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Anonymous Tue 07 Jan 22:40:09 2020 No.11285430 View Reply Original Report

Quoted By: >>11285433 >>11285441

A cylindrical tube full of water is stirred until the water starts rotating 1 rad/s. Then the stirring is stopped. How long will it take for the rotation speed to slow down by a factor of 100? The cylinder's radius is 4 cm and it's height is much longer than it's radius. The viscosity of water is 1,0 mPa * s and the density of water is 1 kg/dm^3.

- Taken from the Estonian physics competition "Kajakas" meant for middle and high schoolers.

Anonymous

Anonymous Tue 07 Jan 2020 22:43:56 No.11285433 Report

Quoted By:

>>11285430
it's trivial

Anonymous

Anonymous Tue 07 Jan 2020 22:47:32 No.11285441 Report

Quoted By:

>>11285430
what's the competition? who can implement a formula the fastest?

Anonymous

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Anonymous Tue 07 Jan 2020 23:16:10 No.11285494 Report

Quoted By: >>11285504

We have given $\tau,\ R,\ \omega_i,\ \omega_f=\omega_i/100,\ \rho$ . The height is not given but will be represented by $H$
The angular momentum of the fluid at any time is given by $L=I\omega=\frac{1}{2}mR^2=\frac{1}{2}\pi\rho HR^4$ . Therefore, the magnitude of angular impulse on the water is $\Delta L=|L_f-L_i|=\frac{99}{200}\pi\rho HR^4\ \ \ (*)$
By the impulse momentum theorem, we have
$\Delta L=\int_0^tT\text{ d}t'=\int_0^t\tau(2\pi RH)R\text{ d}t=2\pi\tau R^2Ht$
Combining with the star, rearranging and plugging in we get
$t=\frac{99\rho R^2\omega_i}{400\tau}=39.6 \text{ seconds}$

Probably made a mistake, but that's the idea~

Anonymous

Anonymous Tue 07 Jan 2020 23:19:52 No.11285504 Report

Quoted By:

>>11285494
Oops, there's suppose to be an omega in that second couple qualities of that first expression for angular momentum (obviously). Also forgot to replace the differential in the second integral with a dummy variable. Kiss my ass.

Anonymous

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Anonymous Tue 07 Jan 2020 23:46:51 No.11285568 Report

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We are given $\mu,\ R,\ \omega_i,\ \omega_f=\omega_i/100,\ \rho$ . The height is not given, but will be represented by $<br />H$ . First of all, we need to determine the torque on the cylinder to get the equation of motion. Shear stress is given by
$\tau=\mu\frac{\Delta v}{\Delta r}=\mu\frac{R\omega}{R}=\mu\omega$
Torque is (stress*area*moment arm). So torque is given by
$T=\tau A R=2\pi R^2H\mu\omega$
But, the torque on a body is the rate of change of angular momentum. And angular momentum is given by
$L=I\omega=\frac{1}{2}mR^2\omega=\frac{1}{2}\rho\pi HR^4\omega\ \implies\ T=\dot{L}=\frac{1}{2}\rho\pi HR^4\dot{\omega}$
So,
$\dot{\omega}=-\frac{4\mu}{\rho R^2}\omega$
where the negative sign appear because torque opposes motion. Solve the ODE and set proper boundary conditions to get
$t=\frac{\rho R^2}{4\mu}\ln 100$
where ln is the natural logarithm. Now you can plug in values. I probably made a mistake. Kiss my ass.

Anonymous

Anonymous Wed 08 Jan 2020 02:23:17 No.11285878 Report

Quoted By:

gimme (You)s. I spent too much time TeXing

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