A ladder of length L is slipping with its ends against a vertical wall and a horizontal floor. At a certain moment, the speed of the end in contact with the horizontal floor is `v` and the ladder makes an angle `theta=30^@` with horizontal. Then, the speed of the ladder's centre of mass must be

Options

A. (a) `sqrt3/2v`
B. (b) `v/2`
C. (c) `v`
D. (d) `2v`

Explanation

To solve the problem, we need to find the speed of the center of mass of a ladder that is slipping against a wall and a floor. Let's break down the solution step by step.

### Step 1: Understand the Geometry
We have a ladder of length \( L \) making an angle \( \theta = 30^\circ \) with the horizontal. The end of the ladder on the ground is moving with a speed \( v \). We denote the points as follows:
- Point A (bottom of the ladder on the ground)
- Point B (top of the ladder against the wall)

### Step 2: Set Up Coordinates
Let:
- The coordinates of point A (on the ground) be \( (x, 0) \).
- The coordinates of point B (against the wall) be \( (0, y) \).

From the geometry of the situation, we can use trigonometric relationships to express \( x \) and \( y \) in terms of the ladder length \( L \):
- \( x = L \cos(30^\circ) = L \cdot \frac{\sqrt{3}}{2} \)
- \( y = L \sin(30^\circ) = L \cdot \frac{1}{2} \)

### Step 3: Determine the Velocity Components
The velocity of point A (bottom of the ladder) is given as \( v \) in the horizontal direction. We need to find the vertical velocity \( v_1 \) of point B (top of the ladder).

Since the ladder is rigid, the velocities along the length of the ladder must be equal. We can express this as:
- The component of the velocity of point A along the ladder: \( v \cos(30^\circ) = v \cdot \frac{\sqrt{3}}{2} \)
- The component of the velocity of point B along the ladder: \( v_1 \cos(60^\circ) = v_1 \cdot \frac{1}{2} \)

Setting these equal gives us:
\[ v \cdot \frac{\sqrt{3}}{2} = v_1 \cdot \frac{1}{2} \]

### Step 4: Solve for \( v_1 \)
From the equation above, we can solve for \( v_1 \):
\[ v_1 = v \cdot \sqrt{3} \]

### Step 5: Find the Center of Mass Velocity
The position of the center of mass \( (x_{cm}, y_{cm}) \) of the ladder can be expressed as:
\[ x_{cm} = \frac{x}{2}, \quad y_{cm} = \frac{y}{2} \]

The velocities of the center of mass \( (v_{cm,x}, v_{cm,y}) \) can be derived from the velocities of points A and B:
\[ v_{cm,x} = \frac{1}{2} v, \quad v_{cm,y} = \frac{1}{2} (-v_1) = -\frac{1}{2} v \sqrt{3} \]

### Step 6: Calculate the Magnitude of the Center of Mass Velocity
The magnitude of the velocity of the center of mass is given by:
\[ v_{cm} = \sqrt{(v_{cm,x})^2 + (v_{cm,y})^2} \]
Substituting the values:
\[ v_{cm} = \sqrt{\left(\frac{1}{2} v\right)^2 + \left(-\frac{1}{2} v \sqrt{3}\right)^2} \]
\[ = \sqrt{\frac{1}{4} v^2 + \frac{3}{4} v^2} = \sqrt{v^2} = v \]

### Conclusion
The speed of the center of mass of the ladder is \( v \).