Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 5 - Dynamics of Uniform Circular Motion - Problems - Page 140: 47


The maximum speed the cycle can have is $21m/s$.

Work Step by Step

The normal force $F_N$ helps to keep the motorcycle to the road. Therefore, the motorcycle will lose contact with the road when $F_N=0$. $F_N=0$ means only gravitational force contributes to the centripetal force as the cycle passes the hill. In other words, $$mg=F_c=\frac{mv^2}{r}$$ $$\frac{v^2}{r}=g$$ $$v=\sqrt {gr}$$ We have $r=45m$ and $g=9.8m/s^2$. So $$v_{max}=21m/s$$
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