Let's say, the wrench is attached and rotates around a fixed point $O$. As an idealization, the wrench is represented as a rod of length $L$, represented as a segment from point $O$ to its other end, and mass $m$ (although the shape of the wrench is not exactly a rod and the mass is not so evenly distributed and is probably more concentrated in one end). The moment of inertia of the wrench, relative to the point $O$, is then $I = \frac{mL^2}{3}$. The equation of motion for the wrench, driven by a torque $T$ generated by the motor and transmitted by the blue arm, should be
$$I \, \frac{d^2\theta}{dt^2} \, = \, - \, \frac{m g L}{2} \, \sin(\theta) \, + \, T$$ where $\theta \, \in \, (-\pi, \pi]$ is the angle between the wrench and the vertical axis, measured in counter-clockwise direction, where the axis oriented from point $O$ downwards, so that, when the torque of the motor is zero $T=0$, the equilibrium of the wrench is at $\theta = 0$. In other words, $\theta = 0$ is your six o'clock.
First, decide what motion you would expect from the wrench and write it as $\theta = \theta_s(t)$. Then by referring to control theory, a good choice for torque that guarantees a stable uniform rotation $\theta = \theta_s(t)$ is one that would make the equation look like an equation that (1) has $\theta = \theta_s(t)$ as a solution and (2) all solutions starting near $\theta = \theta_s(t)$ would stay very close and even converge to the desired one $\theta = \theta_s(t)$ (phenomenon called asymptotic stability). In this case, such target equation, with the properties (1) and (2) is
$$I \, \frac{d^2\theta}{dt^2} \, = \, I \, \frac{d^2\theta_s}{dt^2}(t) \, - \, K_1\left(\frac{d\theta}{dt} - \frac{d\theta_s}{dt}(t)\right) \, - \, K_0\, \big(\theta - \theta_s(t) \big)$$
for some positive constants $K_0$ and $K_1$. The larger they are, the more stable and the more attractive the desired motion $\theta = \theta_s(t)$ is but the bigger the necessary torque could be. Now, if you want to find what torque $T$ turns the first differential equation into the second one, you simply need to equate the two and express $T$, yileding
$$T\left(\theta,\, \frac{d\theta}{dt},\, t \right) = \,\frac{m g L}{2} \, \sin(\theta) \, + \, I \, \frac{d^2\theta_s}{dt^2}(t) \, - \, K_1\left(\frac{d\theta}{dt} - \frac{d\theta_s}{dt}(t)\right) \, - \, K_0\, \big(\theta - \theta_s(t) \big)$$
To make things a bit easier to analyze, you would want to estimate the torque based on the difference $$\varphi = \theta - \theta_s(t)$$ In other words, if you make the change of variable $\varphi = \theta - \theta_s(t)$, then the original equation becomes
$$I\frac{d^2\varphi}{dt^2} \, = \, - \, \frac{m g L}{2} \, \sin\Big(\,\varphi + \theta_s(t)\,\Big) \, - \, I \, \frac{d^2\theta_s}{dt^2}(t) \, + \, T\left(\varphi,\, \frac{d\varphi}{dt},\, t \right)$$
In this new variable $\varphi$, the original desired solution $\theta = \theta_s(t)$ is represented by the solution $\varphi(t) \equiv 0$ and just like before, a good control $T$ would turn the previous equation into
$$I \frac{d^2\varphi}{dt^2} \, = \, -\, K_1 \, \frac{d\varphi}{dt} \, - \, K_0 \, \varphi$$ which means that, since $\frac{d\varphi}{dt} = \frac{d\theta}{dt} - \frac{d\theta_s}{dt}(t)$, the controlling torque, expressed in terms of $\varphi$, is
$$T = T\left(\varphi,\, \frac{d\varphi}{dt},\, t \right) = \,\frac{m g L}{2} \, \sin\big(\varphi + \theta_s(t)\big) \, + \, I \, \frac{d^2\theta_s}{dt^2}(t) \, - \, K_1\,\frac{d\varphi}{dt} \, - \, K_0\, \varphi$$
As an example (or maybe that's what you actually want), let us assume you want to make the wrench rotate uniformly counter-clockwise, so that it makes one full rotation in time $\tau$.
Then, set $\omega = \frac{2\pi}{\tau}$ and thus you want for the angle $\theta = \theta_s(t) = \omega \, t$ with respect to time $t$.
Because of the uniform nature of $\theta_s(t) = \omega \, t$, it is easy to see that $\frac{d^2\theta_s}{dt^2}(t) = 0$ for all $t$, so the torque becomes
$$T \, = \,\frac{m g L}{2} \, \sin\big(\varphi + \omega\, t\big) \, - \, K_1\,\frac{d\varphi}{dt} \, - \, K_0\, \varphi$$
Realistically, initially the wrench will be in position $\theta_0 = \theta(0) = 0$ at time $t=0$ with velocity $ = \frac{d\theta}{dt}(0) = 0$ at time $t=0$.
