# Estimating the Characteristic Time of Spinning Tea

Suppose I have a cup of tea and stir it so it gains some angular velocity $$\omega_0$$. Could you estimate the characteristic time $$\tau$$ it takes for the tea to stop rotating (or to lose half its mechanical energy)?

When trying to estimate, I assumed that the tea was in a very long cylindrical glass and that most of its energy was lost to its viscosity $$\eta$$. This got me to find the angular velocity as a decreasing exponential function in the form of $$\omega (t) = \omega_0 \exp (-bt)$$ and got around $$60$$ seconds.

However, when doing an experiment, I found the time to be much bigger at around $$130$$ seconds. Perhaps my experimentation was faulty, but I am wondering if there are some other factors at play here. How would you estimate the characteristic time?

Edit (My calculations for the estimate)

Let us model the tea like a solid cylinder. We can take the axis to be through the center of the cylinder. Let $$r$$ be the distance from the central axis and let's have $$\omega_0$$ be the angular velocity near the center. For the sake of simplicity, the angular velocity profile, $$\omega (r)$$ will be linear, or $$\omega R = \omega_0 (R - r) \implies \omega (r) = \frac{\omega_0 (R - r)}{R}.$$ A viscous force will inhibit the motion of the water from the walls of the glass. The viscous force is given from Newton's law as $$F = -\eta A \frac{\text{d}v}{\text{d}x}.$$ So the torque acting on the walls is $$\tau = \vec r \times \vec F = \eta (2\pi R^2 H) \left(\frac{\text{d} \omega}{\text{d}r}\right)_{r = R} = \eta 2\pi R^2 H \omega.$$ There will also be a torque by the change in angular momentum, or $$\tau = \Delta \vec L/\Delta t$$. We can integrate to find the infintesimal moment of inertia. Note that $$\text{d}I = \rho \cdot 2\pi R\text{d}r H \cdot r^2.$$ We also know the angular velocity profile $$\omega (r)$$, so $$\text{d}L = \text{d}I \omega (r).$$ This translates to $$\int_{0}^{L}\text{d}L = \int_{0}^{R} \frac{2\pi \rho H \omega_0}{R} (Rr^3 - r^4)\text{d}r ,$$ $$L = \frac{\rho 2\pi H \omega_0}{R} \left(\frac{R^5}{4} - \frac{R^5}{5}\right) = \frac{\rho \pi H \omega_0 R^4}{10}.$$ The change in $$L$$ can now be equated to viscous torque, or $$\frac{\text{d}L}{\text{d}t} = \tau \implies \frac{\rho \pi H R^4}{10}\frac{\text{d}\omega}{\text{d}t} = -2\pi \eta \omega R^2 H.$$Separating and integrating once again using the fact that $$\int 1/x = \ln x + c$$ tells us that $$\omega_f$$ is a decreasing exponential function, or $$\omega_f = \omega_i \exp \left(-\frac{20\eta}{\rho R^2}t\right).$$For the characteristic time, we require $$\omega_f = \omega_0/2$$. Then we would see that $$t = \frac{\rho R^2 \ln 2}{20\eta} \approx 60\;\mathrm{s}$$ assuming $$R = 4\;\mathrm{cm}$$.

• Viscosity is a function of temperature. Was the tea hot? Did the spin down time constant decrease as it cooled down? Other than that the ratio between observation and theory is suspiciously close to a factor of two... Commented Nov 6, 2022 at 1:32
• Oh you are completely right. For the experiment as well as the calculations, I used room temperature water as a substitute. Though even if the tea was hot, I think since the time scale was so short, we could say the temperature and viscosity remains constant. Also I have added a derivation with my calculations if you want to check if there is a factor of 2 missing. Commented Nov 6, 2022 at 1:43
• I will leave the checking to a theoretician, but I might do the experiment with tea, later. I wonder if English Breakfast behaves differently from Earl Grey? Commented Nov 6, 2022 at 1:51
• I have a bit of a doubt about the result because if we take the same geometry with ten times the radius, then the spin down time constant suddenly becomes over an hour. That does not gel with my intuition about how water in a large drum behaves. The dimensional analysis is correct, though. I guess I need to get myself a really large tub of water... Commented Nov 6, 2022 at 2:17
• Yes that is a fair point. Though one thing to note is that I took the assumption of $H\gg R$ so there would be a large amount of water with a lot of initial kinetic energy. Perhaps there is some other force/phenomenon taking play in larger fluids though. Commented Nov 6, 2022 at 2:25

Assuming that the fluid flow is laminar, we can get an exact solution.

The Navier-Stokes Equation in Cylindrical Coordinates for given problem:

$$\frac{\partial ω }{\partial t}=\nu \left ( \frac{\partial^2 ω}{\partial r^2}+\frac{3}{r}\frac{\partial ω}{\partial r} \right )$$

$$0\leqslant r \leqslant R$$

Here $$ω=ω\left ( r,t \right )$$

and $$\nu$$ denotes kinematic viscosity. OP uses dynamic viscosity $$\eta$$.

These are related:

$$\nu=\frac{\eta}{\rho}$$

Boundary condition:

$$ω\left ( R,t \right )=0$$

Initial condition:

$$ω\left ( r,0 \right )=ω_{0}$$

The rest is pure mathematics. I skip math to concentrate on physics only.

The solution:

$$ω\left ( r,t \right )=-\frac{2ω_{0}R}{r}\sum_{k=1}^{\infty}\frac{J_{1}\left ( \frac{r}{R}\mu_{k} \right )}{\mu_{k}J_{0}\left ( \mu_{k}\right )}e^{-\frac{\nu \mu_{k}^{2}}{R^{2}}t}$$

where $$J_{0}\left ( x \right )$$ and $$J_{1}\left ( x \right )$$ are the Bessel functions of the first kind and $$\mu_{k}$$ are zeros of $$J_{1}\left ( x \right )$$.

The first few zeros:

$$\mu_{1}=3.832$$ $$\mu_{2}=7.016$$ $$\mu_{2}=10.173$$

We see that after a short time the first term of the series begins to dominate. So we can write

$$ω\left ( r,t \right )\approx -\frac{2ω_{0}R}{r}\frac{J_{1}\left ( \frac{r}{R}\mu_{1} \right )}{\mu_{1}J_{0}\left ( \mu_{1}\right )}e^{-\frac{t}{\tau_{1}}}$$

where

$$\tau_{1}=\frac{R^{2}}{\nu \mu_{1}^{2}}$$

is the characteristic time constant of the problem.

OP said that he used water at room temperature.

Kinematic viscosity of water at $$20$$ Celsius is about $$0.01\frac{cm^{2}}{s}$$.

Assuming $$R = 4\;\mathrm{cm}$$ we get:

$$\tau_{1}=\frac{4^{2}}{0.01\cdot 3.83^{2}}=110\;\mathrm{s}$$

I think, this result is in satisfactory agreement with the experimental result.