# Is it possible to find the exact value of the dynamic viscosity of a fluid by applying Stokes' law?

I am a high school student who is investigating the dynamic viscosity of a fluid. Typically, high school physics model the dropping of a spherical object vertically into a viscous fluid with the assumption it is at terminal velocity. Purposefully, I have omitted this reasonable assumption to gain a more in-depth understanding of the physics involved. Assuming a small Reynolds number, consider the free body diagram (FBD) below:

Forces acting upon the spherical object in the fluid.

In which, $$m$$ is the mass of the object, $$β$$ is the buoyancy force upon the object, $$v$$ is the velocity of the object with respect to the fluid and $$α=6πηR$$ is a component of Stokes' drag ($$η$$ - dynamic viscosity, $$R$$ radius of the spherical object).

Applying Newton's second law with the initial velocity of the object being zero, $$v_0=0$$, at $$t=0$$ and $$mg > β$$ yielded the following results:

The acceleration of the object, $$a$$, as a function of time, $$t$$:

$$a(t) = \left(g-\frac{β}{m}\right)e^\frac{-αt}{m}$$

The velocity of the object, $$v$$, as a function of time, $$t$$:

$$v(t) = \frac{mg-β}{α}\left(1-e^\frac{-αt}{m}\right)$$

The downwards vertical displacement of the object, $$s$$, as a function of time, $$t$$:

$$s(t) = \frac{mg-β}{α}\left[\frac{m}{α}\left(e^\frac{-αt}{m}-1\right)+t\right]$$

An exact solution for the dynamic viscosity of the fluid can be obtained assuming that the object is at terminal velocity, $$v_T$$.

$$v_T = \lim_{t\to ∞} v(t) = \lim_{t\to ∞}\frac{mg-β}{α}\left(1-e^\frac{-αt}{m}\right) = \frac{mg-β}{α}$$

However, I seek to obtain an exact solution for the dynamic viscosity of the fluid for which the object is not at terminal velocity, $$v_T$$. Moreover, I require a solution for the dynamic viscosity in which the displacement, $$s$$, of the object is known and the time, $$t$$, it has taken to achieve this displacement is also known. Specifically, I want to find the non-zero root of the following function, $$f(α)$$:

$$f(α) = \frac{1}{m(mg-β)}\left[α^2s-(mg-β)t\right]-e^\frac{-αt}{m}+1=0$$

An approximation for the value of $$α$$ which satisfies $$f(α)=0$$ can be determined using the Newton-Raphson method: $$α_{n+1} = α_n - \frac{f(α_n)}{f'(α_n)}$$.

But... Is it possible to find an exact solution of $$f(α)=0$$?

• For the equation you provided, no. For a lot of problems in Engineering, no. That is why we use numerical methods. To quote Chapra et al, in the book Numerical Methods for Engineers, "Numerical methods are extremely powerful problem-solving tools. They are capable of handling large systems of equations, nonlinearities, and complicated geometries that are not uncommon in engineering practice and that are often impossible to solve analytically" – Thales Aug 15 at 19:44
• Why make your life complicated by avoiding the very convenient terminal velocity? – Gert Aug 15 at 20:30