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Jan
22
revised How to apply a force vector to a position vector?
typeset `\sin` and `\cos`
Jan
21
answered Center of Instantaneous velocity in two degree of freedom problem
Jan
21
revised What orientation will let me drive the fastest with a tray of coffee on the floor?
edited tags
Jan
20
comment Momentum of slowly spinning (viscous) fluid
I need to chew on this a bit.
Jan
20
awarded  Benefactor
Jan
20
comment Momentum of slowly spinning (viscous) fluid
Any series of the form $\Omega r + 2 \Omega R \sum \left( \exp(-\beta n^2 t) K \sin(n \kappa r) \right)$ solves the equation when $\beta = \frac{\mu}{\rho} \kappa^2$. Why did you choose $\kappa = \frac{\pi}{R}$ for the shape function. Intuitively it makes sense to fit the BC, but I wonder if there is any other reason. What about $\kappa = \frac{\pi}{2 R}$ such that $\sin()=1$ when $r=R$?
Jan
20
comment Momentum of slowly spinning (viscous) fluid
I did not arbitrarily drop terms. I plugged $\tau$ and $\tau'$ into the equation of motion $A \tau' {\rm d}r + \tau {\rm d}A = \dot{v} A {\rm d}r$ to get $$ \left. r \tau' + \tau = \rho \dot{v} r \right\} \mu r v'' = r \rho \dot{v} $$
Jan
20
revised Momentum of slowly spinning (viscous) fluid
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Jan
20
revised Momentum of slowly spinning (viscous) fluid
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Jan
20
comment Momentum of slowly spinning (viscous) fluid
I don't like seeing time $t$ in the denominator, as in $\frac{1}{\sqrt{\pi \nu t}}$. I feel this approach gives the natural response (disturb the fluid and let it rest) instead of the forced response of moving the outside boundary.
Jan
20
comment Momentum of slowly spinning (viscous) fluid
If look at with separation of variables I find a solution of the $v(r,t) = \exp(-\nu \kappa^2 t) \left( V_1 \sin(\kappa r) + V_2 \cos(\kappa r)\right)$ form.
Jan
20
comment Momentum of slowly spinning (viscous) fluid
I agree, it is very tricky to marry the boundary conditions to the solution.
Jan
20
comment Momentum of slowly spinning (viscous) fluid
With a cursory look it does not seem this meets the natural boundary conditions of $v_\theta(r=0,t)=0$ and $v_\theta(r=R,t)=\Omega R$. Also the initial conditions are $v_\theta(r<R,0)=0$ I think.
Jan
20
revised Momentum of slowly spinning (viscous) fluid
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Jan
19
revised energy of a ball
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Jan
18
revised Momentum of slowly spinning (viscous) fluid
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Jan
17
comment Four momentum in particle physics
With the second definition the units are correct.
Jan
17
revised Inertial forces and centre of mass
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Jan
17
revised Inertial forces and centre of mass
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Jan
15
comment How to calculate the moment of inertia of a 2 point mass system
Are you looking for the MMOI about the rod center, or the center of mass? The former is going to be larger than the latter.