A car is moving along a circular track with tangential acceleration of magnitude `a_(0)`. It just start to slip at speed `v_(0)` then find radius of circle (Coeffecient of friction is `mu`) ?

Options

A. `(V_(0)^(2))/(sqrt((mug)^(2)+a_(0)^(2)))`
B. `(V_(0)^(2))/(sqrt((mug)^(2)-a_(0)^(2)))`
C. `(V_(0)^(2))/(sqrt((mug)^(2)-2a_(0)^(2)))`
D. `(V_(0)^(2))/(sqrt((mug)^(2)+2a_(0)^(2)))`

Answer

To solve the problem of finding the radius of the circular track on which a car is moving, we can follow these steps:

### Step 1: Understand the Forces Acting on the Car
When the car is moving in a circular path, it experiences two types of acceleration:
- **Tangential Acceleration (a₀)**: This is the acceleration along the direction of motion.
- **Centripetal Acceleration (a_c)**: This is the acceleration directed towards the center of the circular path, given by the formula \( a_c = \frac{v^2}{r} \), where \( v \) is the speed of the car and \( r \) is the radius of the circular path.

### Step 2: Write the Expression for Net Acceleration
The net acceleration of the car can be expressed as the vector sum of the tangential and centripetal accelerations:
\[ a_{net} = \sqrt{a_0^2 + a_c^2} \]
Substituting for \( a_c \):
\[ a_{net} = \sqrt{a_0^2 + \left(\frac{v_0^2}{r}\right)^2} \]

### Step 3: Relate Net Acceleration to Friction
The maximum frictional force that can act on the car, which prevents it from slipping, is given by:
\[ F_{friction} = \mu mg \]
This frictional force provides the necessary centripetal force for circular motion, which can also be expressed in terms of acceleration:
\[ F_{friction} = ma_{net} \]
Thus, we can equate the two expressions:
\[ ma_{net} = \mu mg \]
Dividing both sides by \( m \):
\[ a_{net} = \mu g \]

### Step 4: Set Up the Equation
Now we can set the expression for net acceleration equal to the maximum static friction:
\[ \sqrt{a_0^2 + \left(\frac{v_0^2}{r}\right)^2} = \mu g \]

### Step 5: Square Both Sides
To eliminate the square root, we square both sides:
\[ a_0^2 + \left(\frac{v_0^2}{r}\right)^2 = \mu^2 g^2 \]

### Step 6: Rearrange to Find Radius
Rearranging the equation to solve for \( r \):
\[ \left(\frac{v_0^2}{r}\right)^2 = \mu^2 g^2 - a_0^2 \]
Taking the square root:
\[ \frac{v_0^2}{r} = \sqrt{\mu^2 g^2 - a_0^2} \]
Now, solving for \( r \):
\[ r = \frac{v_0^2}{\sqrt{\mu^2 g^2 - a_0^2}} \]

### Final Answer
Thus, the radius of the circular track is given by:
\[ r = \frac{v_0^2}{\sqrt{\mu^2 g^2 - a_0^2}} \]