A ball of mass 100 g is suspended by a string `40 cm` long. Keeping the string taut, the ball describes a horizontal circle of radius 10 cm. Find the angular speed.

Answer

To find the angular speed of the ball suspended by a string while it describes a horizontal circle, we can follow these steps:

### Step 1: Understand the Problem
We have a ball of mass \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) suspended by a string of length \( L = 40 \, \text{cm} = 0.4 \, \text{m} \). The ball is moving in a horizontal circle with a radius \( r = 10 \, \text{cm} = 0.1 \, \text{m} \).

### Step 2: Identify the Forces
The forces acting on the ball are:
1. The gravitational force \( F_g = mg \) acting downwards.
2. The tension \( T \) in the string acting at an angle \( \theta \) from the vertical.

### Step 3: Set Up the Geometry
From the geometry of the situation, we can relate the radius of the circle \( r \), the length of the string \( L \), and the angle \( \theta \):
- The vertical component of the tension must balance the weight of the ball:
\[ T \cos(\theta) = mg \]
- The horizontal component of the tension provides the centripetal force necessary for circular motion:
\[ T \sin(\theta) = \frac{mv^2}{r} \]

### Step 4: Use Trigonometric Relationships
From the right triangle formed by the string, we can express \( \sin(\theta) \) and \( \cos(\theta) \):
- The vertical leg is \( L \cos(\theta) \) and the horizontal leg is \( L \sin(\theta) \).
- The radius \( r = L \sin(\theta) \) gives us:
\[ \sin(\theta) = \frac{r}{L} = \frac{0.1}{0.4} = \frac{1}{4} \]
- Using \( \cos^2(\theta) + \sin^2(\theta) = 1 \):
\[ \cos(\theta) = \sqrt{1 - \left(\frac{1}{4}\right)^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \]

### Step 5: Substitute into the Equations
Substituting \( \sin(\theta) \) and \( \cos(\theta) \) into the equations:
1. For the vertical force balance:
\[ T \cdot \frac{\sqrt{15}}{4} = mg \]
\[ T = \frac{mg \cdot 4}{\sqrt{15}} \]
2. For the horizontal force balance:
\[ T \cdot \frac{1}{4} = \frac{mv^2}{r} \]

### Step 6: Equate the Tension
Setting the two expressions for \( T \) equal:
\[ \frac{mg \cdot 4}{\sqrt{15}} \cdot \frac{1}{4} = \frac{mv^2}{r} \]
This simplifies to:
\[ \frac{mg}{\sqrt{15}} = \frac{mv^2}{r} \]
Cancelling \( m \) and rearranging gives:
\[ v^2 = \frac{g \cdot r}{\sqrt{15}} \]

### Step 7: Calculate Angular Speed
The relationship between linear speed \( v \) and angular speed \( \omega \) is:
\[ v = r \omega \implies \omega = \frac{v}{r} \]
Substituting for \( v^2 \):
\[ \omega^2 = \frac{g \cdot r}{\sqrt{15} \cdot r^2} = \frac{g}{r \sqrt{15}} \]
Thus,
\[ \omega = \sqrt{\frac{g}{r \sqrt{15}}} \]

### Step 8: Plug in the Values
Using \( g \approx 9.81 \, \text{m/s}^2 \) and \( r = 0.1 \, \text{m} \):
\[ \omega = \sqrt{\frac{9.81}{0.1 \cdot \sqrt{15}}} = \sqrt{\frac{9.81}{0.1 \cdot 3.872}} = \sqrt{\frac{9.81}{0.3872}} \approx \sqrt{25.35} \approx 5.035 \, \text{rad/s} \]

### Final Answer
The angular speed \( \omega \) of the ball is approximately \( 5.035 \, \text{rad/s} \).

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