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 of how the tendency to overturn changes when a cyclist doubles his speed while turning around a curve, we can follow these steps:

### Step 1: Understand the Concept of Centrifugal Force
When a cyclist turns around a curve, he experiences a centrifugal force that acts outward from the center of the curve. This force is given by the formula:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( F_c \) is the centrifugal force,
- \( m \) is the mass of the cyclist and the cycle,
- \( v \) is the velocity of the cyclist,
- \( r \) is the radius of the curve.

### Step 2: Calculate the Initial Centrifugal Force
Given that the cyclist's initial speed \( v \) is 15 miles/hour, we can express the initial centrifugal force as:
\[ F_{c1} = \frac{m(15)^2}{r} \]

### Step 3: Determine the New Speed
If the cyclist doubles his speed, the new speed \( v' \) becomes:
\[ v' = 2v = 2 \times 15 = 30 \text{ miles/hour} \]

### Step 4: Calculate the New Centrifugal Force
Now, we can calculate the new centrifugal force at the doubled speed:
\[ F_{c2} = \frac{m(v')^2}{r} = \frac{m(30)^2}{r} \]
\[ F_{c2} = \frac{m(900)}{r} \]

### Step 5: Relate the Two Forces
Now, we can compare the initial and new centrifugal forces:
\[ \frac{F_{c1}}{F_{c2}} = \frac{\frac{m(15)^2}{r}}{\frac{m(30)^2}{r}} \]
This simplifies to:
\[ \frac{F_{c1}}{F_{c2}} = \frac{15^2}{30^2} = \frac{225}{900} = \frac{1}{4} \]

### Step 6: Determine the Change in Tendency to Overturn
From the above ratio, we can see that:
\[ F_{c2} = 4F_{c1} \]
This means that the new centrifugal force (and thus the tendency to overturn) is four times greater than the initial centrifugal force.

### Conclusion
Therefore, when the cyclist doubles his speed, the tendency to overturn increases by a factor of 4, which can be described as "quadrupling."

### Final Answer
The tendency to overturn is quadrupled.

---