A rod of length 1m and mass `0.5 kg` is fixed at one end is initially hanging vertical. The other end is now raised until it makes an angle `60^(@)` with the vertical. How much work is required ?

Answer

To solve the problem of how much work is required to raise the rod from a vertical position to an angle of 60 degrees with the vertical, we can follow these steps:

### Step 1: Identify the mass and length of the rod
- The mass of the rod, \( m = 0.5 \, \text{kg} \)
- The length of the rod, \( L = 1 \, \text{m} \)

### Step 2: Determine the initial position of the center of mass
- When the rod is hanging vertically, the center of mass is located at the midpoint of the rod.
- Therefore, the initial height of the center of mass from the reference level (ground) is:
\[ h_i = \frac{L}{2} = \frac{1}{2} \, \text{m} = 0.5 \, \text{m} \]

### Step 3: Determine the final position of the center of mass
- When the rod is raised to an angle of \( 60^\circ \) with the vertical, we need to find the new height of the center of mass.
- The center of mass will be at a distance of \( \frac{L}{2} \) from the fixed end, which can be calculated using trigonometry:
\[ h_f = \frac{L}{2} \cos(60^\circ) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \, \text{m} = 0.25 \, \text{m} \]

### Step 4: Calculate the change in height of the center of mass
- The change in height \( h \) of the center of mass when the rod is raised is:
\[ h = h_i - h_f = 0.5 \, \text{m} - 0.25 \, \text{m} = 0.25 \, \text{m} \]

### Step 5: Calculate the work done
- The work done \( W \) in raising the center of mass can be calculated using the formula:
\[ W = mgh \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
- Substituting the values:
\[ W = 0.5 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 0.25 \, \text{m} \]
\[ W = 0.5 \cdot 9.81 \cdot 0.25 = 1.22625 \, \text{J} \]

### Final Answer
The work done in raising the rod to an angle of \( 60^\circ \) with the vertical is approximately \( 1.23 \, \text{J} \).

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