Viscosity + Rotation combined

When a solid is moved in a liquid is suffers a viscous force which is given as $F=\eta A \frac {\Delta v}{\Delta z}$ Here, $\eta$ is the coefficient of viscosity, A is the surface area and $\frac {\Delta v}{\Delta z}$ is the velocity gradient. The disc in the problem will also suffer a viscous force that will produce a torque to oppose the rotation of the disc.

The disc can be assumed to be made up of many thin circular rings. Imagine a ring of radius r(<R) and thickness dr. The points on the ring have speed $r\omega$ . The torque on this ring will be.

$d\tau = \left( {\eta 2\pi rdr\frac{r\omega}{h}} \right)r$ The total torque on the disc due to viscous force can be calculated by integrating the torque $\int_0^\tau {d\tau } = \int_0^R {\left( {\eta 2\pi rdr\frac{{r\omega }}{h}} \right)r}$ $\tau = \frac{{\eta 2\pi \omega {R^4}}}{{4h}}$

The torque on the disc is also applied by the force F. $\tau = F\frac{R}{2}$ The total torque equals $\tau = \frac{{\eta 2\pi \omega {R^4}}}{{4h}}+F\frac{R}{2}$

This torque creates an angular retardation ( $\alpha = - \frac{d\omega}{dt}$ ), and the torqe $\tau= I \alpha$ Moment of inertia (I) of disc = $\frac{MR^2}{2}$ Thus, we have $\frac{{\eta 2\pi \omega {R^4}}}{{4h}}+F\frac{R}{2} = -\frac{MR^2}{2}\frac{d\omega}{dt}$ $\begin{gathered} \int_0^t {\frac{{dt}}{{2hMR}}} = - \int_\omega ^0 {\frac{{d\omega }}{{\eta 2\pi \omega {R^4} + 2FR}}} \\ \frac{t}{{2hMR}} = \frac{{\ln \left( {\frac{{\eta 2\pi \omega {R^4} + 2FR}}{{2FR}}} \right)}}{{\eta 2\pi {R^4}}} \\ t = \frac{{MRh}}{{\eta \pi {R^4}}}\ln \left( {1 + \frac{{\eta \pi \omega {R^4}}}{{FR}}} \right) \\ \end{gathered}$ Comparing it with the form in the quesiton we will have a=1, b=1 and c=1 and therefore, a+b+c=3.

Let's write velocity of a small ring at a distance $r$ from axis of rotation, $v(r) = \omega r$

So, the velocity gradient at a distance $r$ from axis of rotation becomes, $\dfrac{dv}{dz} = \dfrac{v(r)}h = \dfrac{\omega r}h$ Force due to viscosity, $F(r) = -\eta S \dfrac{dv}{dz} = -\eta\cdot 2\pi r dr \dfrac{\omega r}h$

So, the small torque acting due to viscosity, $d\tau_1 = rF(r) = -\dfrac{2\pi\eta\omega}h r^3 dr\\\tau_1 =-\dfrac{2\pi\eta\omega}h \int_0^Rr^3 dr = -\dfrac{\pi\eta\omega R^4}{2h}$

Torque due to the external force is, $\tau_2 = -\dfrac{FR}2$

So, the net torque becomes, $\tau = \tau_1+\tau_2 = -\dfrac R2 \left(\dfrac{\pi\eta\omega R^3}{h} + F\right)$

Now, $\dfrac 12 MR^2 \alpha = \tau\\\\\Rightarrow\alpha =- \dfrac{1}{MR}\left(\dfrac{\pi\eta\omega R^3}{h} + F\right)$

Integrating, $\begin{aligned}t &= MR\int_0^\omega \dfrac1{\left(\dfrac{\pi\eta\omega R^3}{h} + F\right)}d\omega\\&= \dfrac{Mh}{\pi \eta R^2}\ln \left( 1 + \dfrac{\pi\eta \omega R^3}{Fh} \right)\end{aligned}$

Thus, $\color{#3D99F6}{a+b+c =\boxed 3}$