A way to model the forces in the bar is staying in its non inertial frame. For this frame, the bar is in equilibrium.

In a small slice with length $\Delta x$ at a distance $x$ from the COM, the balance of forces is $F_x - F_{x+\Delta x} + \rho \Delta x \omega^2 x = 0$, where the last term is the fictitious centrifugal force acting in the slice, and the $F$'s are the centripetal forces at each end of the slice.

Dividing by $\Delta x$ and taking the limit when it goes to zero, and knowing that $\omega \frac{l}{2} = v$:

$$\frac{\partial F}{\partial x} = \rho \frac{4v^2}{l^2} x$$

At the ends of the bar ($x = \frac{l}{2}$), there is no forces, because there is no mass to pull to the center. Integrating, we get the expression for the centripetal force:

$$F(x) = \frac{1}{2}\rho v^2(4\frac{x^2}{l^2} - 1)$$

A way to model the forces in the bar is staying in its non inertial frame. For this frame, the bar is in equilibrium.

In a small slice with length $\Delta x$ at a distance $x$ from the COM, the balance of forces is $F_x - F_{x+\Delta x} + \rho \Delta x \omega^2 x = 0$, where the last term is the fictitious centrifugal force acting in the slice, and the $F$'s are the centripetal forces at each end of the slice.

Dividing by $\Delta x$ and taking the limit when it goes to zero, and knowing that $\omega \frac{l}{2} = v$, where $v$ is the modulus of the velocity of the ends of the bar:

$$\frac{\partial F}{\partial x} = \rho \frac{4v^2}{l^2} x$$

At the ends of the bar ($x = \frac{l}{2}$), there is no forces, because there is no mass to pull to the center. Integrating, we get the expression for the centripetal force:

$$F(x) = \frac{1}{2}\rho v^2(4\frac{x^2}{l^2} - 1)$$

A way to model the forces in the bar is staying in its non inertial frame. For this frame, the bar is in equilibrium.

In a small slice with length $\Delta x$ at a distance $x$ from the COM, the balance of forces is $F_x - F_{x+\Delta x} + \rho \Delta x \omega^2 x = 0$, where the last term is the fictitious centrifugal force acting in the slice, and the $F$'s are the centripetal forces at each end of the slice.

Dividing by $\Delta x$ and taking the limit when it goes to zero, and knowing that $\omega \frac{l}{2} = v$:

$$\frac{\partial F}{\partial x} = \rho \frac{4v^2}{l^2} x$$

At the ends of the bar ($x = \frac{l}{2}$), there is no forces, because there is no mass to pull to the center. Integrating, we get the expression for the centripetal force:

$$F(x) = \frac{1}{2}\rho v^2(4\frac{x^2}{l^2} - 1)$$

A way to model the forces in the bar is staying in its non inertial frame. For this frame, the bar is in equilibrium.

In a small slice with length $\Delta x$ at a distance $x$ from the COM, the balance of forces is $F_x - F_{x+\Delta x} + \rho \Delta x \omega^2 x = 0$, where the last term is the fictitious centrifugal force acting in the slice, and the $F$'s are the centripetal forces at each end of the slice.

Dividing by $\Delta x$ and taking the limit when it goes to zero, and knowing that $\omega \frac{l}{2} = v$:

$$\frac{\partial F}{\partial x} = \rho \frac{4v^2}{l^2} x$$

At the ends of the bar ($x = \frac{l}{2}$), there is no forces, because there is no mass to pull to the center. Integrating, we get the expression for the centripetal force:

$$F(x) = \frac{1}{2}\rho v^2(4\frac{x^2}{l^2} - 1)$$