A weightless thread can support tension up to `30N.A` particle of mass `0.5kg` is tied to it and is revolved in a circle of radius `2m` in a verticle plane. If `g=10m//s^(2)`, then the maximum angular velocity of the stone will be

Options

A. `5rad//s`
B. `sqrt(30)rad//s`
C. `sqrt(60)rad//s`
D. `10rad//s`

Explanation

To find the maximum angular velocity of the particle being revolved in a vertical circle, we can follow these steps:

### Step 1: Identify the forces acting on the particle at the lowest point of the circular motion.
At the lowest point, the forces acting on the particle are:
- The tension (T) in the thread acting upwards.
- The weight (W) of the particle acting downwards.

The weight of the particle can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 0.5 \, \text{kg} \) (mass of the particle)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 0.5 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 5 \, \text{N} \]

### Step 2: Write the equation for the forces at the lowest point.
At the lowest point, the net force acting on the particle provides the centripetal force required for circular motion. The equation can be written as:
\[ T - W = \frac{m v^2}{r} \]
where:
- \( T \) is the tension in the thread (maximum tension = 30 N)
- \( W \) is the weight of the particle (5 N)
- \( m \) is the mass of the particle (0.5 kg)
- \( v \) is the linear velocity of the particle
- \( r \) is the radius of the circular motion (2 m)

Substituting the known values into the equation:
\[ 30 \, \text{N} - 5 \, \text{N} = \frac{0.5 \, \text{kg} \cdot v^2}{2 \, \text{m}} \]

### Step 3: Simplify the equation.
This simplifies to:
\[ 25 = \frac{0.5 v^2}{2} \]
Multiplying both sides by 2 to eliminate the fraction:
\[ 50 = 0.5 v^2 \]

### Step 4: Solve for \( v^2 \).
Now, divide both sides by 0.5:
\[ v^2 = 100 \]

### Step 5: Calculate the linear velocity \( v \).
Taking the square root of both sides gives:
\[ v = \sqrt{100} = 10 \, \text{m/s} \]

### Step 6: Relate linear velocity to angular velocity.
The relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:
\[ v = r \omega \]
where \( r = 2 \, \text{m} \).

Rearranging the formula to find \( \omega \):
\[ \omega = \frac{v}{r} = \frac{10 \, \text{m/s}}{2 \, \text{m}} = 5 \, \text{rad/s} \]

### Final Answer:
The maximum angular velocity of the stone is \( \omega = 5 \, \text{rad/s} \).

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