Is there a rotational analogue to reduced mass, i.e. reduced moment of inertia? Is there a rotational analogue to reduced mass? Is there anything called reduced moment of inertia?
Can we apply the concept of reduced moment of inertia to calculate the change in rotational kinetic energy in a collision where kinetic energy is purely rotational namely
$$\Delta KE =1/2\frac{I_1I_2}{I_1+I_2}(1-e^2){\omega_{rel}}^2$$
(I just came up with this formula by drawing parallels to formula for loss in translational K.E which is
$$\Delta K={\frac {1}{2}}\mu v_{\rm {rel}}^{2}(e^{2}-1)$$
)
For instance a rod ($M$,$L$) is hinged at its end and lies on a horizontal table, a point mass $m$ strikes it at its end perpendicular to the rod at a velocity $v$ and sticks to it .
Could we use the above formula to find the heat evolved?
P.S this is just an example I came up with to demonstrate my point and give some context you need not answer it if not necessary but please address my first query.
 A: To see that your thesis is true , let calculate the collision case of this example .

The two pendulums  are start with initial  angular velocity and collide.
I start with writing the equation of motion shortly after the collision , you have to deals just with the constraint force.
$$I_1\,\ddot \varphi_1=L\,F_c$$
$$I_2\,\ddot \varphi_2=-L\,F_c$$
where $~I_1\,,I_2~$ are the inertia about the suspension points,
~$F_c~$ is the constraint force shortly after the collision.
with :
$$\ddot \varphi=\frac{d}{dt}\,\dot \varphi$$
you obtain
$$I_1\,\frac{d}{dt}\,\dot \varphi_1=L\,F_c$$
$$I_2\, \frac{d}{dt}\,\dot \varphi_2=-L\,F_c$$
$\Rightarrow$
$$I_1\,\int_{\dot\varphi_{1i}}^{\dot \varphi_{1f}}\,d\dot \varphi_1=L\,\int_{t_i}^ {t_f}\,F_c\,dt=L\,p\tag 1$$
$$I_2\,\int_{\dot\varphi_{2i}}^{\dot \varphi_{2f}}\,d\dot \varphi_2=-L\,\int_{t_i}^ {t_f}\,F_c\,dt=-L\,p\tag 2$$
where  $~i~$ stay for "initial state" and $~f~$ the "final  state", p is the linear momentum.
from Eq. (1) and (2) you get:
$$I_{{1}} \left( \dot\varphi _{{1f}}-\dot\varphi _{{1i}} \right) =L\,p\tag 3$$
$$I_{{2}} \left( \dot\varphi _{{2f}}-\dot\varphi _{{2i}} \right) =-L\,p\tag 4$$
you have now two equations for three unknowns, the third equation is, that during the collision the relative velocity is zero, thus
$$\dot\varphi _{{1f}}=\dot\varphi _{{2f}}\tag 5$$
solve Eq. (3),(4) and (5) you obtain the final velocity's and the linear momentum.
$$\dot{\varphi}_{1f}=\frac{I_1\,\dot{\varphi}_{1i}+
I_2\,\dot{\varphi}_{2i}}{I_1+I_2}$$
$$\dot{\varphi}_{2f}=\dot{\varphi}_{1f}$$
$$p=\frac{I_1\,I_2}{L(I_1+I_2)}\,(\dot{\varphi}_{2i}-\dot{\varphi}_{1i})$$
with those results, you can obtain the kinetic energy differenz
$$\Delta T=T_i-T_f=\frac 12\,I_1\,\dot{\varphi}_{1i}^2
+\frac 12\,I_2\,\dot{\varphi}_{2i}^2-\left(\frac 12\,I_1\,\dot{\varphi}_{1f}^2
+\frac 12\,I_2\,\dot{\varphi}_{2f}^2\right)$$
$$\boxed{\Delta T=\frac 12 \frac{I_1\,I_2}{I_1+I_2}\,\left((\dot{\varphi}_{2i}-\dot{\varphi}_{1i})\right)^2}$$
thus, the result are analog to head to head balls collision
