I'll throw my hat into the ring too (and hope I don't live to regret it).

Let the bar have length $l$ and mass $m$ and negligible thickness. Given that the couple is applied momentarily (per the edit to the question), we will assume that the bar acquires an angular velocity of $\omega$ as a result of the momentary application of the force couple.

Consider a differential mass $dm$ located on the right side of the bar a distance $r$ from O. Since the bar is uniform, we have

$$dm=\frac{m}{l}dr$$

The differential centripetal force due to $dm$ is then

$$dF_{C}=\frac{(dm)v^2}{r}$$

Substituting angular velocity $\omega$ for linear velocity $v$ where $v=r\omega$

$$dF_{C}=\frac{(dm)r^{2}\omega^{2}}r=(dm)r\omega^{2}$$

Substituting for $dm$

$$dF_{C}=\frac{m}{l}\omega^{2}rdr$$

The centripetal force on the right side of the bar is then

$$F_{C}=\frac{m}{l}\omega^{2}\int_0^{l/2}rdr=\frac{m\omega^{2}l}{8}$$

where $F_{C}$ acts to the left on point O.

The centripetal force on the left side of the bar is equal in magnitude to that on the right, but acts towards the right on point O, for a net force of zero at point O. Considering the entire bar as a system, the centripetal forces are internal to the system, so the net external force on the system remains zero.

  The source of the centripetal force is the tension in the bar. The tension is equal in magnitude and opposite in direction to the centripetal force. Whereas the centripetal force is zero at O and a maximum at the ends of the bar, the tension is zero at the ends and maximum at the center.

Hope this helps.

I'll throw my hat into the ring too (and hope I don't live to regret it).

Let the bar have length $l$ and mass $m$ and negligible thickness. Given that the couple is applied momentarily (per the edit to the question), we will assume that the bar acquires an angular velocity of $\omega$ as a result of the momentary application of the force couple.

Consider a differential mass $dm$ located on the right side of the bar a distance $r$ from O. Since the bar is uniform, we have

$$dm=\frac{m}{l}dr$$

The differential centripetal force due to $dm$ is then

$$dF_{C}=\frac{(dm)v^2}{r}$$

Substituting angular velocity $\omega$ for linear velocity $v$ where $v=r\omega$

$$dF_{C}=\frac{(dm)r^{2}\omega^{2}}r=(dm)r\omega^{2}$$

Substituting for $dm$

$$dF_{C}=\frac{m}{l}\omega^{2}rdr$$

The centripetal force on the right side of the bar is then

$$F_{C}=\frac{m}{l}\omega^{2}\int_0^{l/2}rdr=\frac{m\omega^{2}l}{8}$$

where $F_{C}$ acts to the left on point O.

The centripetal force on the left side of the bar is equal in magnitude to that on the right, but acts towards the right on point O, for a net force of zero at point O. Considering the entire bar as a system, the centripetal forces are internal to the system, so the net external force on the system remains zero. The source of the centripetal force is the tension in the bar.

Hope this helps.

I'll throw my hat into the ring too (and hope I don't live to regret it).

Let the bar have length $l$ and mass $m$ and negligible thickness. Given that the couple is applied momentarily (per the edit to the question), we will assume that the bar acquires an angular velocity of $\omega$ as a result of the momentary application of the force couple.

Consider a differential mass $dm$ located on the right side of the bar a distance $r$ from O. Since the bar is uniform, we have

$$dm=\frac{m}{l}dr$$

The differential centripetal force due to $dm$ is then

$$dF_{C}=\frac{(dm)v^2}{r}$$

Substituting angular velocity $\alpha$ for linear velocity $v$ where $v=r\omega$

$$dF_{C}=\frac{(dm)r^{2}\omega^{2}}r=(dm)r\omega^{2}$$

Substituting for $dm$

$$dF_{C}=\frac{m}{l}\omega^{2}rdr$$

The centripetal force on the right side of the bar is then

$$F_{C}=\frac{m}{l}\omega^{2}\int_0^{l/2}rdr=\frac{m\omega^{2}l}{8}$$

where $F_{C}$ acts to the left on point O.

The centripetal force on the left side of the bar is equal in magnitude to that on the right, but acts towards the right on point O, for a net force of zero at point O. Considering the entire bar as a system, the centripetal forces are internal to the system, so the net external force on the system remains zero.

The source of the centripetal force is the tension in the bar. The tension is equal in magnitude and opposite in direction to the centripetal force. Whereas the centripetal force is zero at O and a maximum at the ends of the bar, the tension is zero at the ends and maximum at the center.

Hope this helps.

I'll throw my hat into the ring too (and hope I don't live to regret it).

Let the bar have length $l$ and mass $m$ and negligible thickness. Given that the couple is applied momentarily (per the edit to the question), we will assume that the bar acquires an angular velocity of $\omega$ as a result of the momentary application of the force couple.

Consider a differential mass $dm$ located on the right side of the bar a distance $r$ from O. Since the bar is uniform, we have

$$dm=\frac{m}{l}dr$$

The differential centripetal force due to $dm$ is then

$$dF_{C}=\frac{(dm)v^2}{r}$$

Substituting angular velocity $\omega$ for linear velocity $v$ where $v=r\omega$

$$dF_{C}=\frac{(dm)r^{2}\omega^{2}}r=(dm)r\omega^{2}$$

Substituting for $dm$

$$dF_{C}=\frac{m}{l}\omega^{2}rdr$$

The centripetal force on the right side of the bar is then

$$F_{C}=\frac{m}{l}\omega^{2}\int_0^{l/2}rdr=\frac{m\omega^{2}l}{8}$$

where $F_{C}$ acts to the left on point O.

The centripetal force on the left side of the bar is equal in magnitude to that on the right, but acts towards the right on point O, for a net force of zero at point O. Considering the entire bar as a system, the centripetal forces are internal to the system, so the net external force on the system remains zero.

The source of the centripetal force is the tension in the bar. The tension is equal in magnitude and opposite in direction to the centripetal force. Whereas the centripetal force is zero at O and a maximum at the ends of the bar, the tension is zero at the ends and maximum at the center.

Hope this helps.