A weightless thread can bear tension upto `3.7 kg` wt A stone of mass `500g` is tied to it and revolves in a verticle circle of radius `4m` What will be the maximum angular velocity of the stone if `g = 10 m//s^(2)` .

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert the maximum tension from kg to Newtons
The maximum tension that the thread can bear is given as `3.7 kg wt`. To convert this to Newtons, we use the formula:
\[ T_{max} = m \cdot g \]
Where:
- \( m = 3.7 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)

Calculating:
\[ T_{max} = 3.7 \, \text{kg} \times 10 \, \text{m/s}^2 = 37 \, \text{N} \]

### Step 2: Identify the forces acting on the stone
When the stone is at the lowest point of the vertical circle, the forces acting on it are:
- The gravitational force \( Mg \) acting downwards.
- The centripetal force required to keep the stone moving in a circle, which is provided by the tension in the thread.

### Step 3: Write the equation for tension at the lowest point
At the lowest point, the tension in the thread can be expressed as:
\[ T = \frac{Mv^2}{R} + Mg \]
Where:
- \( T \) is the tension in the thread (which we set to \( T_{max} \)),
- \( M \) is the mass of the stone (0.5 kg),
- \( v \) is the linear velocity of the stone,
- \( R \) is the radius of the circle (4 m).

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:
\[ 37 = \frac{0.5 \cdot v^2}{4} + 0.5 \cdot 10 \]

### Step 5: Simplify the equation
Calculating the gravitational force:
\[ 0.5 \cdot 10 = 5 \]
So the equation becomes:
\[ 37 = \frac{0.5 \cdot v^2}{4} + 5 \]

### Step 6: Rearranging the equation
Subtract 5 from both sides:
\[ 37 - 5 = \frac{0.5 \cdot v^2}{4} \]
\[ 32 = \frac{0.5 \cdot v^2}{4} \]

### Step 7: Solve for \( v^2 \)
Multiply both sides by 4:
\[ 32 \cdot 4 = 0.5 \cdot v^2 \]
\[ 128 = 0.5 \cdot v^2 \]

Now, divide by 0.5:
\[ v^2 = \frac{128}{0.5} = 256 \]

### Step 8: Calculate \( v \)
Taking the square root of both sides:
\[ v = \sqrt{256} = 16 \, \text{m/s} \]

### Step 9: Calculate the angular velocity \( \omega \)
The angular velocity \( \omega \) is given by:
\[ \omega = \frac{v}{R} \]
Substituting the values:
\[ \omega = \frac{16}{4} = 4 \, \text{radians/second} \]

### Final Answer
The maximum angular velocity of the stone is:
\[ \omega = 4 \, \text{radians/second} \]