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A condition for an object to return to the xy-plane

  • #1
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Homework Statement


[/B]
An object of m-mass is to be thrown from xy-plane with an initial velocity ##\mathbf v_0 = v_0\mathbf e_z \, (v_0 > 0)## to a force field ##\mathbf F = -F_0 e^{-z/h}\mathbf e_z\,## , where ##F_0, h > 0## are constants. By what condition does the object return to ##xy##-plane? How does the situation change if a velocity dependent drag force affects to the object?

Homework Equations


[/B]
##\mathbf F = -F_0 e^{-z/h}\mathbf e_z##
##\mathbf v_0 = v_0\mathbf e_z##
##F =ma = mv\frac{dv}{dz}##

The Attempt at a Solution


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The sign is positive from xy-plane to upwards as a convention.

There is an acceleration associated with the force field. So Newton's second law should give:
##-F_0 e^{-z/h}\mathbf e_z = m\mathbf a##
##-F_0 e^{-z/h} \mathbf e_z = ma\, \mathbf e_z##
##F_0 e^{-z/h} = -ma##
Integrating ## 0 \to z_0##
[tex]\int_0^{z_0} -F_0 e^{-z/h}\, dz= \int_0^{z_0}m \frac{d^2 z}{dt^2}\, dz[/tex]
[tex]hF_0 \int_0^{z_0} -\frac{1}{h}e^{-z/h}\, dz= m\int_0^{z_0} \frac{d^2 z}{dt^2}\frac{dz}{dt}\, dt[/tex]
[tex]hF_0 \int_0^{z_0} -\frac{1}{h}e^{-z/h}\, dz= m\int_0^{z_0} d \left [\frac{1}{2}\left ( \frac {dz} {dt} \right)^2 \right][/tex]
[tex]hF_0 (e^{-z_0/h}-1)= m\frac{1}{2}(0^2 - \mathbf v{_0}^2)[/tex]
[tex]hF_0 (1-e^{-z_0/h})= m\frac{1}{2}v{_0}^2\mathbf {e_z}^2[/tex]

[tex]v_0 = \sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}}[/tex]
So I think this velocity should be less or equal to this..
[tex]v_0 \leq \sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}}[/tex]

My book does not give any answer to this problem, so I'm not 100% sure this is the correct answer but I assume it is.

Then the second part of the question adds a velocity dependent drag force, so I use ##k \mathbf v = - kv\, \mathbf e_z## to represent this force. The Newton's second law leads to:

##-F_0 e^{-z/h}\, \mathbf e_z + k \mathbf v= m \mathbf a##
##-F_0 e^{-z/h}\, \mathbf e_z - kv\, \mathbf e_z= ma\, \mathbf e_z##
##F_0 e^{-z/h} + kv = -mv\frac{dv}{dz}##
First of all I want to know am I even on the right track for trying to solve this
[tex]m\frac{dv}{dz}+\frac{F_0}{v} e^{-z/h} + k = 0[/tex]

Thanks for reading and any help is appreciated
 
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Answers and Replies

  • #2
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382
So I think this velocity should be less or equal to this..
v0≤√2hF0(1−e−z0/h)m​
You aren’t quite done. To span the space of all possible v0 that will return to the plane, what should you use for z0?
 
  • #3
898
382
hF0(1−e−z0/h)=m12v02ez2hF0(1−e−z0/h)=m12v02ez2​
hF_0 (1-e^{-z_0/h})= m\frac{1}{2}v{_0}^2\mathbf {e_z}^2
Note that the right hand side is kinetic energy. Your left hand integral is work. That doesn’t really change what you had to do, but it’s another way of thinking about the problem. Well, actually, it would have saved you from having to do the right hand integral.
 
  • #4
898
382
First of all I want to know am I even on the right track for trying to solve this
In the first part, by integrating the left side and the right side separately you accidentally turned this into a “work = energy” problem. For the second part I think you should do that explicitly and not bring ma into the work integral.
 
