A rod of length `L` is held vertically on a smooth horizontal surface. The top end of the rod is given a gentle push. At at certain instant of time, when the rod makes an angle `37^@` with horizontal the velocity of `COM` of the rod of `2m//s`. The velocity of the end of the rod in contact with the surface at that instant is

Options

A. `2ms^(-1)`
B. `1ms^(-1)`
C. `4ms^(-1)`
D. `1.5ms^(-1)`

Explanation

To solve the problem, we need to find the velocity of the end of the rod in contact with the surface when the rod makes an angle of \(37^\circ\) with the horizontal and the velocity of the center of mass (COM) is \(2 \, \text{m/s}\).

### Step-by-Step Solution:

1. **Understand the Geometry of the Rod**:
- The rod is making an angle of \(37^\circ\) with the horizontal.
- The length of the rod is \(L\).
- The center of mass (COM) of the rod is located at a distance of \(L/2\) from either end.

2. **Identify the Velocities**:
- The velocity of the center of mass (COM) is given as \(V_{COM} = 2 \, \text{m/s}\) downward.
- Let the angular velocity of the rod be \(\omega\).

3. **Determine the Velocity Components**:
- The velocity of the point \(P\) (the end of the rod in contact with the surface) has two components:
- A vertical component due to the rotation of the rod.
- A horizontal component due to the motion of the center of mass.

4. **Vertical Component of Velocity**:
- The vertical component of the velocity of point \(P\) can be expressed as:
\[ V_{P, vertical} = \omega \frac{L}{2} \cos(37^\circ) - V_{COM} \]
- Since the rod is in contact with the surface, this vertical component should equal zero:
\[ \omega \frac{L}{2} \cos(37^\circ) - 2 = 0 \]
- Rearranging gives:
\[ \omega \frac{L}{2} \cos(37^\circ) = 2 \]

5. **Horizontal Component of Velocity**:
- The horizontal component of the velocity of point \(P\) is given by:
\[ V_{P, horizontal} = \omega \frac{L}{2} \sin(37^\circ) \]

6. **Relate the Two Components**:
- From the vertical velocity equation, we can express \(\omega\):
\[ \omega = \frac{4}{L \cos(37^\circ)} \]
- Substitute this into the horizontal component:
\[ V_{P, horizontal} = \left(\frac{4}{L \cos(37^\circ)}\right) \frac{L}{2} \sin(37^\circ) = \frac{2 \sin(37^\circ)}{\cos(37^\circ)} = 2 \tan(37^\circ) \]

7. **Calculate the Velocity of Point \(P\)**:
- Since \(\tan(37^\circ) = \frac{3}{4}\):
\[ V_{P, horizontal} = 2 \cdot \frac{3}{4} = \frac{3}{2} \, \text{m/s} = 1.5 \, \text{m/s} \]

### Final Answer:
The velocity of the end of the rod in contact with the surface is \(1.5 \, \text{m/s}\).