If a cyclist moving with a speed of 4.9 m/s on a level can take a sharp circular turn of radius 4 m, then coefficient between the cycle tyres and road is

Options

A. 0.41
B. 0.51
C. 0.61
D. 0.71

Explanation

To find the coefficient of friction (μ) between the cycle tires and the road, we can use the relationship between centripetal force and frictional force. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Speed of the cyclist (v) = 4.9 m/s
- Radius of the circular turn (r) = 4 m
- Acceleration due to gravity (g) = 9.8 m/s² (standard value)

### Step 2: Write the formula for centripetal force
The centripetal force (F_c) required to keep an object moving in a circular path is given by:
\[ F_c = \frac{m v^2}{r} \]
where:
- m = mass of the cyclist and cycle (which will cancel out later)
- v = speed of the cyclist
- r = radius of the circular path

### Step 3: Write the formula for frictional force
The frictional force (F_f) that provides the centripetal force is given by:
\[ F_f = \mu \cdot N \]
where:
- μ = coefficient of friction
- N = normal force

### Step 4: Determine the normal force
Since the cyclist is moving on a level surface, the normal force (N) is equal to the weight of the cyclist and cycle:
\[ N = m \cdot g \]

### Step 5: Set centripetal force equal to frictional force
Since the frictional force provides the necessary centripetal force, we can set them equal to each other:
\[ \frac{m v^2}{r} = \mu \cdot m \cdot g \]

### Step 6: Cancel mass (m) from both sides
Since mass (m) appears on both sides of the equation, we can cancel it out:
\[ \frac{v^2}{r} = \mu \cdot g \]

### Step 7: Solve for the coefficient of friction (μ)
Rearranging the equation gives:
\[ \mu = \frac{v^2}{r \cdot g} \]

### Step 8: Substitute the known values
Now, we can substitute the known values into the equation:
\[ \mu = \frac{(4.9)^2}{4 \cdot 9.8} \]

### Step 9: Calculate the values
Calculating the numerator:
\[ (4.9)^2 = 24.01 \]

Calculating the denominator:
\[ 4 \cdot 9.8 = 39.2 \]

Now substituting these values:
\[ \mu = \frac{24.01}{39.2} \]

### Step 10: Perform the division
Calculating the final value:
\[ \mu \approx 0.612 \]

### Step 11: Round to two decimal places
Rounding gives us:
\[ \mu \approx 0.61 \]

### Conclusion
The coefficient of friction (μ) between the cycle tires and the road is approximately **0.61**.