# Applying conservation of mass to a falling plywood

1. Jan 31, 2012

### polka129

A sheet of plywood is held with one edge resting on the floor, as shown in the figure below. The
plywood surface initially makes a small angle, Θ, to the floor. At time t=0 the held edge is released and
the plywood pivots about the edge on the floor as it falls down. It is observed that the plywood falls
rather softly to the floor, much more so than had the plywood been falling in a vacuum.
1. Explain, using conservation of mass and the momentum equation, why the plywood falls softly.
2. Do you think that the plywood would fall softly if the air were incompressible?
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this is what i have been abl to come up with..

suppose the ply descends with a velocity of u m/s and it displaces the air beneath with a velocity of W m/s.u is downward and W is rightwards.

the downward momentum of ply=the increase of momentum of air towards the right
the ply makes an arc with an angle of phi.so i can find the rate of change of area(of this arc)

area of arc=(phi/360)* pi*r**2
differentiate this area with respect to time..i get dA/dt
then equation of continutiy dm/dt=density * volume flow rate

volume flow rate=dA/dt* velocity
here velocity of air moving rightwards w
so i get dm/dt.

now what more?..how to prove that the plywood falls softly..ie. with a low velocity

2. Jan 31, 2012

### Spinnor

I this case I don't think that the compressibility of air comes into play. If one had an incompresible fluid with the same density as air (don't think there is such a substance) the results would be nearly the same?

3. Feb 1, 2012

### polka129

thank you spinnor for your reply.i believe that one can be very well answered once we have the equations before us.i am struggling to come up with the final equations which describe this phenomenon.

one thing i know is that the downward momentum of the faling plywood equals the momnetum gained by the air when it is being swept by this plywood..

momentum ,p=mass * velocity

taking the derivative...

4. Feb 1, 2012

### Studiot

I suggest considering the setup as a rotating panel 'pump'.

The ply rotates about its bottom like a door against a spring = the compressibility of the air.

You can use thermodynamics to obtain the compressibility if you assume air is a perfect gas

$$\begin{array}{l} V = \frac{{RT}}{P} \\ {\left( {\frac{{\partial V}}{{\partial P}}} \right)_T} = - \frac{{RT}}{{{P^2}}} \\ \kappa = - \frac{1}{V}{\left( {\frac{{\partial V}}{{\partial P}}} \right)_T} = \frac{{RT}}{{{P^2}V}} = \frac{1}{P} \\ \end{array}$$

This gives you an expression for the spring constant and thus the resisting force slowing the ply down.

As volume decreases the pressure increases, and you can use Bernoulli to convert this to outflow of air.

Note if air was incompressible there would still be an increase in pressure under the ply due to the weight of the board. Again Bernoulli would give the outflow, but it would be smaller, so the board would take longer to subside.

Does this help?

5. Feb 2, 2012

### polka129

thanks a lot for this insight...that did not occur to me..but you see the questions requires me to explain using the eqautions of mass conservaton and mometum..there is where i am strugling

6. Feb 2, 2012

### Studiot

Well I suppose you could look at the momentum analysis for a turbine blade and consider the ply as a single blade turbine in reverse.

This seems to me to be the long way round, but it should be in your textbook.

7. Feb 2, 2012

### Spinnor

You need to approximate the velocity and density of air, v(x,y.z,t) and ρ(x,y.z,t), the actual flow is very complicated and without its knowledge your answer will be a guess. If you can argue that only when the door is near impact will the velocity of the air matter then you can assume a simple flow.

Let the approximate volume of air effected by the moving plywood be,

V = LR^2sinθ/2

Work in cylindrical coordinates. We do not have a purely radial flow but as an approximation it should be close for a start. We might also want the volume up to some radius r, V(r,θ) = LRrsinθ/2

where L is the length of the plywood, R is its width, and r is less then or equal to R.

dV(r,θ) = -LRrcosθdθ/2

Assume the flow is radial. When the wood moves the volume changes as given above as a function of r. Assume that volume dV flows through the area A(r) = L*r*θ, we neglect the end flow.

Let the flow velocity at r, θ, and dθ/dt be small v, v(r,θ,dθ/dt)

We can relate the change in volume to the velocity,

dV = v(r,θ,dθ/dt) * dt * L*r*θ = -LRrcosθdθ/2

v(r,θ,dθ/dt) = -[Rcosθ/θ]dθ/dt

r cancels, does that seem right? Velocity with our assumption of no end flow is independent of r.

I think you could argue that dv/dt will only become important for small θ. If so the above can be simplified.

cosθ/θ --> [1-θ^2/2]/θ --> 1/θ for small θ

v(r,θ,dθ/dt) --> -[R/θ]dθ/dt

F = dP/dt = d(mv)/dt = d(ρVv)/dt = ρVdv/dt assume ρ constant.

Edit, I pulled V past the time derivative, not so sure about that now?

One could plug in θ(t) for the case of no air above and get a better value for θ(t) by letting the above force slow down the plywood.

How can such a seemingly simple problem be so complicated, or am I missing something here. Any gross errors here?

Thanks for any corrections.

Last edited: Feb 2, 2012
8. Feb 2, 2012

### Spinnor

I wrote,

v(r,θ,dθ/dt) = -[Rcosθ/θ]dθ/dt

If this is close then maybe energy conservation can be used to get an approximate answer.

MgRcos(θ)/2 = kinetic energy of wood + kinetic energy of air

where θ starts at 90 degrees.

?

Still not very simple.

9. Feb 5, 2012

### LawrenceC

I'll put my \$0.02 for a simple minded approach.

Based on conservation of mass principle you can say that as the plywood falls and nears the ground, small theta, the volume of fluid between the plywood and ground must escape. Therefore a flow must be established.

Write an expression for the volume of fluid trapped beneath the plywood as it is nears the ground (function of theta). Take the derivative with respect to the angle. This represents the change in volume as a function of angle.

Write another expression for the area through which the fluid must escape. It escapes through the radial edge of the board as well as through the sides (3 exits). The dimension of the escape areas are functions of the angle.

Divide the volume derivative by the area to establish an average exit velocity at the angle as it nears the ground. Look at your expression and see what happens to it as the angle gets small.

Based on your result and the momentum equation, you can then conclude something about mean pressures on the face of the board as the board nears the ground.