  • #5
48
2
You aren’t quite done. To span the space of all possible v0 that will return to the plane, what should you use for z0?
Would something like ##z_0 sin(\theta)## suffice? Where ##\theta## is the angle between object's initial trajectory and ##xy##-plane
Note that the right hand side is kinetic energy. Your left hand integral is work. That doesn’t really change what you had to do, but it’s another way of thinking about the problem. Well, actually, it would have saved you from having to do the right hand integral.
That is true, I just fancied putting it there
 
  • #6
898
382
Would something like z0sin(θ)z0sin(θ)z_0 sin(\theta) suffice? Where θθ\theta is the angle between object's initial trajectory and xyxyxy-plane
The motion is all in z, so there is no angle with the xy plane.

Let’s try it this way. z0 is a parameter you made up. It is the height at which the integral of the force over the path (work) has countered the initial energy. In other words it is the height at which the object stops and start falling back toward the plane. The problem has no specific z0. So let’s say you pick a value for z0, say, 100m. Your result says that any v0 less than what you show will turn around and fall back to the plane. That is true. You did that correctly. However the problem wants a criterion that describes ALL cases where the object returns to the plane. If you pick z0=100m, is there not some slightly higher velocity that will rise to 101m violating your criterion, but nevertheless turns around and falls back to the plane?

So, for what value of z0 will your equation give a v0 so that all values less than or equal to v0 will return to the plane (true for all z0) AND all values greater than v0 will NOT return to the plane? (true only for one specific choice of z0)
 
  • #7
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So, for what value of z0 will your equation give a v0 so that all values less than or equal to v0 will return to the plane (true for all z0) AND all values greater than v0 will NOT return to the plane? (true only for one specific choice of z0)
I'm understanding that this should be evaluated with potential and kinetic energies

At the distance ##r## the object has potential energy ##E_{p1}##
[tex]E_{p1}= hF_0 (1-e^{-r/h})[/tex]
This is the limit after the object will not return to the xy-plane. The object returns when its kinetic energy transforms to potential energy ##E_{p1}## in this range:
[tex]0 \lt E_{p1} \leq m\frac{1}{2}v{_0}^2[/tex]
[tex]0 \lt hF_0 (1-e^{-r/h}) \leq m\frac{1}{2}v{_0}^2[/tex]
After ##z_0## it does not return when its kinetic energy transforms to potential energy ##E_{p2}##
[tex]E_{p2} = hF_0 (1-e^{-z_0/h})[/tex]
meaning this inequality
[tex]hF_0 (1-e^{-z_0/h}) \gt m\frac{1}{2}v{_0}^2[/tex]
and finally
[tex]hF_0 (1-e^{-z_0/h}) \gt m\frac{1}{2}v{_0}^2 \geq hF_0 (1-e^{-r/h})[/tex]
[tex]\sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}} \gt v_0 \geq \sqrt {\frac{2hF_0(1-e^{-r/h})}{m}}[/tex]
So when
[tex]\sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}} \gt v_0 [/tex]
the object does not return and when
[tex]v_0 \geq \sqrt {\frac{2hF_0(1-e^{-r/h})}{m}} \gt 0[/tex]
the object always returns
 
  • #8
898
382
So when
√2hF0(1−e−z0/h)m>v02hF0(1−e−z0/h)m>v0​
\sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}} \gt v_0
the object does not return and when
v0≥√2hF0(1−e−r/h)m>0v0≥2hF0(1−e−r/h)m>0​
v_0 \geq \sqrt {\frac{2hF_0(1-e^{-r/h})}{m}} \gt 0
the object always returns
No, I don’t think you understood what I was saying. This appears to be just saying the same thing twice. r is no different than z0 and this result is no different than your first solution.

You don’t need a lot more math. Your answer was right the first time. You just need one further step of logic. Note that the force extends to infinity. There is no finite z0 for which the force is zero. What you have found is the v0 that will reach z0 (or r) in height. However, for a v0 that reaches any FINITE z0 (hint hint) there exists a slightly larger v0 that results in a slightly higher z0 where the object still stops and falls back to the plane. If you want to escape the potential well you have to get all the way out. What should you substitute for z0?
 
  • #9
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No, I don’t think you understood what I was saying. This appears to be just saying the same thing twice. r is no different than z0 and this result is no different than your first solution.

