Finding Velocity from Acceleration

[alex_b93]

Homework Statement

Particle (Mass = m) falling through a viscous liquid due to gravity.

Experiences a drag force that is proportional to its speed.
Magnitude of drag force .
F = ku = +ve Constant x velocity.

At time t = 0, z = position = 0, initial speed = u_0.

Homework Equations

Acceleration found to be

du/dt = g - (k/m)u

The Attempt at a Solution

So to find the velocity this needs to be integrated, the way I attempted to do it was to separate the variables and substitute in
α = mg / k
Meaning
(du/dt) = g(1-u/α)
This lead to the integral

∫(α du/ α-u) = ∫g dt
α∫(1/-u) du = g ∫dt

-α ln (α-u) = gt + c

ln (α-u) = -(gt + c )/α
(α-u) = exp -(gt + c)/α

u = α - exp[-(gt+c)/α]

Would that be correct up to there, because when I carry on to find c and to substitute for α it gets messy, so I'm unsure if I've done it right.

Thanks

 

[LawrenceC]

Suggestion:

du/dt = g - (k/m)u

Above is correct equation. Solve the homogeneous equation by separation of variables. Then visualize a particular solution and add solutions together. The particular solution is 'particularly' easy to see. Evaluate constant by initial condition.

 

[LawrenceC]

For your solution

exp(ax+b)=d*exp(ax)

a,b,d are constants.

 

[alex_b93]

Suggestion:

du/dt = g - (k/m)u

Above is correct equation. Solve the homogeneous equation by separation of variables. Then visualize a particular solution and add solutions together. The particular solution is 'particularly' easy to see. Evaluate constant by initial condition.

Okay so I ended up with

u(t) = [U_0 - (gm/k)] exp(-kt/m) + (gm/k)

Is that anywhere near correct?

 

[LawrenceC]

That is what I got. You can 'test' your solution. Let t go to infinity and the velocity is mg/k which agrees with a force balance on the object. At t=0, the solution is V=V0.

Good work!

 

[alex_b93]

That is what I got. You can 'test' your solution. Let t go to infinity and the velocity is mg/k which agrees with a force balance on the object. At t=0, the solution is V=V0.

Good work!

Awesome, thanks mate.

If I were to need to find the position I would need to integrate that, right?

Just that I need in a specific form, and I can't seem to get it in that form.

 

[LawrenceC]

Yes, the integral of velocity over time provides displacement. Displacement is the area under the velocity-time curve. With an indefinite integral, you'll get another constant of integration that you evaluate from the initial condition at t=0 which the problem stated is z=0.

 

[alex_b93]

Yes, the integral of velocity over time provides displacement. Displacement is the area under the velocity-time curve. With an indefinite integral, you'll get another constant of integration that you evaluate from the initial condition at t=0 which the problem stated is z=0.

Yeah okay, so this is what I ended up with as the position

z(t) = [( m exp(kt /m)(g m-k U_0)) /k + gmt) ]/k + c

Which is quite hard to understand when wrote on here.

But is that correct, trying to work out when I'm going wrong as I can't get it in the right form.
If that is right at least I know I should be able to write it in that form.

Cheers again.

 

[LawrenceC]

You essentially have:

z(t) = integral[((Vo - mg/k)exp(-kt/m) + mg/k)*dt]

Vo-mg/k and mg/k are constants so you would get after integration

z(t) = (-m/k)[Vo - mg/k]exp(-kt/m) + mgt/k + C

with C the constant of integration that is evaluated when t=0.

Check your integration.

Also, your parentheses do not balance.

 

[LawrenceC]

When deriving formulas/functions, it is prudent to plug in the units to ensure you have not made a careless error. If the units are not correct, you have an error.

 

[alex_b93]

You essentially have:

z(t) = integral[((Vo - mg/k)exp(-kt/m) + mg/k)*dt]

Vo-mg/k and mg/k are constants so you would get after integration

z(t) = (-m/k)[Vo - mg/k]exp(-kt/m) + mgt/k + C

with C the constant of integration that is evaluated when t=0.

Check your integration.

Also, your parentheses do not balance.

Don't know how I missed the whole constant thing, feel a bit of an idiot now.

I'm getting

(V0 - mg/k)+ mg/k ∫exp(-kt/m)

[(V0 - mg/k)+ mg/k] [-m/k exp(-kt/m) + c]

Am I heading the right way now?

 

[LawrenceC]

Don't know how I missed the whole constant thing, feel a bit of an idiot now.

I'm getting

(V0 - mg/k)+ mg/k ∫exp(-kt/m)

[(V0 - mg/k)+ mg/k] [-m/k exp(-kt/m) + c]

Am I heading the right way now?

You still have an error. Write it like this and integrate:

z(t)= (V0 - mg/k)*integral(exp(-kt/m))dt + integral(mg/k)dt

 

[alex_b93]

You still have an error. Write it like this and integrate:

z(t)= (V0 - mg/k)*integral(exp(-kt/m))dt + integral(mg/k)dt

Ah right, now I'm getting answers that match up now, I didn't think to do it that way, thank you for the advice.
Problem is now though the form it is wanted in is extremely similar to what I'm getting, or rather we're getting, except the last term is

[1-exp(-kt/m)]

Rather than

[-exp(-kt/m)]

Am I missing something really simple?

 

[LawrenceC]

OK, when you integrate you pick up another constant of integration that must be evaluated. See if that helps.

 

[alex_b93]

OK, when you integrate you pick up another constant of integration that must be evaluated. See if that helps.

Okay I got it, cheers for the help mate.

I have one last question to do, and would appreciate it if you could perhaps give me a nudge in the right direction with it.

It leads on from the last question.

Position of a particle acting under a constant force F in the same z direction is given by

z = u0 t + (F/2m)t^2

With the same initial conditions.
It asks me to show that the position of a particle falling under the influence of gravity through a viscous fluid can be approximated by that of a particle acted upon by a uniform force F, for sufficiently small values of t.
It then asks for an expression for F.

All I can think of is t≈0, and then to substitute this in.

 

[alex_b93]

Okay I got it, cheers for the help mate.

I have one last question to do, and would appreciate it if you could perhaps give me a nudge in the right direction with it.

It leads on from the last question.

Position of a particle acting under a constant force F in the same z direction is given by

z = u0 t + (F/2m)t^2

With the same initial conditions.
It asks me to show that the position of a particle falling under the influence of gravity through a viscous fluid can be approximated by that of a particle acted upon by a uniform force F, for sufficiently small values of t.
It then asks for an expression for F.

All I can think of is t≈0, and then to substitute this in.

Talking to people in class it apparently uses the approximation for exp, I'm so confused.