Rotational Kinetic Energy Conservation

We get the energy equation $$tau cdot theta = frac {I_1 omega_1^2}{2} + frac {I_2 omega_2^2}{2}$$.

As the two gears are attatched to each other, their linear speed at the edges is same. So, we get the equation $omega_1 r_1 = omega_2 r_2$.

Let the moment of inertia of a gear be $I = k m r^2$. Assuming that the gears are made of the same material, their 2-D density - $sigma$ (mass per unit area) is constant. So, $I = k (sigma pi r^2) r^2$, i.e. $I = k' r^4$

Substituting $r_1 = frac {omega_2 r_2}{omega_1}$ and $I_n = k' r_n^4$ in the energy equation, we get $$ tau cdot theta = frac {k' omega_2^2 r_2^2}{2} (r_1^2 + r_2^2)$$ As $tau cdot theta $ is constant in both cases, $$ omega_2 = frac {k''}{r_2 sqrt{r_1^2 + r_2^2}}$$ So, when $r_2$ is increased to $r_3$, it is apparent from the equation that $omega_2$ will reduce to $omega_3$.

The bigger gear will only spin faster if its density is sufficiently lower than the smaller gear to reduce its moment of inertia.

Assuming no other stuff the larger system spins slower due to its larger moment of inertia.

So like the total kinetic energy in the first system assuming a massless chain is given by the angular velocities $omega_{1,2}$ as $$ K=frac12 I_1omega_1^2 +frac12 I_2omega_2^2 $$where the moments of inertia are $I_{1,2}.$ The chain between them forces $r_1omega_1=r_2omega_2$ when it is taut, so this is $$K =frac12left(I_1(r_2/r_1)^2+I_2right)omega_2^2$$ and if the gears are of similar construction (same thickness material but pattern otherwise zoomed-in from one size to the other) you'll have $I_1 =(r_1/r_2)^4 I_2$ from dimensional analysis alone, so you will have a rate $$omega_1=frac{r_2}{r_1} ~frac{taucdotpi/4}{I_2(1+(r_1/r_2)^2)}.$$

So if $s=r_{1,3}/r_2$ you have a term that goes like $1/(s +s^3),$ it decreases as $s$ increases. Bigger systems spin slower with the same energy.