A weightless thread can bear tension upto 3.7 kg wt .A stone of mass 500 gm is tied to it and rotated in a circular path of radius 4 m, in a vertical circle If ` g=10 m//s^(2)` then the maximum angular velocity of the stone will be

Options

A. 2 rad /s
B. 5 rad /s
C. 4rad /s
D. 3 rad /s

Explanation

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

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

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

### Step 2: Convert the mass of the stone to kg
The mass of the stone is given as 500 g. We convert this to kilograms:
\[ m = 500 \, \text{g} = 0.5 \, \text{kg} \]

### Step 3: Identify the forces acting on the stone at the bottom of the vertical circle
At the bottom of the vertical circle, the forces acting on the stone are:
1. The weight of the stone acting downwards: \( W = m \cdot g = 0.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 5 \, \text{N} \)
2. The tension in the thread acting upwards.

### Step 4: Apply the centripetal force equation
At the bottom of the circle, the net force providing the centripetal acceleration is given by:
\[ T - W = \frac{m v^2}{r} \]
Where:
- \( T = 37 \, \text{N} \)
- \( W = 5 \, \text{N} \)
- \( m = 0.5 \, \text{kg} \)
- \( r = 4 \, \text{m} \)

Substituting the values:
\[ 37 \, \text{N} - 5 \, \text{N} = \frac{0.5 \, \text{kg} \cdot v^2}{4 \, \text{m}} \]

### Step 5: Solve for \( v^2 \)
This simplifies to:
\[ 32 \, \text{N} = \frac{0.5 \, v^2}{4} \]
Multiplying both sides by 4:
\[ 128 = 0.5 v^2 \]
Now, divide by 0.5:
\[ v^2 = 256 \]
Taking the square root:
\[ v = \sqrt{256} = 16 \, \text{m/s} \]

### Step 6: Calculate the angular velocity \( \omega \)
The relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:
\[ v = r \cdot \omega \]
Rearranging gives:
\[ \omega = \frac{v}{r} \]
Substituting the values:
\[ \omega = \frac{16 \, \text{m/s}}{4 \, \text{m}} = 4 \, \text{rad/s} \]

### Conclusion
The maximum angular velocity of the stone is:
\[ \omega = 4 \, \text{rad/s} \]