A car round on unbanked curve of radius 92 m without skidding at a speed of `26 ms ^(-1)` . The smallest possible coefficient of static friction between the tyres and the road is

Options

A. 0.75
B. `0.60`
C. 0.45
D. `0.30`

Answer

To find the smallest possible coefficient of static friction between the tyres and the road for a car moving on an unbanked curve, we can follow these steps:

### Step 1: Understand the Forces Acting on the Car
When a car is moving in a circular path, it experiences a centripetal force that keeps it moving in that circle. This force is provided by the friction between the tyres and the road. The frictional force must be equal to the required centripetal force to prevent skidding.

### Step 2: Write the Equation for Centripetal Force
The centripetal force \( F_c \) required to keep the car moving in a circle is given by the formula:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( m \) = mass of the car (which will cancel out later),
- \( v \) = speed of the car (26 m/s),
- \( r \) = radius of the curve (92 m).

### Step 3: Write the Equation for Frictional Force
The frictional force \( F_f \) that prevents the car from skidding is given by:
\[ F_f = \mu mg \]
where:
- \( \mu \) = coefficient of static friction,
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 4: Set the Forces Equal
Since the frictional force provides the necessary centripetal force, we can set the two equations equal to each other:
\[ \frac{mv^2}{r} = \mu mg \]

### Step 5: Cancel Out the Mass
We can cancel the mass \( m \) from both sides of the equation:
\[ \frac{v^2}{r} = \mu g \]

### Step 6: Solve for the Coefficient of Friction
Rearranging the equation to solve for \( \mu \):
\[ \mu = \frac{v^2}{rg} \]

### Step 7: Substitute the Values
Now, substitute the known values into the equation:
- \( v = 26 \, \text{m/s} \)
- \( r = 92 \, \text{m} \)
- \( g = 9.8 \, \text{m/s}^2 \)

\[ \mu = \frac{(26)^2}{(92)(9.8)} \]

### Step 8: Calculate the Value
Calculating the numerator:
\[ (26)^2 = 676 \]
Calculating the denominator:
\[ 92 \times 9.8 = 901.6 \]
Now, substituting these values:
\[ \mu = \frac{676}{901.6} \approx 0.75 \]

### Final Answer
The smallest possible coefficient of static friction between the tyres and the road is approximately \( \mu \approx 0.75 \).

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