You don’t need a lot more math. Your answer was right the first time. You just need one further step of logic. Note that the force extends to infinity. There is no finite z0 for which the force is zero. What you have found is the v0 that will reach z0 (or r) in height. However, for a v0 that reaches any FINITE z0 (hint hint) there exists a slightly larger v0 that results in a slightly higher z0 where the object still stops and falls back to the plane. If you want to escape the potential well you have to get all the way out. What should you substitute for z0?
Okay, so the answer is infinity meaning that the object can only escape if it reaches for infinity. A limit can be taken
[tex]\lim_{z_0 \rightarrow \infty} \sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}} = \sqrt {\frac{2hF_0}{m}} [/tex]
I want to understand this clearly: the object still has some potential energy. Is this the case? So the object does not return when:
[tex]\sqrt {\frac{2hF_0}{m}} \gt v_0 [/tex]
But at some finite point ##z_0## it will return
[tex]v_0 \geq \sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}} \gt 0[/tex]
This makes sense to me, seems correct?
 
  • #10
898
382
Okay, so the answer is infinity meaning that the object can only escape if it reaches for infinity. A limit can be taken
[tex]\lim_{z_0 \rightarrow \infty} \sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}} = \sqrt {\frac{2hF_0}{m}} [/tex]
I want to understand this clearly: the object still has some potential energy. Is this the case? So the object does not return when:
[tex]\sqrt {\frac{2hF_0}{m}} \gt v_0 [/tex]
But at some finite point ##z_0## it will return
[tex]v_0 \geq \sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}} \gt 0[/tex]
This makes sense to me, seems correct?
Yes, I think you’ve got it, but there is just a little oddness in the language here. The phrase “the object still has some potential energy” is a strange way to put it. The object is in a potential well. Think of a gravity well, or maybe one of those vortexes that you roll coins down into, or a roller coaster traveling fast in a valley and zooming up the next hill. The object starts deep in the well and has low potential energy. It starts out with high kinetic energy. The whole time it is climbing and slowing down it is trading kinetic energy for potential energy. The potential energy is getting larger and larger the higher the object goes i.e. the more work could be done by the object falling back into the hole. Think of the rollercoaster getting near the top of the hill. However if it is going fast enough it will get all the way to the top of the hill and the potential energy doesn’t get used unless the roller coaster happens to find its way back to fall down this or some other hill. The difference in this case is that the top of the hill is at infinity.

Another slight logic problem: when you say
“But at some finite point z0"
You should realize that your inequality “v0 >=“ includes all v0 up to infinity including all of those that will escape. The point is that there are v0 between those two values that still return.
 
  • #11
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2
Yes, I think you’ve got it, but there is just a little oddness in the language here. The phrase “the object still has some potential energy” is a strange way to put it. The object is in a potential well. Think of a gravity well, or maybe one of those vortexes that you roll coins down into, or a roller coaster traveling fast in a valley and zooming up the next hill. The object starts deep in the well and has low potential energy. It starts out with high kinetic energy. The whole time it is climbing and slowing down it is trading kinetic energy for potential energy. The potential energy is getting larger and larger the higher the object goes i.e. the more work could be done by the object falling back into the hole. Think of the rollercoaster getting near the top of the hill. However if it is going fast enough it will get all the way to the top of the hill and the potential energy doesn’t get used unless the roller coaster happens to find its way back to fall down this or some other hill. The difference in this case is that the top of the hill is at infinity.

Another slight logic problem: when you say
“But at some finite point z0"
You should realize that your inequality “v0 >=“ includes all v0 up to infinity including all of those that will escape. The point is that there are v0 between those two values that still return.
Yes I hopefully understand this better now. It is more fitting to word it as "when object does not return to xy-plane it has to approach infinity" and refrain to mention potential energy. It is very logical, if it would not approach infinity then the object would be returning back. The clause:
[tex]\sqrt {\frac{2hF_0}{m}} \gt v_0 [/tex]
should actually mean that with velocity ##\sqrt {\frac{2hF_0}{m}}## or higher the object does not return to xy-plane. All velocities below that do return, so better put is:
[tex]\sqrt {\frac{2hF_0}{m}} \gt v_0 \gt 0[/tex]
For the second part I follow your advice to proceed it as work = energy problem and post about it later here
 
  • #12
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Me being busy with other work, this problem got neglected and it is still unsolved. Thus returning to get this done, there is a part that bring difficulties.

