A stone tied to a string of length `l` is whirled in a horizontal circle at a constant angular speed `omega` in a circle of radius r. If the string makes an angle `theta` with vertical then the tension in the string is :

Options

A. `m(g cos theta+omega^ (2)r)`
B. `msqrt(omega^(4)r^(2)+g^(2))`
C. `msqrt(omega^(2)r^(2)+g^(2))`
D. `m(g cos theta -omega^(2)r)`

Answer

To find the tension in the string when a stone is tied to it and whirled in a horizontal circle at a constant angular speed while making an angle θ with the vertical, we can follow these steps:

### Step 1: Understanding the Forces
When the stone is whirled in a horizontal circle, two main forces act on it:
1. The gravitational force (weight) acting downwards: \( F_g = mg \)
2. The tension in the string, which can be resolved into two components:
- \( T \cos \theta \) acting vertically (balancing the weight)
- \( T \sin \theta \) acting horizontally (providing the centripetal force)

### Step 2: Setting Up the Equations
From the forces acting on the stone, we can set up the following equations:
1. For vertical forces (balancing weight):
\[ T \cos \theta = mg \quad \text{(1)} \]
2. For horizontal forces (providing centripetal force):
\[ T \sin \theta = m \frac{v^2}{r} \quad \text{(2)} \]
where \( v \) is the linear velocity of the stone, which can be related to angular speed \( \omega \) by \( v = r \omega \).

### Step 3: Substituting Linear Velocity
Substituting \( v = r \omega \) into equation (2):
\[ T \sin \theta = m \frac{(r \omega)^2}{r} \]
This simplifies to:
\[ T \sin \theta = m r \omega^2 \quad \text{(3)} \]

### Step 4: Dividing the Equations
Now, we can divide equation (3) by equation (1) to eliminate \( T \):
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{m r \omega^2}{mg} \]
This simplifies to:
\[ \tan \theta = \frac{r \omega^2}{g} \]

### Step 5: Finding Tension
From equation (1), we can express \( T \):
\[ T = \frac{mg}{\cos \theta} \quad \text{(4)} \]

### Step 6: Final Expression for Tension
Using the relationship from step 4, we can express \( \cos \theta \) in terms of \( \tan \theta \):
\[ \cos \theta = \frac{g}{\sqrt{g^2 + (r \omega^2)^2}} \]
Substituting this into equation (4):
\[ T = mg \sec \theta = mg \frac{\sqrt{g^2 + (r \omega^2)^2}}{g} \]
Thus, the final expression for tension \( T \) is:
\[ T = m \sqrt{g^2 + (r \omega^2)^2} \]

### Final Answer
The tension in the string is:
\[ T = m \sqrt{g^2 + (r \omega^2)^2} \]

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