A cyclist turns around a curve at $15 ext{ miles/hour}$. If he turns at double the speed, the tendency to ov

Question

A cyclist turns around a curve at $15 	ext{ miles/hour}$. If he turns at double the speed, the tendency to overturn is:

Accepted Answer

1. **Analyze the Physics Principle:** When a cyclist navigates a curve, the tendency to overturn (or the need to lean to prevent overturning) is determined by the moment of the centripetal force required to maintain the circular path. To maintain equilibrium, the cyclist bends inward at an angle $	heta$ with the vertical.
2. **Mathematical Relation:** The angle of bending is given by the formula $	an 	heta = \frac{v^2}{rg}$, where $v$ is the speed and $r$ is the radius of the curve [Source 50]. The 'tendency to overturn' is directly proportional to the required centripetal force ($F_c \propto v^2$) or the tangent of the angle of lean.
3. **Apply Given Conditions:** The speed is doubled, so the new speed $v' = 2v$.
4. **Calculate Change:**
   $$ 	ext{Tendency} \propto v^2 $$
   $$ 	ext{New Tendency} \propto (v')^2 = (2v)^2 = 4v^2 $$
   Since the relationship involves the square of the velocity, doubling the speed increases the tendency to overturn by a factor of 4.
5. **Conclusion:** The tendency to overturn is quadrupled.

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