In terms of $\varphi$ however, at $t=0$ you would have $\varphi_0 = \theta_0 - \omega \, 0 = 0$ and $\dot{\varphi}_0 = \dot{\theta}_0 - \omega = - \omega$.
To get a very good estimate for the toque needed for this kind of motion, we have to calculate the solution to the initial value problem
\begin{align*}
I \frac{d^2\varphi}{dt^2} \, &= \, -\, K_1 \, \frac{d\varphi}{dt} \, - \, K_0 \, \varphi \\
\varphi(0) &= 0\\
\frac{d\varphi}{dt}(0) &= -\omega
\end{align*}
Observe the equation is linear with constant coefficients, so it is explicitly solvable.
After we have calculated this solution, call it $\varphi = \varphi(t)$, we plug it in the expression for the torque above:
\begin{align*}
T \, =& \, T\big( t, \, m, \,L, \, \omega, K_0,\, K_1 \,) \, = \\
=& \, \frac{m g L}{2} \, \sin\big(\varphi(t) + \omega\, t\big) \, - \, K_1\,\frac{d\varphi}{dt}(t) \, - \, K_0\,
\varphi(t)\\
&= \, \frac{m g L}{2} \, \sin\big(\varphi(t) + \omega\, t\big) \, + \, I\, \frac{d^2\varphi}{dt^2}(t)
\end{align*}
The latter equality comes form the fact that $- K_1\frac{d\varphi}{dt}(t) - K_0\varphi(t) = I\, \frac{d^2\varphi}{dt^2}(t)$
because $\varphi(t)$ is a solution of the linear differential equation above. At this point, we can tell, that by design, as $t$ grows,
both $\varphi(t)$ and $\frac{d\varphi}{dt}(t)$ approach $0$ very fast, and therefore, by the linear differential equation,
so does $\frac{d^2\varphi}{dt^2}(t)$,
which allows us to say that for large enough times $t$
$$|T| \leq \frac{m g L}{2} + I = \frac{m g L}{2} + \frac{mL^2}{3} $$ and in fact the torque's maximum is close to $\frac{m g L}{2}$.
However, at this point I am not completely sure what happens to $\frac{d^2\varphi}{dt^2}(t)$ for small $t$, in the beginning of the motion, when
this acceleration could be potentially large, so I would explore further, just to be sure.
Instead of using $K_0$ and $K_1$, define two positive numbers $0 < \lambda_1 < \lambda_2$, such that
$\lambda_1 \lambda_2 = \frac{K_0}{I}$ and $\lambda_1 + \lambda_2 = \frac{K_1}{I}$. This is always possible whenever $K_1^2 - 4\, K_0 \, I \geq 0$
Then the linear differential equation can be written as follows:
\begin{align*}
\frac{d^2\varphi}{dt^2} \, &= \, -\, (\lambda_1 + \lambda_2) \, \frac{d\varphi}{dt} \, - \, (\lambda_1 \lambda_2) \, \varphi \\
\varphi(0) &= 0\\
\frac{d\varphi}{dt}(0) &= -\omega
\end{align*}
It can be directly verified that the general solution to the linear differential equation is
$$\varphi = c_1\,e^{-\lambda_1t} + c_2\,e^{-\lambda_2t}$$ for any constants $c_1$ and $c_2$. In particular,
if we want the solution to satisfy the initial conditions too, then
$$\varphi \, = \,\frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_2 t} \, - \,\frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_1\, t}$$
with first and second derivatives
$$\frac{d\varphi}{dt} \, = \,\frac{- \lambda_2\, \omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_2\, t} \, - \,\frac{- \lambda_1 \,\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_1\, t} $$
$$\frac{d^2\varphi}{dt^2} \, =
\,\frac{\lambda_2^2\, \omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_2\, t} \,
- \,\frac{\lambda_1^2\,\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_1\, t} $$
So, when you plug these expressions into the formula for the torque, using the identity
$$- K_1\frac{d\varphi}{dt}(t) - K_0\varphi(t) = I\, \frac{d^2\varphi}{dt^2}(t)$$
to simplify the calculations, it can be expressed as
\begin{align*}
T \, =& \,\, T\big( t, \, m, \,L, \, \omega, \lambda_1,\, \lambda_2 \,)\\
=& \,\, \frac{m g L}{2} \, \sin\left( \frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_2 t} \, - \,\frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_1 t}\,
+\, \omega\, t \right)\\
&\,\, + \, \frac{m\, L^2}{3}\, \left(\frac{\lambda_2^2\, \omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_2 t} \, -
\,\frac{\lambda_1 ^2\,\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_1 t} \right)
\end{align*}
or alternatively
\begin{align*}
T \, =& \,\, T\big( t, \, m, \,L, \, \omega, \lambda_1,\, \lambda_2 \,)\\
=& \,\, \frac{m g L}{2} \, \sin\left( \frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_2 t} \, - \,\frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_1 t}\,
