• Support PF! Buy your school textbooks, materials and every day products Here!

Strange question in mechanics

  • Thread starter PhMichael
  • Start date
  • #1
134
0

Homework Statement



A particle of mass 1[kg] is moving with a velocity of: [tex]\vec{V}=2 \hat{x}+ \hat{y} -4 \hat{z} [m/s] [/tex]. At the instant t=0, the particle experiences a force which is given by:
[tex] \vec{F}= \vec{A} \times \vec{V} - (\vec{A} \cdot \vec{V}) \hat {A} [/tex] where [tex] \vec{A}= \hat{x} -6 \hat{y} + 2 \hat{z} [/tex].
What is the kinetic energy of the particle after a very long time?

2. The attempt at a solution

What I've done is to do these vectorial multiplications in order to obtain the force and from this expression I obtain the acceleration and by integration I obtain the time dependent velocity. However, if I let t approach infinity, then also the velocity will be infinity and the same thing for the kinetic energy, but this totally wrong.

How can I solve this?
 

Answers and Replies

  • #2
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
683
Please show your work. It's kind of hard to see where you went wrong if we can't see what you did.
 
  • #3
134
0
Ok...

First, I found the force using the formula given:

[tex] \vec{F} = 23.65 \hat{x} -1.89 \hat{y} +16.29 \hat{z} [/tex]

So the acceleration is (m=1kg) the same:

[tex] \vec{a} = 23.65 \hat{x} -1.89 \hat{y} +16.29 \hat{z} [/tex]

Now, I integrate this last expression in order to obtain the time dependent velocity:

[tex] \vec{v} = (23.65 \hat{x} -1.89 \hat{y} +16.29 \hat{z}) t+2 \hat{x}+ \hat{y} -4 \hat{z} [/tex]

So, letting t approch infinity leads to an infinite velocity and hence an infinite kinetic energy, which is wrong because the answer is 8.74 [J]
 
  • #4
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
683
The answer is correct.

Your mistake is that you are assuming the force is constant. It isn't. It is a function of velocity.

Hint: Think of the velocity as comprising components parallel to and normal to [tex]\vec A[/tex]. What does the force equation say will happen to those two components?
 
  • #5
134
0
Aha, I think I got you but i'm still not getting the right answer =/

In the direction parallel to [tex]\vec{A}[/tex], a forces: [tex]-(\vec{A} \cdot \vec{V})[/tex] acts on the body, hence the equation of motion is:

[tex] \frac{dV_{||}}{dt}=\frac{-\vec{A} \cdot \vec{V}}{m} [/tex]

and the solution is a decaying exponential, so this velocity doesn't contribute to the kinetic energy at a very long time.

Now, in the direction perpendicular to [tex]\vec{A}[/tex], no force acts on the body so it moves with a constant speed, which is:

[tex] V=\sqrt{2^{2}+1^{2}+(-4)^{2}} [/tex]

and the kinetic energy is:

[tex] K=0.5mV^{2}=10.5 [J] [/tex]

What's wrong in this solution?
 
  • #6
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
683
You have the exponential decay part correct, but you do not have the big picture yet.

The vector V has components parallel to and normal to A. The force has two components, one is AxV, the other is -(A·V)Â. The first is normal to A and only the normal component of V contributes to this term. The latter is parallel to A and only the parallel of V contributes to this term.

What happens to each of those terms over time?
 
  • #7
134
0
I have no clue. I'm totally confused now :|
 
  • #8
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
683
What are these two components of the velocity vector, initially?
 
  • #9
134
0
Well, at t<0 the velocity vector was: [tex] \vec{V}=2 \hat{x}+ \hat {y}-4 \hat{z} [/tex] and at t=0 , when the force acts on the body, the time dependent velocity which is parallel to [tex] \vec{A} [/tex] is given by (after solving the 1st order ODE):

[tex] v=v_{0} e^{-At/m} [/tex] where [tex]v_{0} = \vec{V} \cdot \hat{A} = -1.8741 [/tex].

Regarding the velocity which is perpendicular to [tex] \vec{A} [/tex], all I know is that it's a constant because no forces act perpendicular to [tex] \vec{A} [/tex].

This is all I know ... I think.
 
Last edited:
  • #10
134
0
anyone?
 
  • #11
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,631
1,267
In symbols, you have

[tex]\vec{V} = \vec{v}_\perp + \vec{v}_\parallel[/tex]

where [tex]\vec{A}[/tex] defines the perpendicular and parallel directions. You've figured out how [itex]\vec{v}_\parallel[/itex] evolves with time. What about [itex]\vec{v}_\perp[/itex]?
 
  • #12
134
0
Well, [tex]\vec{v}_\perp=constant [/tex] because no forces act in this direction, so it should be the given [tex] \vec{V} [/tex]. No?
 
  • #13
134
0
wait a minute. [tex] \vec{v}_\perp= \vec{V} \cdot \hat{A}_{\perp} [/tex]

now, [tex] \vec{A}_{\perp}=a*(6,1,0)+b*(-2,0,1) [/tex] where a and b are constants.

