A vehicle is moving on a banked road `theta` is angle of banking and `mu` is coefcient of friction between tyre of vehicle and road. The vehicle is moving along horizontal circular path of radius (r) velocity of vehicle for which it does not slip is 'v'. Choose the correct alternative

Options

A. V should between 'o' and `sqrt(mugr)`
B. V should be betweem 'o' and `sqrt((rg(mu+ tantheta))/(1-mutantheta))`
C. V should lie between `sqrt((rg(mu+ tantheta))/(1+mutantheta))` and 'o'
D. V should lie between `sqrt((rg(mu+ tantheta))/(1+mutantheta))` and `sqrt((rg(mu+ tantheta))/(1-mutantheta))`

Answer

To solve the problem, we need to analyze the motion of a vehicle on a banked road with friction. Here’s a step-by-step breakdown of how to approach this:

### Step 1: Understand the Forces Acting on the Vehicle
When a vehicle is on a banked road, several forces act on it:
- The gravitational force (mg) acting downwards.
- The normal force (N) acting perpendicular to the surface of the banked road.
- The frictional force (f) acting parallel to the surface of the road, which can either aid or oppose the motion.

### Step 2: Set Up the Equations of Motion
For a vehicle moving in a circular path, the net force providing the centripetal acceleration must be equal to the required centripetal force. The centripetal force (F_c) is given by:
\[ F_c = \frac{mv^2}{r} \]
where:
- m = mass of the vehicle
- v = velocity of the vehicle
- r = radius of the circular path

### Step 3: Resolve Forces
The forces can be resolved into components:
- The vertical component of the normal force balances the weight of the vehicle:
\[ N \cos(\theta) = mg \]
- The horizontal component of the normal force and the frictional force provides the necessary centripetal force:
\[ N \sin(\theta) + f = \frac{mv^2}{r} \]

### Step 4: Express Frictional Force
The frictional force can be expressed in terms of the coefficient of friction (μ) and the normal force:
\[ f = \mu N \]

### Step 5: Substitute and Rearrange
Substituting the expression for friction into the centripetal force equation gives:
\[ N \sin(\theta) + \mu N = \frac{mv^2}{r} \]
Factoring out N:
\[ N (\sin(\theta) + \mu) = \frac{mv^2}{r} \]

### Step 6: Substitute Normal Force
From the vertical force balance, we have:
\[ N = \frac{mg}{\cos(\theta)} \]
Substituting this into the centripetal force equation:
\[ \frac{mg}{\cos(\theta)} (\sin(\theta) + \mu) = \frac{mv^2}{r} \]

### Step 7: Cancel Mass and Rearrange for Velocity
Canceling m from both sides (assuming m ≠ 0):
\[ \frac{g(\sin(\theta) + \mu)}{\cos(\theta)} = \frac{v^2}{r} \]
Rearranging gives:
\[ v^2 = rg \frac{\sin(\theta) + \mu}{\cos(\theta)} \]

### Step 8: Final Expression for Velocity
Taking the square root gives the velocity:
\[ v = \sqrt{rg \frac{\sin(\theta) + \mu}{\cos(\theta)}} \]

### Step 9: Conclusion
The velocity 'v' for which the vehicle does not slip can be expressed as:
\[ v = \sqrt{rg \left(\frac{\mu + \tan(\theta)}{1 - \mu \tan(\theta)}\right)} \]