The second part is treated as a energy = work problem, where the change in kinetic energy is ##\Delta E_z = -\frac {1}{2}mv_0^2##

Then the work is ##W_z = \int_0^{z_0} (\mathbf F + k \mathbf v)d \mathbf s##
,where ##d \mathbf s = e_z dz##

Since ##\Delta E_z = W_z## this leads to

$$-\frac {1}{2}mv_0^2 = hF_0 \int_0^{z_0} -\frac{1}{h}e^{-z/h} \,dz + \int_0^{z_0} k v \,dz$$

there is ##dz = v\, dt## which might help..?

$$-\frac {1}{2}mv_0^2 = hF_0 (1- e^{-z_0/h}) + \int_0^{z_0} k v^2 \,dt$$

it rather feels that I made a major mistake earlier, since I cannot get past that integral . Thanks for reading
 
  • #13
ZapperZ
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Shouldn't this be similar to finding the escape velocity?

If you know the description of the force field, are you able to find the potential energy field? After that, what is the PE at z=0? Equate this to the KE at that position and you should be able to find the escape velocity, meaning that any velocity smaller than this will cause the object to drop back.

So far, you are basically doing the same math, but you're getting stuck somewhere.

Zz.
 
  • #14
haruspex
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Then the work is ##W_z = \int_0^{z_0} (\mathbf F + k \mathbf v)d \mathbf s##
you are only told that the drag is a function of velocity, not that it is proportional to velocity.
 
  • #15
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you are only told that the drag is a function of velocity, not that it is proportional to velocity.
The original text is not in English and my translation was not carefully done. The second part asks:

"How does the situation change if the object is also affected by drag force that is proportional to velocity?"

Previously I said it was "a velocity dependent drag force" which, yeah I see it does not mean it has to be proportional to velocity. My mistake, sorry about that
 
  • #16
48
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Shouldn't this be similar to finding the escape velocity?

If you know the description of the force field, are you able to find the potential energy field? After that, what is the PE at z=0? Equate this to the KE at that position and you should be able to find the escape velocity, meaning that any velocity smaller than this will cause the object to drop back.

So far, you are basically doing the same math, but you're getting stuck somewhere.

Zz.
The potential energy field ##U## is as far as I understand

$$U = hF_0 \int_0^{z} -\frac{1}{h}e^{-z/h} \,dz = hF_0 (e^{-z/h} - 1)$$
##((e^{-z/h} - 1)## not ##(1 - e^{-z/h})##, made a mistake earlier##)##

To be honest, I am just confused right now. U = 0 when z = 0, but I thought finding escape velocity was all about the object travelling to infinity

EDIT: Wait, the sign of the force field matters. I need to think about this for a moment and hopefully I understand your point
 
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  • #17
haruspex
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The second part asks:

"How does the situation change if the object is also affected by drag force that is proportional to velocity?"
Ok.
As you observed, you cannot perform that integral because you do not have v as a function of t. This means you need to go back to a diiferential equation form. However, I think you will find that insoluble too.
The trick here is to extract an inequality from the differential equation and see what that says about the asymptotic behaviour.
Post the ODE and see if you can the spot the inequality.
 
  • #18
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ODE can be made to look like this:

##k\, z'(t) = -F_0e^{-z(t)/h}-m\,z''(t)##

I lack experience in studying asymptotic behaviour, so this is what I gathered. The time approaches infinity;
##t \to \infty##, so velocity approaches some constant ##v_1## and ##z \to \infty##. The deceleration ceases, meaning:
as ##t \to \infty:##
##z'(t) \to v_1##
##e^{-z(t)/h} \to 1##
##m\,z''(t) \to 0##

I can proceed from here, if this has been the right way to do it
 
  • #19
haruspex
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ODE can be made to look like this:

##k\, z'(t) = -F_0e^{-z(t)/h}-m\,z''(t)##
The signs don't look right.
 

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