+\, \omega\, t \right)\\
&\,\, + \, \frac{ \omega \, m\, L^2 }{ 3(\lambda_2 - \lambda_1) }\, \left(\lambda_2^2 \, e^{- \lambda_2 t} \, - \, \lambda_1^2 \, e^{- \lambda_1 t}\right)
\end{align*}
If the mass of the wrench is $m = 1.2$ kg and its length is $L = 0.475$ meters, the torque is
\begin{align*}
T \, =& \, T\big( t, \, \omega, \lambda_1,\, \lambda_2 \,)\\
=& \,\, 2.79585 \, \sin\left( \frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_2 t} \, - \,\frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_1 t}\,
+\, \omega\, t \right)\\
&\,\,+ \, \frac{0.09025\, \omega}{ 3(\lambda_2 - \lambda_1) }\,
\left(\lambda_2^2 \, e^{- \lambda_2 t} \, - \, \lambda_1^2 \, e^{- \lambda_1 t}\right)
\end{align*}
If you analyze the function $f(t) = \lambda_2^2 \, e^{- \lambda_2 t} \, - \, \lambda_1^2 \, e^{- \lambda_1 t}$ you will see that for $t \geq 0$
it starts from its positive maximum at $t = 0$, then at some point reaches a much smaller, in absolute value,
negative minimum and then very rapidly asymptotically converges to $0$ from below. Thus,
$$|f(t)| \leq f(0) = \lambda_2^2 \, - \, \lambda_1^2 = (\lambda_2 - \lambda_1)(\lambda_2 + \lambda_1)$$ and we can estimate, given some reasonable values for
$\lambda_2 > \lambda_1 > 0$
\begin{align*}
|\, T \,| \leq & \,\, 2.79585 \, \left|\, \sin\left( \frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_2 t} \, - \,\frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_1 t}\,
+\, \omega\, t \right) \,\right|\\
&\,\,+ \, \frac{0.09025\, \omega}{ 3}\,(\lambda_2 + \lambda_1)
\end{align*}
Now, we see that the torque would most likely not exceed
$$
|\, T \,| \leq \,\, 2.79585 \,+ \, 0.09025\,\frac{\omega}{ 3}\,(\lambda_2 + \lambda_1) \,
= \, \, 2.79585 \,+ \, 0.09025\,\frac{2\, \pi}{3\, \tau}\,(\lambda_2 + \lambda_1)$$
$$|\, T \,| \leq \,\, 2.79585 \,+ \, 0.189 \,\frac{(\lambda_2 + \lambda_1)}{\tau}\,
$$
But it will definitely reach the value $2.79585$ (in fact it will keep reaching it during the rotation when the wrench is near horizontal position).
Thus for the maximal torque required, we get the estimates
$$2.79585 \, \leq \, \max|\, T \,| \, \leq \, 2.79585 \,+ \, 0.189 \,\frac{(\lambda_2 + \lambda_1)}{\tau}\,$$
The larger you pick $\lambda_1$ and $\lambda_2$, the faster the wrench will enter the rotation mode $\theta = \omega \, t$. But then, more torque is required.
The larger the period of rotation $\tau$, the smaller the maximal torque that is required.
For example, if you pick $\lambda_2 = 3$ and $\lambda_1 = 1$ and $\tau = 12$ seconds, the max torque's estimate is
$$2.79585 \, \leq \, \max|\, T \,| \, \leq \, 2.79585 \,+ \, 0.063\, = 2.85885$$
I guess, to get your wrench to move uniformly in a full circle, you would aim to find a motor that can generate maximal torque somewhere between $2.8$ and $2.85885$ at least.
$$$$
For further exploration, you could take the formula
\begin{align*}
T \, =& \, T\big( t, \, \omega, \lambda_1,\, \lambda_2 \,)\\
=& \,\, 2.79585 \, \sin\left( \frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_2 t} \, - \,\frac{\omega}{\lambda_2 - \lambda_1}\, e^{- \lambda_1 t}\,
+\, \omega\, t \right)\\
&\,\,+ \, \frac{0.09025\, \omega}{ 3(\lambda_2 - \lambda_1) }\,
\left(\lambda_2^2 \, e^{- \lambda_2 t} \, - \, \lambda_1^2 \, e^{- \lambda_1 t}\right)
\end{align*}
go to the website of Desmos https://www.desmos.com/calculator write the formula there, using $t$ as a variable and $\omega, \lambda_1, \lambda_2$ as parameters,
and you can see how the demand for torque changes with time, if you want your wrench to rotate uniformly like $\theta = \omega \, t$.
You can even simulate the more general formula
\begin{align*}
T \, =& \, T\big( t, \, \omega, \lambda_1,\, \lambda_2, \, c_1,\, c_2 \,)\\
=& \,\, 2.79585 \, \sin\left( c_1\, e^{- \lambda_2 t} \, - \,c_2\, e^{- \lambda_1 t}\,
+\, \omega\, t \right)\\
&\,\,+ \, 0.09025\,
\left(\lambda_2^2 c_1 \, e^{- \lambda_2 t} \, - \, \lambda_1^2 c_2 \, e^{- \lambda_1 t}\right)
\end{align*}
where $c_1$ and $c_2$ correspond to different initial positions and velocities of the wrench, not just the equilibrium one.
