Question #271431

A bike successfully made a turn around a circular path whose radius is 13 m. At what angle is the road banked if it’s velocity was 12.0 m/s?


1
Expert's answer
2021-11-29T11:38:02-0500

Let's apply the Newton's Second Law of Motion in projections on xx- and yy-axis:


Nsinθ=mv2r,Nsin\theta=\dfrac{mv^2}{r},Ncosθ=mg.Ncos\theta=mg.

Dividing the first equation by the second one, we get:


tanθ=v2gr,tan\theta=\dfrac{v^2}{gr},θ=tan1(v2gr),\theta=tan^{-1}(\dfrac{v^2}{gr}),θ=tan1((12.0 ms)29.8 ms2×13 m)=48.5.\theta=tan^{-1}(\dfrac{(12.0\ \dfrac{m}{s})^2}{9.8\ \dfrac{m}{s^2}\times13\ m})=48.5^{\circ}.

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