Suppose that the feed spindle has outer radius $r_f(t)$ and the intake spindle has outer radius $r_i(t)$, similarly let the spindle's angular velocity be given by $\omega_f(t)$ and $\omega_i(t)$. Note that both $r_f$ and $r_i$ lie in the interval $[r_0, R_0]$, where $r_0= .8$ cm and $R_0 = 1.8$ cm.
The only constraint is that the tape must pass between the two spindles at a constant rate of $s = 5$ cm/s. For the tape to remain taught under this condition the tangential velocity of each spindle must be $s$. Thus,
$$s = r_f(t)\cdot\omega_f(t) = r_i(t)\cdot\omega_i(t).$$
So
$$\omega(t) = \frac{s}{r(t)}$$
for either spool. So now we simply need to find a description of the outer radius of the spindle. Consider that the tape has thickness $a \ll r_0$ and width $w$, thus the volume of tape transfered per unit time is
$$ \frac{dV}{dt} = \pm a s w$$
and since $V = \pi w r(t)^2$ then we also have
$$ \frac{dV}{dt} = 2\pi r(t) \dot{r}(t) w$$
and then by equating these expressions and solving the IVP for the feeding spindle yields,
$$r_f(t) = \sqrt{R_0^2 - \frac{as}{\pi} t}.$$
Similarly, the IVP for the intake spindle yields
$$r_i(t) = \sqrt{r_0^2 + \frac{as}{\pi} t}.$$
From those expressions it is easy to see that the proper speeds to drive the spindles are approximately
$$\omega_f(t) = \left(\sqrt{\left(\frac{R_0}{s}\right)^2 - \frac{a}{s\pi} t}\right)^{-1}$$
and
$$\omega_i(t) = \left(\sqrt{\left(\frac{r_0}{s}\right)^2 + \frac{a}{s\pi} t}\right)^{-1}.$$