Find the maximum velocity for skidding for a car moved on a circular track of radius 100 m . The coefficient of friction between the road and tyre is 0.2

Options

A. `0.14 m//s`
B. `140 m//s`
C. `1.4 km//s`
D. `14 m//s`

Answer

To find the maximum velocity for skidding for a car moving on a circular track, we can follow these steps:

### Step 1: Understand the forces involved
When a car moves in a circular path, it experiences a centripetal force that keeps it moving in that path. This centripetal force is provided by the friction between the tires and the road. If the speed of the car exceeds a certain limit, the frictional force will not be enough to provide the required centripetal force, and the car will skid.

### Step 2: Write down the equations
The maximum frictional force (F_max) can be expressed as:
\[ F_{\text{max}} = \mu \cdot N \]
where:
- \( \mu \) = coefficient of friction (given as 0.2)
- \( N \) = normal force (which is equal to the weight of the car, \( mg \), where \( m \) is the mass of the car and \( g \) is the acceleration due to gravity, approximately 10 m/s²).

Thus,
\[ F_{\text{max}} = \mu \cdot mg \]

### Step 3: Set up the centripetal force equation
The centripetal force required to keep the car moving in a circle of radius \( r \) is given by:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( v \) = velocity of the car
- \( r \) = radius of the circular track (given as 100 m).

### Step 4: Equate the forces
For the car to avoid skidding, the maximum frictional force must be equal to or greater than the centripetal force:
\[ F_{\text{max}} \geq F_c \]
Substituting the expressions we have:
\[ \mu mg \geq \frac{mv^2}{r} \]

### Step 5: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g \geq \frac{v^2}{r} \]
Rearranging gives:
\[ v^2 \leq \mu g r \]

### Step 6: Substitute the values
Now, substituting the known values:
- \( \mu = 0.2 \)
- \( g = 10 \, \text{m/s}^2 \)
- \( r = 100 \, \text{m} \)

We get:
\[ v^2 \leq 0.2 \cdot 10 \cdot 100 \]
\[ v^2 \leq 200 \]

### Step 7: Solve for \( v \)
Taking the square root of both sides:
\[ v \leq \sqrt{200} \]
\[ v \leq 14.14 \, \text{m/s} \]

### Conclusion
The maximum velocity for skidding for the car on a circular track of radius 100 m, with a coefficient of friction of 0.2, is approximately:
\[ v \approx 14.1 \, \text{m/s} \]