The simple pendulum bob is given a horizontal velocity `sqrt(gl)` at the bottom. The angle with the vertical through which the ball swings before its velocity becomes zero is

Options

A. `120^(@)`
B. `60^(@)`
C. `45^(@)`
D. `30^(@)`

Answer

To solve the problem, we will use the principle of conservation of mechanical energy. Let's break down the solution step by step:

### Step 1: Understand the Initial Conditions
The pendulum bob is given a horizontal velocity of \( v = \sqrt{gl} \) at the lowest point of its swing. At this point, all the energy is kinetic energy.

### Step 2: Write the Expression for Kinetic Energy
The kinetic energy (KE) at the lowest point can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the value of \( v \):
\[ KE = \frac{1}{2} m (\sqrt{gl})^2 = \frac{1}{2} m (gl) = \frac{mgl}{2} \]

### Step 3: Write the Expression for Potential Energy
When the pendulum reaches its maximum height, its velocity becomes zero, and all the kinetic energy is converted into potential energy (PE). The potential energy at height \( h \) is given by:
\[ PE = mgh \]
At the maximum height, the height \( h \) can be expressed in terms of the pendulum length \( l \) and the angle \( \theta \):
\[ h = l - l \cos \theta = l(1 - \cos \theta) \]
Thus, the potential energy can be expressed as:
\[ PE = mg(l(1 - \cos \theta)) = mgl(1 - \cos \theta) \]

### Step 4: Apply Conservation of Energy
According to the conservation of mechanical energy:
\[ KE_{initial} = PE_{final} \]
Substituting the expressions we derived:
\[ \frac{mgl}{2} = mgl(1 - \cos \theta) \]

### Step 5: Simplify the Equation
We can cancel \( mgl \) from both sides (assuming \( m \neq 0 \) and \( l \neq 0 \)):
\[ \frac{1}{2} = 1 - \cos \theta \]
Rearranging gives:
\[ \cos \theta = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 6: Solve for \( \theta \)
The angle \( \theta \) for which \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = 60^\circ \]

### Final Answer
The angle with the vertical through which the ball swings before its velocity becomes zero is \( \theta = 60^\circ \).

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