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

Options

A. Doubled
B. Quadrupled
C. Halved
D. unchanged

Answer

To solve the problem, we need to analyze the relationship between the speed of the cyclist and the tendency to overturn while turning around a curve. The tendency to overturn is related to the centrifugal force acting on the cyclist.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The cyclist is initially moving at a speed \( v = 15 \) miles/hour.
- The radius of the curve is denoted as \( r \).
- The mass of the cyclist and the bicycle is denoted as \( m \).

2. **Centrifugal Force Calculation**:
- The centrifugal force \( F_c \) acting on the cyclist while turning is given by the formula:
\[ F_c = \frac{mv^2}{r} \]
- Here, \( v \) is the speed of the cyclist.

3. **Doubling the Speed**:
- If the cyclist doubles his speed, the new speed \( v' \) becomes:
\[ v' = 2v = 2 \times 15 \text{ miles/hour} = 30 \text{ miles/hour} \]

4. **Calculating the New Centrifugal Force**:
- The new centrifugal force \( F'_c \) when the cyclist is moving at the new speed \( v' \) is:
\[ F'_c = \frac{m(v')^2}{r} = \frac{m(2v)^2}{r} = \frac{m \cdot 4v^2}{r} = 4 \cdot \frac{mv^2}{r} \]
- This shows that the new centrifugal force is four times the initial centrifugal force:
\[ F'_c = 4F_c \]

5. **Conclusion**:
- The tendency to overturn, which is proportional to the centrifugal force, increases by a factor of 4 when the cyclist doubles his speed.

### Final Answer:
The tendency to overturn is **4 times greater** when the cyclist turns at double the speed.

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