Is there a closed form formula of rotation matrix for objects with constant angular acceleration?

Question

If an object is of constant angular velocity, then we can derived the closed form formula of rotation matrix thanks to the theory of linear ODEs (and the closed form of $e^{[\omega]}$):

$$R' = [\omega] R \Rightarrow R = e^{[\omega]t}R_0$$

However, if the things got a little more complicated, say, the object is of constant angular acceleration, namely $\omega(t) = \omega_0 + \alpha t$, is a closed form formula for $R$ also possible? Because the ODE $$R' = [\omega_0 + \alpha t]R$$ fails to be linear.

EDIT: this system is indeed linear, but not autonomous. Anyway, I want to emphasis that this is not an ODE of the form $X' = AX$ where A is constant.

Answer 1

You're basically asking how to find the equation to the equation $$ \frac{d\mathbf{R}}{dt} = M(t) \mathbf{R}. $$ It can be shown that the answer can always be written as $$ \mathbf{R}(t) = U(t) \mathbf{R}(0), $$ for a linear operator $U(t)$, but the form of $U(t)$ depends on whether $M(t_1)$ commutes with $M(t_2)$ for all $t_1$ & $t_2$. If $[M(t_1), M(t_2)] = 0$, then it can be shown that we have $$ U(t) = \exp \left[ \int_0^t dt'\, M(t') \right]. \tag{1} $$ In particular, the case where $M$ is independent of $t$ means that $U(t) = e^{M t}$.

If, on the other hand, $[M(t_1), M(t_2)] \neq 0$, then we have to define the evolution operator in terms of a Dyson series: $$ U(t) = 1 + \sum_{n = 1}^\infty \int_0^t dt_1 \int_0^{t_1} dt_2 \cdots \int_0^{t_{n-1}} dt_{n} \, M(t_1) M(t_2) \cdots M(t_n) \tag{2} $$ In other words, this is an infinite sum whose first term is a single integral, whose second term is a double integral, whose third term is a triple integral, etc. This turns out to be equivalent to $$ U(t) = 1 + \sum_{n = 1}^\infty \frac{1}{n!} \int_0^t dt_1 \int_0^t dt_2 \cdots \int_0^t dt_{n} \, \mathcal{T} \left[ M(t_1) M(t_2) \cdots M(t_n) \right] $$ where $\mathcal{T} \left[ M(t_1) M(t_2) \cdots M(t_n) \right]$ is the time-ordered product of the operators, i.e., the operators are reordered so that the arguments decrease from left to right in the string. In this form, it is a bit easier to see how this might be equivalent to the form (1) above.

In particular, if we are talking about 2-D rotations, then the 2-D rotation group is Abelian, i.e., every rotation commutes with every other rotation and we're in the case where we can just write $U(t)$ in the form from (1) above. If, on the other hand, we're talking about 3-D rotations, then it is no longer the case that two rotations at different times will commute (because 3-D rotations generally do not commute), and so we would have to deal with the form (2) for $U(t)$ instead.

Answer 2

Sure. The shape of the matrix that produces the rotation is the same. It's the standard [cos - sin sin cos] thing. The angle simply changes. For constant angular velocity the angle is just $wt$. For constant acceleration it's just $\frac{1}{2} a t^2 + wt$. So you just put that in where you had $wt$.