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On an ideally banked, frictionless curve, the banking angle theta needed for speed v and radius r satisfies which relationship?
Atan(theta) = v^2 / (r*g)
Bsin(theta) = v / (r*g)
Ctan(theta) = r*g / v^2
Dcos(theta) = v^2 * r * g
Answer & Solution
Correct answer: A. tan(theta) = v^2 / (r*g)
1. On a frictionless banked curve, the horizontal component of the normal force supplies the centripetal force and the vertical component balances the car's weight.
2. Dividing the horizontal equation by the vertical equation eliminates both the normal force and the mass, leaving tan(theta) = v^2 / (r*g).
3. This form correctly shows that a larger speed or a smaller radius requires a steeper banking angle.
4. Inverting the ratio, or replacing tangent with sine or cosine, would not follow from dividing the two force-balance equations and gives the wrong dependence on v and r.
_Source: OpenStax College Physics (CC BY 4.0), Ch 6 "Uniform Circular Motion and Gravitation", section 6.3 Centripetal Force_
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