A ball of mass `0.1kg` is suspended by a string `30 cm` long Keeping the string always taut the ball describe a horizontal circle of radius `15cm` Calculate the angular speed of the ball .

Explanation

To calculate the angular speed of the ball, we can follow these steps:

### Step 1: Understand the Geometry of the Problem
The ball is suspended by a string of length \( L = 30 \, \text{cm} = 0.3 \, \text{m} \) and describes a horizontal circle of radius \( r = 15 \, \text{cm} = 0.15 \, \text{m} \). The string makes an angle \( \theta \) with the vertical.

### Step 2: Use Trigonometry to Find the Angle
Using the right triangle formed by the string, the vertical line, and the radius of the circle:
- The vertical side (adjacent to the angle \( \theta \)) is \( L \cos \theta \).
- The horizontal side (opposite to the angle \( \theta \)) is \( r \).

From the geometry, we can write:
\[ \cos \theta = \frac{r}{L} \]
Substituting the values:
\[ \cos \theta = \frac{0.15}{0.3} = \frac{1}{2} \]
Thus, \( \theta = 60^\circ \).

### Step 3: Apply Forces in the Vertical Direction
The forces acting on the ball are:
- The tension \( T \) in the string acting upwards.
- The weight \( mg \) acting downwards.

From the vertical component of the forces, we have:
\[ T \cos \theta = mg \]

### Step 4: Apply Forces in the Horizontal Direction
In the horizontal direction, the tension provides the centripetal force required for circular motion:
\[ T \sin \theta = m \frac{v^2}{r} \]
Where \( v \) is the linear speed of the ball.

### Step 5: Relate Linear Speed to Angular Speed
The relationship between linear speed \( v \) and angular speed \( \omega \) is given by:
\[ v = r \omega \]
Substituting this into the horizontal force equation gives:
\[ T \sin \theta = m \frac{(r \omega)^2}{r} = m r \omega^2 \]

### Step 6: Solve for Tension
From the vertical force equation, we can express \( T \):
\[ T = \frac{mg}{\cos \theta} \]

### Step 7: Substitute Tension into the Horizontal Equation
Substituting \( T \) into the horizontal equation:
\[ \frac{mg}{\cos \theta} \sin \theta = m r \omega^2 \]
Cancelling \( m \) from both sides:
\[ \frac{g \sin \theta}{\cos \theta} = r \omega^2 \]
Using \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ g \tan \theta = r \omega^2 \]

### Step 8: Solve for Angular Speed \( \omega \)
Rearranging gives:
\[ \omega^2 = \frac{g \tan \theta}{r} \]
Taking the square root:
\[ \omega = \sqrt{\frac{g \tan \theta}{r}} \]

### Step 9: Substitute Known Values
Using \( g = 9.81 \, \text{m/s}^2 \) and \( \tan 60^\circ = \sqrt{3} \):
\[ \omega = \sqrt{\frac{9.81 \cdot \sqrt{3}}{0.15}} \]

### Step 10: Calculate the Result
Calculating the above expression:
\[ \omega = \sqrt{\frac{9.81 \cdot 1.732}{0.15}} \approx \sqrt{113.14} \approx 10.65 \, \text{rad/s} \]

### Final Answer
The angular speed of the ball is approximately \( 10.65 \, \text{rad/s} \).

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