A car of maas M is moving on a horizontal circular path of radius r. At an instant its speed is v and is increasing at a rate a.

Options

A. The acceleration of the car is towards the centre of the path
B. The magnitude the frictional force on the car is greater than `(mv^2)/r`.
C. The friction coefficient between the ground and the car is not less than `a/g`.
D. The friction coefficient between the ground and the car is `mu=tan^-1 v^2/rg`

Answer

To solve the problem, we need to analyze the forces acting on the car as it moves in a circular path while accelerating. Here’s a step-by-step solution:

### Step 1: Identify the Forces Acting on the Car
The car experiences two main accelerations:
1. **Tangential Acceleration (a)**: This is due to the increase in speed of the car.
2. **Centripetal Acceleration**: This is directed towards the center of the circular path and is given by the formula \( \frac{v^2}{r} \).

### Step 2: Write the Equations for Forces
The frictional force \( F_s \) provides the necessary centripetal force and also supports the tangential acceleration. We can decompose the frictional force into two components:
- \( F_s \cos \theta = m a \) (for tangential acceleration)
- \( F_s \sin \theta = \frac{m v^2}{r} \) (for centripetal acceleration)

### Step 3: Solve for the Frictional Force
To find the frictional force \( F_s \), we can use the Pythagorean theorem since the two components are perpendicular to each other:
\[ F_s = \sqrt{(m a)^2 + \left(\frac{m v^2}{r}\right)^2} \]

### Step 4: Relate Frictional Force to Normal Force
The frictional force can also be expressed in terms of the coefficient of friction \( \mu \) and the normal force \( N \):
\[ F_s = \mu N \]
Since the car is on a horizontal surface, the normal force \( N \) is equal to the weight of the car:
\[ N = mg \]
Thus, we have:
\[ F_s = \mu mg \]

### Step 5: Set the Two Expressions for Frictional Force Equal
Equating the two expressions for \( F_s \):
\[ \mu mg = \sqrt{(m a)^2 + \left(\frac{m v^2}{r}\right)^2} \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \mu g = \sqrt{a^2 + \left(\frac{v^2}{r}\right)^2} \]

### Step 6: Solve for the Coefficient of Friction
To find \( \mu \), we rearrange the equation:
\[ \mu = \frac{1}{g} \sqrt{a^2 + \left(\frac{v^2}{r}\right)^2} \]

### Conclusion
The coefficient of friction \( \mu \) is given by:
\[ \mu = \frac{\sqrt{a^2 + \left(\frac{v^2}{r}\right)^2}}{g} \]