Question

race track

A car of mass M = 1500 kg traveling at 65.0 km/hour enters a banked turn covered with ice. The road is banked at an angle theta, and there is no friction between the road and the car's tires. Use g = 9.80 m/s^2 throughout this problem.What is the radius r of the turn if theta = 20.0^circ (assuming the car continues in uniform circular motion around the turn)?

Solutions

Expert Solution

$$ \begin{aligned} &v=\sqrt{\operatorname{Rgtan} \theta}\\ &v=65 \mathrm{~km} / \mathrm{hr}=18.05 \mathrm{~m} / \mathrm{s} \text { appr } \mathrm{x}\\ &\text { squaring eq(1), we get }\\ &\mathrm{v}^{2}=\operatorname{Rgtan} \theta\\ &\mathrm{R}=\frac{v^{2}}{\operatorname{gtan} \theta}=\frac{18.05^{2}}{9.8 \tan 20}=326 / 3.567=91.39 \mathrm{~m} \text { apprx }(\text { radius }) \end{aligned} $$

answered by: Kinkform


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