The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of `sqrt(3gl)`. Find the angle rotated by the string before it becomes slack.

Explanation

To solve the problem of finding the angle rotated by the string of a pendulum before it becomes slack after being given a horizontal speed of \( \sqrt{3gl} \), we can follow these steps:

### Step 1: Understand the Initial Conditions
The pendulum bob is initially at rest and is given a horizontal speed of \( v = \sqrt{3gl} \). We need to find the angle \( \theta \) at which the tension in the string becomes zero.

### Step 2: Apply Conservation of Energy
At the initial point (point A), the bob has kinetic energy and no potential energy (since we take the reference level at the height of the bob). The kinetic energy \( KE_A \) at point A is given by:
\[ KE_A = \frac{1}{2} mv^2 = \frac{1}{2} m (\sqrt{3gl})^2 = \frac{3}{2} mgl \]
At point B, when the bob has rotated to angle \( \theta \), it has both kinetic energy and potential energy. The potential energy \( PE_B \) at point B is:
\[ PE_B = mgh = mg(l - l \cos \theta) = mg(l(1 - \cos \theta)) \]
The kinetic energy \( KE_B \) at point B is:
\[ KE_B = \frac{1}{2} mv^2 \]

By conservation of energy, we have:
\[ KE_A = KE_B + PE_B \]
Substituting the expressions we found:
\[ \frac{3}{2} mgl = \frac{1}{2} mv^2 + mg(l(1 - \cos \theta)) \]

### Step 3: Simplify the Energy Equation
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{3}{2} gl = \frac{1}{2} v^2 + g(l(1 - \cos \theta)) \]
Rearranging gives:
\[ \frac{3}{2} gl - g(l - l \cos \theta) = \frac{1}{2} v^2 \]
This simplifies to:
\[ \frac{3}{2} gl - gl + gl \cos \theta = \frac{1}{2} v^2 \]
\[ \frac{1}{2} gl + gl \cos \theta = \frac{1}{2} v^2 \]

### Step 4: Express \( v^2 \) in Terms of \( \theta \)
From the centripetal force requirement when the tension is zero, we have:
\[ mg \cos \theta = \frac{mv^2}{l} \]
Cancelling \( m \) gives:
\[ g \cos \theta = \frac{v^2}{l} \]
Thus,
\[ v^2 = gl \cos \theta \]

### Step 5: Substitute \( v^2 \) into the Energy Equation
Substituting \( v^2 \) back into the energy equation:
\[ \frac{1}{2} gl + gl \cos \theta = \frac{1}{2} (gl \cos \theta) \]
This leads to:
\[ \frac{1}{2} gl + gl \cos \theta = \frac{1}{2} gl \cos \theta \]
Rearranging gives:
\[ \frac{1}{2} gl = \frac{1}{2} gl \cos \theta - gl \cos \theta \]
\[ \frac{1}{2} gl = -\frac{1}{2} gl \cos \theta \]
\[ 1 = -\cos \theta \]
This gives:
\[ \cos \theta = \frac{1}{3} \]

### Step 6: Find the Angle Rotated
The angle \( \theta \) is given by:
\[ \theta = \cos^{-1} \left(\frac{1}{3}\right) \]
The angle rotated before the string becomes slack is:
\[ \text{Angle rotated} = \pi - \theta = \pi - \cos^{-1} \left(\frac{1}{3}\right) \]

### Final Answer
Thus, the angle rotated by the string before it becomes slack is:
\[ \pi - \cos^{-1} \left(\frac{1}{3}\right) \]