An unbanked curve has a radius of 60m. The maximum speed at which a car can make a turn if the coefficient of static friction is `0.75` , is

Options

A. `2.1m//s`
B. `14m//s`
C. `21m//s`
D. `7m//s`

Explanation

To find the maximum speed at which a car can make a turn on an unbanked curve, we can use the following steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Radius of the curve (R) = 60 m
- Coefficient of static friction (μ) = 0.75
- Acceleration due to gravity (g) = 9.8 m/s² (we will use this value for more accurate calculations)

2. **Understand the Forces at Play**:
- When a car is turning, the frictional force provides the necessary centripetal force to keep the car moving in a circular path.
- The maximum frictional force (F_friction) can be calculated using the formula:
\[ F_{\text{friction}} = \mu \cdot N \]
- Here, N (the normal force) is equal to the weight of the car (Mg), where M is the mass of the car.

3. **Set Up the Equation**:
- The centripetal force required to keep the car moving in a circle is given by:
\[ F_{\text{centripetal}} = \frac{M \cdot V_{\text{max}}^2}{R} \]
- At maximum speed, the frictional force equals the centripetal force:
\[ \mu \cdot Mg = \frac{M \cdot V_{\text{max}}^2}{R} \]

4. **Cancel Out the Mass (M)**:
- Since M appears on both sides of the equation, we can cancel it out:
\[ \mu g = \frac{V_{\text{max}}^2}{R} \]

5. **Rearrange the Equation to Solve for V_max**:
- Rearranging gives us:
\[ V_{\text{max}}^2 = \mu g R \]
- Taking the square root of both sides:
\[ V_{\text{max}} = \sqrt{\mu g R} \]

6. **Substitute the Known Values**:
- Now, substituting the known values into the equation:
\[ V_{\text{max}} = \sqrt{0.75 \cdot 9.8 \cdot 60} \]

7. **Calculate the Value**:
- First calculate the product:
\[ 0.75 \cdot 9.8 \cdot 60 = 441 \]
- Now take the square root:
\[ V_{\text{max}} = \sqrt{441} = 21 \text{ m/s} \]

8. **Final Answer**:
- The maximum speed at which the car can make a turn is **21 m/s**.