So there are infinite vectors which are perpendicular to [tex]\vec{A} [/tex] ! How can I figure out which is the right one?
 
  • #14
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,631
1,267
wait a minute. [tex] \vec{v}_\perp= \vec{V} \cdot \hat{A}_{\perp} [/tex]

now, [tex] \vec{A}_{\perp}=a*(6,1,0)+b*(-2,0,1) [/tex] where a and b are constants.

So there are infinite vectors which are perpendicular to [tex]\vec{A} [/tex] ! How can I figure out which is the right one?
Think of [tex]\vec{v}_\perp[/tex] as the part of [tex]\vec{V}[/tex] that's not parallel to [tex]\vec{A}[/tex]. In other words,

[tex]\vec{v}_\perp = \vec{V}-\vec{v}_\parallel[/tex]
 
  • #15
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,631
1,267
wait a minute. [tex] \vec{v}_\perp= \vec{V} \cdot \hat{A}_{\perp} [/tex]

now, [tex] \vec{A}_{\perp}=a*(6,1,0)+b*(-2,0,1) [/tex] where a and b are constants.

So there are infinite vectors which are perpendicular to [tex]\vec{A} [/tex] ! How can I figure out which is the right one?
Just wanted to add that this approach would work, but you have to add the constraint that [tex]\hat{A}_\perp[/tex] lies in the plane defined by [tex]\vec{A}[/tex] and [tex]\vec{V}[/tex].
 
  • #16
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,631
1,267
Well, [tex]\vec{v}_\perp=constant [/tex] because no forces act in this direction, so it should be the given [tex] \vec{V} [/tex]. No?
To be more precise, its magnitude is constant. The vector, however, isn't, because its direction changes with time.
 
  • #17
134
0
well, in that case after a very long time the parallel component drops downs and we're left with,

[tex] \vec{v}_{\perp} = \vec{V} [/tex]

so the perpendicular component equals the velocity given.

Something stinks here for sure ... I can feel it ... but what it that? :D
 
  • #18
134
0
Think of [tex]\vec{v}_\perp[/tex] as the part of [tex]\vec{V}[/tex] that's not parallel to [tex]\vec{A}[/tex]. In other words,

[tex]\vec{v}_\perp = \vec{V}-\vec{v}_\parallel[/tex]
[tex] \hat{A}=\frac{1}{\sqrt{41}} (1,-6,2) [/tex]

[tex] \vec{v}_{||}=\frac{v_{0}e^{-At/m}}{\sqrt{41}}(1,-6,2) [/tex]

[tex] \vec{v}_{\perp} = (2,1,-4) - \frac{v_{0}e^{-At/m}}{\sqrt{41}}(1,-6,2) [/tex]

so as [tex] t \to infinity [/tex]

[tex] \vec{v}_{\perp} = (2,1,-4) [/tex]

Is this, so far, right?
 
  • #19
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
683
No! You do not know what the velocity vector is as t gets large. The reason is simple: It isn't constant.

Fortunately, the velocity vector does not have to be constant to solve this problem.
 
  • #20
134
0
Then how am I supposed to solve it? :|
 
  • #21
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,631
1,267
The particle has velocity [itex]\vec{v}(t) = \vec{v}_\perp(t)+\vec{v}_\parallel(t)[/itex]. The mistake you're making is assuming [tex]\vec{v}(t) = \vec{V}[/tex] for t>0. It's only true for t=0. After t=0, the applied force causes the velocity of the particle to change so it's no longer equal to [tex]\vec{V}[/tex].
 
  • #22
134
0
Let me arrange this stuff in my head:

As you guys told me, the expression for [tex] \vec{v}_{||} [/tex] is correct. Now, we all agree that [tex] \vec{v}_{\perp} [/tex] is a constant ... right?

Now from this point, what am I supposed to do? ... I'm not assuming anything, it's just that as far as I know, when no forces exist in some direction then the corresponding velocity is constant.
 
  • #23
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
683
Now, we all agree that [tex] \vec{v}_{\perp} [/tex] is a constant ... right?
No!

How can it possibly be constant? Look at the force equation.
 
  • #24
134
0
No!

How can it possibly be constant? Look at the force equation.
I'm looking and going nuts ...

The second part of the force equation is already used for the parallel velocity component so I guess you're refering to the first one: [tex] \vec{A} \times \vec{V} [/tex].

[tex]\vec{A} \times \vec{V} = \vec{A} \times (\vec{v}_{||} + \vec{v}_{\perp} ) \to \vec{A} \times \vec{V} = \vec{A} \times \vec{v}_{\perp} [/tex]

all I can say is that the result is a vector which is perpendicular to the plane formed by [tex]\vec{A}[/tex] and [tex]\vec{v}_{\perp}[/tex] ... WHAT AM I MISSING HERE?!?!?!?!
 
  • #25
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,631
1,267
What does a force that's perpendicular to velocity do?
 

Related Threads on Strange question in mechanics

  • Last Post
Replies
13
Views
4K
Replies
10
Views
1K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
4
Views
925
  • Last Post
Replies
1
Views
964
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
4
Views
1K
Top