A verical thin rod is hinged at point P, due to slight push rod starts to rotate.The velocity of centre of mass of rod just after turning angle 0 is [L is length of rod] (## AAK_TST_09_NEET_19_PHY_E09_041_Q001 ##)

Options

A. `sqrt((6g)/L) sin`0/2`
B. `sqrt((3gL)/2) sin`0/2`
C. `sqrt((6gL)/5 (1+cos theta))`
D. `sqrt((3gL)/2).cos`theta/2`

Answer

To solve the problem of finding the velocity of the center of mass of a vertical thin rod hinged at point P after it has been pushed slightly and rotated through an angle θ, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**:
The system consists of a vertical thin rod of length L, hinged at point P. When the rod is pushed, it rotates about the hinge.

2. **Change in Potential Energy**:
When the rod rotates through an angle θ, the center of mass (which is at a distance L/2 from the hinge) rises. The change in height (h) of the center of mass can be expressed as:
\[ h = \frac{L}{2}(1 - \cos \theta) \]
Therefore, the change in potential energy (ΔPE) is given by:
\[ \Delta PE = mg \cdot h = mg \cdot \frac{L}{2}(1 - \cos \theta) \]

3. **Kinetic Energy**:
The change in potential energy is converted into kinetic energy (KE) of the rod. The moment of inertia (I) of the rod about the hinge is:
\[ I = \frac{1}{3}mL^2 \]
The kinetic energy of the rod when it rotates with angular velocity ω is:
\[ KE = \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{1}{3}mL^2 \cdot \omega^2 = \frac{1}{6}mL^2 \omega^2 \]

4. **Equating Potential Energy and Kinetic Energy**:
Setting the change in potential energy equal to the kinetic energy gives:
\[ mg \cdot \frac{L}{2}(1 - \cos \theta) = \frac{1}{6}mL^2 \omega^2 \]
We can cancel m from both sides:
\[ g \cdot \frac{L}{2}(1 - \cos \theta) = \frac{1}{6}L^2 \omega^2 \]

5. **Solving for ω**:
Rearranging the equation to solve for ω²:
\[ \omega^2 = \frac{3g(1 - \cos \theta)}{L} \]

6. **Velocity of the Center of Mass**:
The velocity (v) of the center of mass of the rod can be expressed in terms of ω:
\[ v = \omega \cdot \frac{L}{2} \]
Substituting ω² into this equation:
\[ v^2 = \left(\sqrt{\frac{3g(1 - \cos \theta)}{L}}\right)^2 \cdot \left(\frac{L}{2}\right)^2 = \frac{3g(1 - \cos \theta)}{L} \cdot \frac{L^2}{4} = \frac{3gL}{4}(1 - \cos \theta) \]

7. **Using the Trigonometric Identity**:
We can use the identity \(1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right)\) to express the velocity:
\[ v^2 = \frac{3gL}{4} \cdot 2 \sin^2\left(\frac{\theta}{2}\right) = \frac{3gL}{2} \sin^2\left(\frac{\theta}{2}\right) \]
Therefore, the final expression for the velocity of the center of mass is:
\[ v = \sqrt{\frac{3gL}{2}} \cdot \sin\left(\frac{\theta}{2}\right) \]

### Final Answer:
The velocity of the center of mass of the rod just after turning an angle θ is:
\[ v = \sqrt{\frac{3gL}{2}} \cdot \sin\left(\frac{\theta}{2}\right) \]