1. Initial Equilibrium State
Initially, the rod is at rest. The total downward force is gravity ($mg$), balanced by the tension ($T_0$) in the two cords.

$$ 2T_0 = mg \implies T_0 = \frac{mg}{2} $$

2. Motion Constraints
When the rod rotates with angular velocity $\omega$, the ends of the rod move in a horizontal circle of radius $R = L/2$. The tangential speed of the rod’s tips is:

$$ v = \omega R = \frac{\omega L}{2} $$

Since the cord length $l$ is fixed, the tip of the rod is constrained to move on the surface of a sphere of radius $l$. The center of this sphere is the suspension point on the ceiling.

3. Vertical Acceleration
For a particle moving on a spherical surface, the acceleration normal to the surface (towards the center) is the centripetal acceleration. At the bottom-most point, this acceleration is directed vertically upwards:

$$ a_{up} = \frac{v^2}{l} $$

Substituting $v = \omega L / 2$:

$$ a_{up} = \frac{(\frac{\omega L}{2})^2}{l} = \frac{\omega^2 L^2}{4l} $$

Because the rod is rigid, the Center of Mass (CM) must rise with this same acceleration $a_{cm} = a_{up}$.

4. Newton’s Second Law
Applying $\sum F_y = m a_{cm}$ where $T$ is the new tension:

$$ 2T – mg = m \left( \frac{\omega^2 L^2}{4l} \right) $$

The increment in tension is defined as the total force increase required to produce this acceleration. Since $2T_0 = mg$, the change in tension $\Delta T = T – T_0$ is:

$$ 2(T – T_0) = \frac{m \omega^2 L^2}{4l} $$ $$ \Delta T = \frac{m \omega^2 L^2}{8l} $$

This method analyzes the vertical displacement $h$ as a function of the rotation angles, derived from the geometry.

1. Geometric Relations
Let $\phi$ be the angle of rod rotation in the horizontal plane.
Let $\theta$ be the angle the string makes with the vertical.
For a small rotation, the horizontal displacement of the rod’s tip can be expressed in two ways:

• Arc length of rod: $x \approx \frac{L}{2} \phi$
• Horizontal projection of string: $x = l \sin \theta \approx l \theta$ (for small $\theta$)

Equating these gives the relationship between $\theta$ and $\phi$:

$$ l \theta = \frac{L}{2} \phi \implies \theta = \frac{L}{2l} \phi $$

2. Vertical Kinematics ($\dot{h}$ and $\ddot{h}$)
The vertical rise of the rod, $h$, is determined by the string angle:

$$ h = l(1 – \cos\theta) $$

Differentiating with respect to time to find velocity ($\dot{h}$):

$$ \dot{h} = l(\sin\theta) \dot{\theta} $$

Differentiating again to find acceleration ($\ddot{h}$):

$$ \ddot{h} = l [ (\cos\theta)\dot{\theta}^2 + (\sin\theta)\ddot{\theta} ] $$

3. Evaluating at $t=0$
At the instant motion starts ($t=0$):

• The displacement is zero, so $\theta = 0, \sin\theta = 0, \cos\theta = 1$.
• The rod has angular velocity $\omega$, so $\dot{\phi} = \omega$.

From our geometric relation $\theta = \frac{L}{2l}\phi$, the angular velocity of the string is:

$$ \dot{\theta} = \frac{L}{2l} \dot{\phi} = \frac{L}{2l} \omega $$

Substituting these values into the expression for $\ddot{h}$:

$$ \ddot{h} = l [ (1)(\frac{L\omega}{2l})^2 + (0)\ddot{\theta} ] $$ $$ \ddot{h} = l \left( \frac{L^2 \omega^2}{4l^2} \right) = \frac{L^2 \omega^2}{4l} $$

4. Force Equation
Using Newton’s Second Law for vertical motion ($2\Delta T = m\ddot{h}$):

$$ \Delta T = \frac{m}{2} \ddot{h} = \frac{m}{2} \left( \frac{L^2 \omega^2}{4l} \right) $$ $$ \Delta T = \frac{m \omega^2 L^2}{8l} $$