A circular race track is banked at 45o and has a radius of 40m. If the coefficient of friction between the wheels and the track is 12, find the maximum speed (in m/s) at which the car can travel around the track without skidding.

Normal to plane, N−mgcos45o+mv2rcos45o−m2(g+v2r)∫−μN−12m2(g+v2r) Along the plane, friction + mg sin45o+mv2rcos45o−m2(g+v2r) m22(g+v2r)+mg2−mv22r⇒g2+v22r+g−v2r⇒v22r−3g2 ⇒v2−3gr−3×10×40−1200 ⇒vm:n−1200 m/s

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A circular race track is banked at $45^{o}$ and has a radius of 40m. If the coefficient of friction between the wheels and the track is $\frac{1}{2}$ , find the maximum speed (in m/s) at which the car can travel around the track without skidding.

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Detailed Solution

Normal to plane, $N - m g \cos 45^{o} + \frac{m v^{2}}{r} \cos 45^{o} - \frac{m}{\sqrt{2}} (g + \frac{v^{2}}{r})$
$\int - μ N - \frac{1}{2} \frac{m}{\sqrt{2}} (g + \frac{v^{2}}{r})$
Along the plane, friction + mg $\sin 45^{o} + \frac{m v^{2}}{r} \cos 45^{o} - \frac{m}{\sqrt{2}} (g + \frac{v^{2}}{r})$
$\frac{m}{2 \sqrt{2}} (g + \frac{v^{2}}{r}) + \frac{m g}{\sqrt{2}} - \frac{m v^{2}}{\sqrt{2} r} \Rightarrow \frac{g}{2} + \frac{v^{2}}{2 r} + g - \frac{v^{2}}{r} \Rightarrow \frac{v^{2}}{2 r} - \frac{3 g}{2} \Rightarrow v^{2} - 3 g r - 3 \times 10 \times 40 - 1200 \Rightarrow v_{m : n} - \sqrt{1200} m / s$

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