9. When solving dynamical problems involving Newton\'s laws, we of course usuall

Question

9. When solving dynamical problems involving Newton's laws, we of course usually end up with first order ordinary differential equations (ODEs) with respect to time governing velocities, and/or second- order differential equations for positions. We normally presume nice existence and uniqueness properties for these differential equations, supposing a unique solution assuming a sufficient number of constants of integration (one for each order of time derivative), often taken to be initial conditions But existence and uniqueness are only guaranteed for suficiently well-behaved ODEs. Here we explore a simple differential equation where uniqueness can fail. Consider a vertical, cylindrical vat of cross sectional area A, near sea level. At time t, the vat contains water up to a height y(t), but the water can leak out slowly through a small hole of cross sectional area at the very bottom of the vat. Assume a 4, and take the water to be incompressible and inviscid, yet with laminar flow. (These assumptions are not entirely plausible physically, but we are just trying to generate a simple example). (a) Using Bernoulli's Principle (hopefully still vaguely familiar from Physics 5/7), write down a differen- tial equation for the height y(t) of the fluid remaining in the vat as a function of elapsed time t. HINT: you should obtain a simple-looking, first-order ODE. (b) If y(0)-yo at timet0, what is the time T at which the vat just empties? (c) Find a different solution y(t) valid for all t 2 0 which agrees with your previous solution for all t 2 T, but disagrees for all 0 St

Explanation / Answer

9. consider a vertical cylinder

cross section area = A

at sea level

contains water upto height y(t) at t = t

area of hole = a

a. now

speed of flowing water from the hole = v

then

-Ady/dt = av ( from continuity equaiton)

and from bernoullis equation

rho*gy + 0.5*rho*(dy/dt)^2 = 0.5*rho*v^2 = 0.5*rho*A^2(dy/dt)^2/a^2

2gy + (dy/dt)^2 = (A/a)^2*(dy/dt)^2

dy/dt = -sqrt(2gy/((A/a)^2 - 1))

b. y(0) = yo > 0

then

dy/dqrt(y) = -sqrt(2g/((A/a)^2 - 1))dt

iontegrating

2[sqrt(y) - sqrt(yo)] = -sqrt(2g/((A/a)^2 - 1))t

y = 0

t = T

T = 2sqrt(yo((A/a)^2 - 1))/sqrt(2g)

c. we can have any solution of y

like

y = {ln(t), 0 < t < T}

{0, 0 < T < = T}

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