A cyclist on a level road takes a sharp tun of radius 3 m (`g = 10 ms^(-2)`). If the coefficient of between the cycle tyres and the road is 0.2 following speeds will the cyclist not skid while turn?

Options

A. `14.4 km h^(-1)`
B. `8.8 km h^(-1)`
C. `9 km h^(-1)`
D. `10.8 km h^(-1)`

Explanation

To solve the problem of determining the maximum speed at which a cyclist can take a sharp turn without skidding, we can follow these steps:

### Step 1: Identify the given values
- Radius of the turn, \( R = 3 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Coefficient of friction, \( \mu = 0.2 \)

### Step 2: Understand the forces involved
When the cyclist takes a turn, the required centripetal force to keep the cyclist moving in a circular path is provided by the frictional force between the tires and the road. The centripetal force \( F_c \) can be expressed as:
\[ F_c = \frac{mv^2}{R} \]
where \( m \) is the mass of the cyclist and \( v \) is the speed.

### Step 3: Calculate the maximum frictional force
The maximum frictional force \( F_f \) that can act without skidding is given by:
\[ F_f = \mu mg \]
where \( m \) is the mass of the cyclist and \( g \) is the acceleration due to gravity.

### Step 4: Set the forces equal for no skidding
For the cyclist not to skid, the maximum frictional force must be equal to or greater than the required centripetal force:
\[ \mu mg \geq \frac{mv^2}{R} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g \geq \frac{v^2}{R} \]

### Step 5: Rearrange the equation to find \( v \)
Rearranging the equation gives:
\[ v^2 \leq \mu g R \]
Taking the square root of both sides, we find:
\[ v \leq \sqrt{\mu g R} \]

### Step 6: Substitute the values
Now, substituting the known values into the equation:
\[ v \leq \sqrt{0.2 \times 10 \, \text{m/s}^2 \times 3 \, \text{m}} \]
Calculating the right-hand side:
\[ v \leq \sqrt{0.2 \times 10 \times 3} = \sqrt{6} \]

### Step 7: Calculate the maximum speed
Calculating \( \sqrt{6} \):
\[ v \leq 2.45 \, \text{m/s} \]

### Step 8: Convert to km/h
To convert the speed from m/s to km/h, we use the conversion factor \( 1 \, \text{m/s} = 3.6 \, \text{km/h} \):
\[ v \leq 2.45 \, \text{m/s} \times \frac{18}{5} \approx 8.81 \, \text{km/h} \]

### Final Answer
The maximum speed at which the cyclist can take the turn without skidding is approximately **2.45 m/s** or **8.81 km/h**.

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