A circular road of radius `50m` has the angle of banking equal to `30^(@)` . At what should a vehicle go on this road so that the friction is not used?

Explanation

To solve the problem of finding the speed at which a vehicle should travel on a circular road of radius 50 m with a banking angle of 30 degrees, we can follow these steps:

### Step 1: Understand the Forces Acting on the Vehicle
When a vehicle is moving on a banked circular road, the forces acting on it include:
- The gravitational force (Mg) acting downwards.
- The normal force (N) acting perpendicular to the surface of the road.
- The centripetal force required for circular motion, which is provided by the horizontal component of the normal force.

### Step 2: Resolve the Forces
We need to resolve the forces into two components:
1. **Perpendicular to the surface (vertical direction)**:
\[ N \cos(\theta) = Mg \]
2. **Parallel to the surface (horizontal direction)**:
\[ N \sin(\theta) = \frac{Mv^2}{R} \]
where \( v \) is the speed of the vehicle, \( R \) is the radius of the circular path, and \( \theta \) is the angle of banking.

### Step 3: Express Normal Force in Terms of Gravitational Force
From the first equation, we can express the normal force \( N \):
\[ N = \frac{Mg}{\cos(\theta)} \]

### Step 4: Substitute Normal Force into the Second Equation
Substituting \( N \) into the second equation gives us:
\[ \frac{Mg}{\cos(\theta)} \sin(\theta) = \frac{Mv^2}{R} \]

### Step 5: Simplify the Equation
We can cancel \( M \) from both sides (assuming \( M \neq 0 \)):
\[ \frac{g \sin(\theta)}{\cos(\theta)} = \frac{v^2}{R} \]
This simplifies to:
\[ g \tan(\theta) = \frac{v^2}{R} \]

### Step 6: Solve for Speed \( v \)
Rearranging the equation gives:
\[ v^2 = g R \tan(\theta) \]
Taking the square root to find \( v \):
\[ v = \sqrt{g R \tan(\theta)} \]

### Step 7: Substitute Known Values
We know:
- \( g = 9.8 \, \text{m/s}^2 \)
- \( R = 50 \, \text{m} \)
- \( \theta = 30^\circ \) (where \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \))

Substituting these values into the equation:
\[ v = \sqrt{9.8 \times 50 \times \tan(30^\circ)} \]
\[ v = \sqrt{9.8 \times 50 \times \frac{1}{\sqrt{3}}} \]
\[ v = \sqrt{9.8 \times 50 \times 0.577} \approx \sqrt{288.6} \]
Calculating this gives:
\[ v \approx 17 \, \text{m/s} \]

### Final Answer
The speed at which the vehicle should go on this road so that friction is not used is approximately **17 m/s**.

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