A horizontal heavy uniform bar of weight W is supported at its ends by two men. At the instant, one of the men lets go off his end of the rod, the other feels the force on his hand changed to

The correct option is W/4.

To understand why the force changes to this value at the instant one man lets go, we can analyze the rotational and translational dynamics of the uniform bar.

Let:
- $W$ be the weight of the bar,
- $M$ be the mass of the bar (where $W=Mg$),
- $L$ be the length of the bar,
- $F$ be the vertical force exerted by the remaining support (the other man) at the instant of release.

Step 1: Identify the Pivot and Torque
At the exact instant one man releases his end, the other end (where the second man is holding the bar) acts as an instantaneous pivot point. The bar begins to rotate about this pivot.
The force of gravity (weight $W$) acts at the center of mass of the bar, which is at a distance of $\frac{L}{2}$ from the pivot.
The torque $\tau$ about this pivot is given by:

$\tau =Mg\left(\frac{L}{2}\right)$

Step 2: Relate Torque to Angular Acceleration
Using the rotational analog of Newton's second law:

$\tau =I\alpha$

where $I$ is the moment of inertia of the uniform bar about one of its ends, and $\alpha$ is the angular acceleration.
For a uniform rod of mass $M$ and length $L$ rotating about one end:

$I=\frac{1}{3}M{L}^2$

Substituting $\tau$ and $I$ into the torque equation:

$Mg\frac{L}{2}=\left(\frac{1}{3}M{L}^2\right)\alpha$

Step 3: Calculate the Angular Acceleration
Solving for $\alpha$ by simplifying the equation:

$\alpha =\frac{3g}{2L}$

Step 4: Find the Linear Acceleration of the Center of Mass
The linear downward acceleration of the center of mass, $a_{\text{cm}}$, is related to the angular acceleration $\alpha$ by:

$a_{\text{cm}}=\alpha \left(\frac{L}{2}\right)$

Substituting the value of $\alpha$:

$a_{\text{cm}}=\left(\frac{3g}{2L}\right)·\frac{L}{2}=\frac{3}{4}g$

Step 5: Apply Newton's Second Law for Translation
Now we write the equation for the vertical translational motion of the center of mass:

$Mg-F=M{a}_{\text{cm}}$

Substituting $a_{\text{cm}}$ into the equation:

$F=Mg-M\left(\frac{3}{4}g\right)$

$F=\frac{1}{4}Mg=\frac{W}{4}$

Thus, at the instant the rod is released, the force supported by the remaining hand instantaneously changes from the initial static value of $\frac{W}{2}$ to $\frac{W}{4}$.