# Problems classical mechanic

1. Oct 18, 2012

### Dragster

http://img27.imageshack.us/img27/1158/physich.png [Broken]

I really do not see how I can solve the problems. All I need is hint how to start the problems most of the time. I know I should read more, ask questions to teacher, etc. I have 5 classes, 5 homeworks to do every week. On top of that I have to revise old courses, remember the old stuff I used to know and study for the courses of the following week. It is really hard with my schedule to see the demonstrators or the teachers.
Add to that I am not an english native speaker and my teacher gave us an english book despite the course being in french and our first physic course.

I am not asking you to do the problems, just give me some hint where I should start and what I should use.

For 2.2 (2.3) is flin = bv (2.5) value is equal to 1.6E-4 N*s/m^2

β=3πη which gives me the equation flin= 3πηDv = βDv = bv

2.4 What is cross-sectional area A? fquad= cv^2 c=γD^2 where γ= 0.25N*s^2/m^4
I guess that ∂Av^2 is actually cv^2 where c=∂A in this medium but I don't know how to prove the projectile encounters the fluid at the rate of ∂Av and all the more so ∂Av^2

2.11 m dy/dt = -mg -bv
a)
vy= -gt -bvt/m +v0 (1st integration gives the speed)

y= -gt^2 -bvt^2/m +v0t (2nd integration) gives the position

b) When v0= 0 we have ymax

y= -gt^2 -bvt^2/m +v0t
ymax= -gtmax^2 -bvtmax^2/m

tmax^2= ymax/(-(gm+bv)/2m)

c) no idea...

2.12 Again no idea how to begin this

2.13 With 2.12 solved I should be able to do it.

+karma for any help

Last edited by a moderator: May 6, 2017
2. Oct 19, 2012

### Simon Bridge

There are 5 problems on that sheet - lets focus on one at a time so you can use what you learn doing one problem to help solve the others.

You need to show that $$\frac{dm}{dt}=\rho v A$$ ...I would have started with $dm=\rho dV$...where V is a volume element encountered by the projectile. So $$\frac{dm}{dt}=\rho\frac{dV}{dt}=\cdots$$... so what is dV?

3. Oct 20, 2012

### Dragster

http://img12.imageshack.us/img12/1158/physich.png [Broken]

My question is probably stupid but what is dm?

V is π*r^2*v since v=dx/dt the volume increases over time. So dV/dt = π*r^2*dx/dt

The object encounter the medium at the rate of ∂Av because Av is the volume of fluid and ∂ is the density.

Also, if it doesn't pose problem, I like to work on all problems at the time. I would greatly appreciate to have a hint for every problems beside 2.2 :)

In words I can explain why the projectile encounters the fluid at the rate of ∂Av but why does the drag force is v^2? What is the equation proving this? I know fquad is proportional to v.

Thank you for you help!

Last edited by a moderator: May 6, 2017
4. Oct 20, 2012

### Simon Bridge

It is the amount of mass in volume dV at position r.
Ah ... density would be that funny g-shape symbol your text book uses and I have not seen very often or a greek letter rho "ρ". That "∂" symbol is for a partial - used in partial differential equations. I wondered why you seemed to be constructing one.

It would be innefficient.

The exercise leads you through the calculation in a particular order. The point of doing that is to answer exactly the questions you have asked. Getting ahead of the structure is, as you have discovered, counter-productive.

We could just throw you in the deep-end and just ask for the proof cold. But then you'd be lost right at the start with no idea where to begin. The exercise is giving you the starting point.

Your reasoning is fine, you need to be able to express it in math: you want to find an expression for dm/dt in terms of initial assumptions and show that it comes to ρAv

dm = ρ(x,y,z)dV = ? (in terms of coordinate elements)

dm/dt = ρdV/dt = ?

hint: if the motion is in the z direction, then v=dz/dt.

All this is crucial to your understanding of the other problems.

5. Oct 21, 2012

### Dragster

Okay right the mass of the object being the liquid accelerated increases as the volume increases but what does the integration of dm gives?

If I start from the beginning with 2nd law,
F = ma = cv^2
m = cv^2/a
dm/dt = cv^2/a
dm = (cv^2/a)*dt

I know that the mass of the liquid is equal to p*V so the variation of mass shall be p*dV as you said.

so dm = pdV

pdV = cv^2/a * dt
dV = cv^2/pa *dt

How can I use this to end with pAv? I tried many things like replacing dV by A*dx, x being the position of the projectile.

Thank you for your time! :)

6. Oct 21, 2012

### Simon Bridge

depends on how you do the integration - you are getting ahead of yourself again. Relax. It is OK to start a calculation without having a clear idea where it will end up. In this case you know where you need to go, it just isn't immediately clear exactly how you'll get there. You have to set out on your journey and hope matters will become clearer as you go. Remember what you have to find?

You are looking for dm/dt encountered by the projectile - this will involve adding up all the mass elements directly in front of it (hence the choice of calculus notation) and including some time-dependence (that's the unclear part)

... in your case, the mass is not being accelerated until after your object encounters it. You want to know the rate your object encounters the mass. The acceleration of the object is zero. F=ma wouldn't work anyway because it assumes that the mass is a constant ... here it depends on density, which may vary with position. The force law is, properly, F=dp/dt: p=mv. (note the difference between p and ρ?)
In rectangular components, dV=dxdydz.
In cylindrical coordinates, dV=rdrdθdz.
(In both cases taking motion along the z axis.)

So - substituting the one you like best into dm/dt=ρdV you should get something suggestive of a way to the solution you expect.

This is the style of thinking that seems to be expected throughout - you need to get used to it.

Your problem is that it is pretty much intuitively reasonable that dm/dt = ρAv and if you were just asked what rate the object encounters mass that is what you'd have written down right away and not given it a second thought. The exercise forces you to give it that thought and a chance to go, "Oh hey, it's not that obvious after all!"