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. `2.1ms^(-1)`
B. `14ms^(-1)`
C. `21ms^(-1)`
D. `7ms^(-1)`

Answer

To find the maximum speed at which a car can make a turn on an unbanked curve, we can use the relationship between centripetal force and friction. Here’s a step-by-step solution:

### Step 1: Identify the forces acting on the car
On an unbanked curve, the only force providing the necessary centripetal force to keep the car moving in a circle is the frictional force.

### 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:
\[ F_c = \frac{mV^2}{R} \]
where:
- \( m \) is the mass of the car,
- \( V \) is the speed of the car,
- \( R \) is the radius of the curve.

### Step 3: Write the equation for frictional force
The maximum static frictional force \( F_f \) that can act on the car is given by:
\[ F_f = \mu m g \]
where:
- \( \mu \) is the coefficient of static friction,
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 4: Set the centripetal force equal to the frictional force
For the car to make the turn without skidding, the frictional force must be equal to the centripetal force:
\[ \mu m g = \frac{mV^2}{R} \]

### Step 5: Cancel out the mass \( m \)
Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ \mu g = \frac{V^2}{R} \]

### Step 6: Rearrange the equation to solve for \( V^2 \)
Rearranging gives us:
\[ V^2 = \mu g R \]

### Step 7: Substitute the known values
We know:
- \( \mu = 0.75 \)
- \( g = 10 \, \text{m/s}^2 \)
- \( R = 60 \, \text{m} \)

Substituting these values into the equation:
\[ V^2 = 0.75 \times 10 \times 60 \]
\[ V^2 = 0.75 \times 600 \]
\[ V^2 = 450 \]

### Step 8: Take the square root to find \( V \)
Now, taking the square root of both sides:
\[ V = \sqrt{450} \]
Calculating this gives:
\[ V \approx 21.21 \, \text{m/s} \]

### Final Answer
The maximum speed at which the car can make the turn is approximately \( 21.21 \, \text{m/s} \).

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