$\def\ph{\varphi}\def\l{\left}\def\r{\right}\def\nR{\mathbb{R}}\def\hF{\hat{F}}\def\rmL{{\mathrm{L}}}\def\vr{\vec{r}}\def\vF{\vec{F}}\def\m#1{{\mathbf{#1}}}\def\det{\operatorname{det}}\def\vomega{{\vec{\omega}}}\def\vph{{\vec{\varphi}}}\def\vom{{\vec{\omega}}}\def\valpha{{\vec{\alpha}}}\def\vv{{\vec{v}}}\def\ddt{\frac{d}{dt}}$
We assume that the rod can be regarded as a straight line section with the rotation angle $\ph$ as its only degree of freedom. We use the pivot as origin. With these assumptions the bar can be described as
$$
z(s,t) = s\exp(i\ph(t)).
$$
in the complex plane with $s\in[0,l]$.
From this we can easily compute the velocity and the acceleration of the bar points:
\begin{align}
\dot z(s,t) &= i s \exp(i\ph(t))\dot\ph(t)\\
\ddot z(s,t) &= s \exp(i\ph(t))\bigl(-\dot\ph(t)^2 + i\ddot\ph(t)\bigr)
\end{align}
As you say, we know the force applied to the outer bar end. Let us name it $F_l(t)$.
Since you do not want to deal with elasticities you have constraints and you need to apply one of the principles of mechanics for constrained systems. This is as close as you get to Newton's principles for a system with constraints. It has the interpretation that Newton's law must be valid in the direction of the degree of freedom. In all the other directions there are constraint forces keeping the system within the constraints. If you do not want to apply one of the principles of mechanics you have to consider the full 3d-beam with elastic forces and maybe take the limit to a rigid bar. You find a good description about that procedure in Arnolds book for point mass systems. I apply d'Alembert's principle here. Luckily, we do not need to consider the pivot force since there the virtual displacement $\delta z(s,\ph) = s\exp(i\ph(t))i\delta\ph$ is zero because of $s=0$ there.
\begin{align*}
0&=\int_{s=0}^l \l\langle\delta z(s,\ph(t))\mid -\ddot z(s,\ph(t))\r\rangle\frac{m}{l}\,ds + \l\langle\delta z(l,\ph(t))\mid F_l(t)\r\rangle
\end{align*}
Thereby, $\langle\bullet\mid\bullet\rangle$ is the normal scalar product of $\nR^2$ which can be calculated as $\langle a\mid b\rangle=\Re(a^*\cdot b)=a_x b_x + a_y b_y$.
\begin{align}
0&= \Re\l(\int_0^l \l(s\exp(-i\ph(t))(-i)\delta\ph\cdot \l(-s\exp(i\ph(t))(-\dot\ph(t)^2+i\ddot\ph(t))\r)\r)\frac ml\,ds + \l( l\exp(-i\ph(t))(-i)\delta\ph\cdot F_l(t)\r)\r)\\
0&= \delta\ph\cdot\l(-\ddot\ph(t)\frac {ml^2}3 + l\Re\l(-i\exp(-i\ph(t))F_l(t)\r)\r)
\end{align}
This equation must be valid for all virtual displacements $\delta\ph$, e.g. $\delta\ph=1$ which gives us the equation of motion
$$
\ddot\ph(t)\frac {ml^2}3 = l\Re\l(-i\exp(-i\ph(t))F_l(t)\r)
$$
Let us consider a force $F_l(t)$ that always acts orthogonal to the bar and has constant absolute value $\hF$. We choose the orientation of the force such that it drives the bar in mathematically positive direction. It can be represented as
$$
F_l(t) = i\hF\exp(i\ph(t))
$$
Therewith we get
$$
\ddot\ph(t)\frac {ml^2}3 = l\Re\l(\hF\r) = l\hF.
$$
The bar has been modeled as a line segment to simplify the calculation.
The restriction of the motion to a plane is a further simplification.
The general rigid body model is a domain $B\subset\nR^3$ embeded through a rigid-body motion
\begin{align}
\vr(\vr^\rmL,t) &= \vr_0(t) + R(t)\cdot \vr^\rmL
\end{align}
with $\vr^\rmL\in B$. Thereby, $\vr^\rmL$ are point coordinates in the rigid body reference frame and (for simplicity) $R$ is a rotation matrix (with $R R^T = \m1$ and $\det(R)=1$).
Let $\vF_k$ be external forces imprinted to the body at points $\vr_k^\rmL$ $(k=1,\ldots,n)$. The corresponding points in space are $\vr_k = \vr_0 + R\vr_k^\rmL$.
