizontal component of N doesn't equal m*g*tan(theta) then net force wont be directed towards the center of the circle
 
  • #28
Andrew Mason
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beginner16 said:
I don't understand your reply . N*sin(theta) has same in magnitude as m*g*tan(theta).
Only in the case that [itex]mv^2/r = mg tan\theta[/itex]. The point here is that, as a general rule, [itex]Nsin\theta \ne mv^2/r[/itex]. They are equal only where [itex]v = \sqrt{rg tan\theta}[/itex]

If horizontal component of N doesn't equal m*g*tan(theta) then net force wont be directed towards the center of the circle
That is right. It points to the centre of the circle only where the car actually moves in the circular path defined by the curvature of the road. That only occurs for one speed: [itex]v = \sqrt{rg tan\theta}[/itex].

AM
 

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