Let's say that the torque equation about the pivot (in the pivot's frame of reference) is shown below: $$k\sin\phi=I\ddot\phi$$ where $k$ is some arbitrary constant.
However, the pivot point is accelerating at $\textbf{a}$ in the inertial reference frame. Then how can we convert the torque equation from the pivot's frame of reference to the inertial reference frame? Thank you.
First off the equation presented is not correct because the torque about the pivot must contain terms that relate to the translational acceleration of the center of mass. The torque due to inertial forces is missing. It is better to just state that the torque of the joint at the pivot is known (and it is a function of the configuration of the system). Also, it is easier to treat a planar problem as a 3D problem and only consider the in-plane components in the end.
So I understand the problem as follows:
A planar rigid body with the center of mass at point C (see below) contains a "pivot point" (point A) below, with prescribed translational acceleration $\boldsymbol{a}_A = \pmatrix{ \ddot{x}_A & \ddot{y}_A & 0}^\top$, but free to rotate about this point. The axis of rotation coming out of the plane is $\boldsymbol{\hat{k}}$, and the angle of rotation is $\phi$. This pivot has known torque $ \boldsymbol{\tau}_A = \pmatrix{ 0 & 0 & \tau(\phi)}^\top$, which affects the motion of the body. Write the equations of motion of such a system.
For convenience, I use the vector $\boldsymbol{c}_1 = \boldsymbol{r}_1 - \boldsymbol{r}_A = \pmatrix{c_x & c_y & 0}^\top$ to describe the position of the center of mass of body 1, relative to point A.
One way to solve this problem is to consider the following steps:
Kinematics - given the motion of the pivot, transform it as motion of the center of mass. $$ \begin{aligned} \boldsymbol{\omega}_1 &= \boldsymbol{\hat{k}}\,\dot \phi \\ \boldsymbol{\alpha}_1 &= \boldsymbol{\hat{k}}\, \ddot \phi \\ \boldsymbol{a}_C &= \boldsymbol{a}_A + \boldsymbol{\alpha}_1 \times \boldsymbol{c}_1 + \boldsymbol{\omega}_1 \times ( \boldsymbol{\omega}_1 \times \boldsymbol{c}_1) \end{aligned}$$
Forces and Torques - Account for all external & constraint forces and torques applied to the body. The pivot must supply a constraint force $\boldsymbol{F}_A = \pmatrix{A_x & A_y & 0}^\top$ as well as the known torque $\boldsymbol{\tau}_A$. If gravity is present, we can account for it with a force $\boldsymbol{W} = m_1 \boldsymbol{g} = \pmatrix{0 & -m_1 g & 0}^\top$ applied top the center of mass
Dynamics - Use the sum of forces and torques about the center of mass to relate to the motion of the center of mass. $$ \left. \begin{aligned} \boldsymbol{F}_A + \boldsymbol{W} & = m_1 \boldsymbol{a}_C \\ \\ \boldsymbol{\tau}_A -\boldsymbol{c}_1 \times \boldsymbol{F}_A &= \mathrm{I}_1 \boldsymbol{\alpha}_1 \end{aligned} \right\} \begin{aligned} A_x & = m_1 (\ddot{x}_A -c_y \ddot{\phi} -c_x \dot{\phi}^2) \\ A_y -m_1 g & = m_1 (\ddot{y}_A+c_x \ddot{\phi}-c_y \ddot{\phi}^2) \\ \tau(\phi) + c_y \,A_x -c_x \, A_y &= I_1 \ddot{\phi} \end{aligned} $$
which is three equations to be solved for three unknowns, the two reaction forces $A_x$, $A_y$, and the rotational acceleration $\ddot{\phi}$.
The correct equation relating rotational acceleration to applied torque $\tau_A = k \sin \phi$ involves the mass moment of inertia about the pivot $I_A = I_1 + m_1 (c_x^2+c_y^2)$ and it is $$ \boxed{ \tau_A + c_y m_1 \ddot{x}_A - c_x m_1 (g+\ddot{y}_A) = I_A \ddot \phi} $$
NOTE: A lot of out-of-plane terms have been neglected which should be present for a 3D system. For example the rotational acceleration vector in 3D is $\boldsymbol{\alpha}_1 = \boldsymbol{\hat{k}}\,\ddot{\phi} + \boldsymbol{\omega}_1 \times \boldsymbol{\hat{k}}\,\dot{\psi}$ but the last part is ingored because $\boldsymbol{\omega}_1$ is always parallel to $\boldsymbol{\hat{k}}$ in planar systems.
