A weightless thread can bear tension up to 3.7 kg weight. A stone of mass 500 g is tied at its one end revolved in a vertical circular path of radius 4m. If `g=10m//s^(2)`, then the maximum angular velocity of the stone is (rad/s) is

Options

A. `3`
B. `4`
C. `5`
D. `6`

Answer

To solve the problem step by step, we will follow the concepts of circular motion and tension in a string.

### Step 1: Understand the given data
- Maximum tension the thread can bear: \( T_{max} = 3.7 \, \text{kg} \)
- Mass of the stone: \( m = 500 \, \text{g} = 0.5 \, \text{kg} \)
- Radius of the circular path: \( r = 4 \, \text{m} \)
- Acceleration due to gravity: \( g = 10 \, \text{m/s}^2 \)

### Step 2: Convert maximum tension to Newtons
The maximum tension in Newtons can be calculated as:
\[ T_{max} = 3.7 \, \text{kg} \times g = 3.7 \times 10 = 37 \, \text{N} \]

### Step 3: Write the equation for tension in circular motion
In vertical circular motion, the tension in the string at the lowest point can be expressed as:
\[ T = mg + \frac{mv^2}{r} \]
Where:
- \( T \) is the tension,
- \( m \) is the mass of the stone,
- \( v \) is the tangential velocity,
- \( r \) is the radius of the circular path.

### Step 4: Rearrange the equation to find \( v^2 \)
We can rearrange the equation to find \( v^2 \):
\[ T_{max} = mg + \frac{mv^2}{r} \]
\[ T_{max} - mg = \frac{mv^2}{r} \]
\[ v^2 = \frac{r(T_{max} - mg)}{m} \]

### Step 5: Substitute the values into the equation
First, calculate \( mg \):
\[ mg = 0.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 5 \, \text{N} \]
Now substitute \( T_{max} \), \( mg \), \( r \), and \( m \) into the equation:
\[ v^2 = \frac{4 \, \text{m} \times (37 \, \text{N} - 5 \, \text{N})}{0.5 \, \text{kg}} \]
\[ v^2 = \frac{4 \times 32}{0.5} = \frac{128}{0.5} = 256 \, \text{m}^2/\text{s}^2 \]

### Step 6: Calculate \( v \)
Now, take the square root to find \( v \):
\[ v = \sqrt{256} = 16 \, \text{m/s} \]

### Step 7: Relate tangential velocity to angular velocity
The relationship between tangential velocity \( v \) and angular velocity \( \omega \) is given by:
\[ v = r \omega \]
Thus,
\[ \omega = \frac{v}{r} = \frac{16 \, \text{m/s}}{4 \, \text{m}} = 4 \, \text{rad/s} \]

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

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