Discussion
Find the maximum velocity for the overturn of a car moving on a circular track of radius 100 m. The co-efficient of friction between the road and tyre is 0.2
Answer: D
The maximum speed is 14 m/s.
On an unbanked track the friction supplies the centripetal force: \[\frac{mv^2}{r} \le \mu mg \Rightarrow v_{max} = \sqrt{\mu r g}\]
\[v_{max} = \sqrt{0.2 \times 100 \times 9.8} = \sqrt{196} = 14\ \mathrm{m\,s^{-1}}\]
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