Particle thrown straight up, need to find velocity

[Mugged]

Ok so I have this particle of mass m that is thrown upward with initial velocity $$v_{0}$$. Now there is a downward force due to gravity as well as a square-law drag force described by b*v^2 where b is some constant. I need to find the velocity as a function of time.

Any help is appreciated!

 

[gabbagabbahey]

Ok so I have this particle of mass m that is thrown upward with initial velocity $$v_{0}$$. Now there is a downward force due to gravity as well as a square-law drag force described by b*v^2 where b is some constant. I need to find the velocity as a function of time.

Any help is appreciated!

Open your textbook and read the section on Newton's 2nd law

 

[Mugged]

heres what i have so far:

F = ma = mg + bv^2

a = g + (b*v^2)/(m)

then i have the kinematic equation:
V = V0 +at

so i plug in my acceleration and then i get the following:

V = V0 + (g + (b*v^2)/(m) ) t

but i need to solve for V and I am not sure how to algebraically move V to one side? I am stuck

 

[gabbagabbahey]

heres what i have so far:

F = ma = mg + bv^2

a = g + (b*v^2)/(m)

Do the +/- signs make sense for that? When the particle is moving upwards, the drag force should be downwards and vice versa.

then i have the kinematic equation:
V = V0 +at

That equation is for constant acceleration problems. Your acceleration depends on velocity (thanks to the drag force) and isn't constant. What is the more general relationship between acceleration and velocity?

]

 

[Mugged]

oh ok,

so then i should have a = g - (b*v^2)/(m) with g = -9.8 m/s^2

this is while the particle is moving upwards.

as for a general acceleration to velocity relationship, I am not quite sure what you mean. The only other one i can think of is if i differentiated velocity with respect to time i would get acceleration with respect to time.

so since i have acceleration above, you're saying i should integrate to get velocity? if that's correct, i still have no idea how to integrate the right hand side of my above equation

well if i was doing differential equations, it would look something like this:

V' = g - (b*V^2)/(m)

but, I am terrible at dif eq, so id need help here too

 

[gabbagabbahey]

oh ok,

so then i should have a = g - (b*v^2)/(m) with g = -9.8 m/s^2

this is while the particle is moving upwards.

If your convention is that v is positive when the particle is moving upwards and negative when it is moving downwards, then yes, that's correct. Similarly, once the particle starts moving downwards, you will have $a=g+\frac{bv^2}{m}$. You can combine these into one expression by using the sign function $\text{sgn}(v)\equiv \frac{v}{|v|}$ :

$$a= g - \text{sgn}(v)\frac{bv^2}{m}$$

The only other one i can think of is if i differentiated velocity with respect to time i would get acceleration with respect to time.

Yes, in other words you have a differential equation:

$$\frac{dv}{dt}= g - \text{sgn}(v)\frac{bv^2}{m}$$[/tex]

i still have no idea how to integrate the right hand side of my above equation

This ODE is separable, surely you know how to solve one of those?

 

[Mugged]

$$dv= g - \text{sgn}(v)\frac{bv^2}{m} * dt$$

$$v = (g - \text{sgn}(v)\frac{bv^2}{m} ) t$$

is this what you meant? but id still like to get v on one side

 

[gabbagabbahey]

$$dv= g - \text{sgn}(v)\frac{bv^2}{m} * dt$$

$$v = (g - \text{sgn}(v)\frac{bv^2}{m} ) t$$

is this what you meant? but id still like to get v on one side

Divide both sides by $$g - \text{sgn}(v)\frac{bv^2}{m}$$ before you integrate.

 

[Mugged]

hey gab, i think i got the answer in the end after i separated, thanks.

by chance, do you have any idea how to find velocity as a function of the displacement?

 

[gabbagabbahey]

hey gab, i think i got the answer in the end after i separated, thanks.

by chance, do you have any idea how to find velocity as a function of the displacement?

Use your expression for v(t) to integrate and get y(t) and then eliminate "t" from the two expressions.

 

[Mugged]

that can't be possible..i have something extraordinarily ugly for my velocity function

$$v(t) = \sqrt{\frac{mg}{b}} tan ( tan^-1( \sqrt{\frac{b}{mg}}V_{0}) - gt \sqrt{\frac{b}{mg}} )$$

oh by the way gab, this function is only valid for when velocity is in the positive upward direction, i realized in my homework that i don't need to account for the velocity going down.

 

[vela]

That only looks complicated. It's actually just a shifted tan function. Let

$$\alpha = \sqrt{\frac{mg}{b}}$$

Note $\alpha$ is just a convenient combination of constants. You can write v(t) as

$$v(t) = \alpha \tan(\phi_0-\frac{gt}{\alpha})$$

where $\phi_0 = \tan^{-1}(v_0/\alpha)$, which is also a constant.