The variation of $\vr_0$ is $\delta\vr_0$.
We look at the variation of $R$ somewhat more closely. The derivative of $R R^T = \m1$ gives $$\m0=\delta(\m1)=\delta(R\cdot R^T) = \delta R \cdot R^T + R \cdot \delta R^T.$$
That means the matrix
$$
\delta\Phi:=\delta R \cdot R^T = -R \cdot \delta R^T = -(\delta R\cdot R^T)^T
$$
is skew-symmetric and therefore only has 3 relevant components $\delta\ph_1:=\delta\Phi_{32},\delta\ph_2:=\delta\Phi_{13},\delta\ph_3:=\delta\Phi_{21}$. With the vector $\delta\vph:=(\delta\ph_1,\delta\ph_2,\delta\ph_3)$ of these three relevant components the product of $\delta\Phi$ with any vector $\vec a$ can be represented as cross product
$$
\delta\Phi\cdot \vec a = \delta\vph \times \vec a.
$$
Now, we are ready to calculate the virtual displacement of the rigid body motion
$$
\delta\vr(\vr^\rmL)=\delta\vr_0 + \delta R \vr^\rmL
$$
Augmenting with the factor $\m{1}=R^TR$ gives
\begin{align}
\delta\vr(\vr^\rmL)&=\delta\vr_0 + \delta R R^T R \vr^\rmL\\
&=\delta\vr_0 + \delta\vph\times R\vr^\rmL
\end{align}
In the same way we can compute the velocity of the points of the body
\begin{align}
\vv(\vr^\rmL,t):=\dot\vr(\vr^\rmL,t) &= \dot\vr_0 + \dot R \vr^\rmL\\
&= \vv_0 + \vom \times R\vr^\rmL
\end{align}
with $\vom\times \l(R\vr^\rmL\r) := \dot R R^T \l(R\vr^\rmL\r)$.
Alembert's principle gives
\begin{align}
0 &= \ddt\int_{\vr^\rmL\in B} \delta\vr(\vr^\rmL)^T \cdot (-\rho \vv(\vr^\rmL,t)) \cdot d V + \sum_{k=1}^n \delta\vr_k \cdot \vF_k
\end{align}
Putting in all the bits gives:
\begin{multline}
\ddt\int_{\vr^\rmL\in B} \l(\delta\vr_0 + \delta \vph\times R\vr^\rmL\r)^T \cdot \l(\vv_0 +\vom \times R\vr^\rmL\r) \cdot \rho d V \\
- \sum_{k=1}^n \l(\delta\vr_0 + \delta \vph\times R\vr^\rmL_k\r)^T \cdot \vF_k = 0
\end{multline}
Considering that in general $(\vec a\times \vec b) \cdot \vec c = \vec a\cdot \l(\vec b \times \vec c\r)$ one obtains
\begin{multline}
\delta\vr_0^T\cdot\l(\ddt\int_{\vr\in B} \l(\vv_0 + \vom \times R\vr^\rmL\r) \cdot \rho d V -\sum_k \vF_k \r)\\
+\delta\vph^T \cdot\l(\ddt\int_{\vr\in B}
\l(R\vr^\rmL\r)\times\l(\vv_0 - \l(R\vr^\rmL\r)\times \vom\r)\rho d V\\
-\sum_k\l(R\vr^\rmL_k\r)\times\vF_k
\r)
= 0
\end{multline}
and with $J(t):=\int_{\vr\in B} -(R\vr^\rmL)\times(R\vr^\rmL)\rho d V$ and $m:=\int_{\vr\in B} \rho d V$ one obtains
$$
\delta\vr_0^T\cdot\l(\ddt \l(\vv_0 + \vom \times R\vr^\rmL\r)m -\sum_k \vF_k \r)\\
+\delta\vph^T \cdot\l(
\ddt\l(m\l(R\vr^\rmL\r)\times\vv_0 + J(t)\times \vom\r)
-\sum_k\l(R\vr^\rmL_k\r)\times\vF_k
\r)
= 0
$$
Since the variations of $\delta\vr$ and $\delta\vph$ are mutual independent their factors must vanish separately and one obtains the well known balance equations
\begin{align}
\ddt \l(\vv_0 + \vom \times R\vr^\rmL\r)m &= \sum_k \vF_k \\
\ddt \l(m\l(R\vr^\rmL\r)\times\vv_0 + J(t)\times \vom\r)
&= \sum_k\l(R\vr^\rmL_k\r)\times\vF_k
\end{align}
