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, we need to find the radius of the circular track on which a car is moving, given its tangential acceleration, the speed at which it starts to slip, and the coefficient of friction.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Car:**
- The car experiences two types of acceleration: tangential acceleration (\(a_0\)) and centripetal (normal) acceleration, which is given by \(\frac{v^2}{R}\), where \(v\) is the speed of the car and \(R\) is the radius of the circular path.

2. **Condition for Slipping:**
- The car starts to slip when the frictional force is at its maximum. The maximum frictional force can be expressed as:
\[ f_{\text{max}} = \mu m g \]
- Here, \(m\) is the mass of the car, \(\mu\) is the coefficient of friction, and \(g\) is the acceleration due to gravity.

3. **Net Acceleration at the Point of Slipping:**
- At the point of slipping, the net acceleration \(a_{\text{net}}\) can be expressed as the vector sum of the tangential acceleration and the centripetal acceleration:
\[ a_{\text{net}} = a_0 + \frac{v_0^2}{R} \]

4. **Setting Up the Equation:**
- At the point of slipping, the net acceleration must equal the maximum frictional acceleration:
\[ a_0 + \frac{v_0^2}{R} = \mu g \]

5. **Rearranging the Equation:**
- Rearranging the equation gives:
\[ \frac{v_0^2}{R} = \mu g - a_0 \]

6. **Solving for the Radius \(R\):**
- Rearranging to isolate \(R\):
\[ R = \frac{v_0^2}{\mu g - a_0} \]

### Final Expression for Radius:
The radius of the circular track is given by:
\[ R = \frac{v_0^2}{\mu g - a_0} \]