# Strange question in mechanics

1. Aug 27, 2010

### PhMichael

1. The problem statement, all variables and given/known data

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

2. Aug 27, 2010

### D H

Staff Emeritus
Please show your work. It's kind of hard to see where you went wrong if we can't see what you did.

3. Aug 27, 2010

### PhMichael

Ok...

First, I found the force using the formula given:

$$\vec{F} = 23.65 \hat{x} -1.89 \hat{y} +16.29 \hat{z}$$

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

$$\vec{a} = 23.65 \hat{x} -1.89 \hat{y} +16.29 \hat{z}$$

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

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

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. Aug 28, 2010

### D H

Staff Emeritus

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 $$\vec A$$. What does the force equation say will happen to those two components?

5. Aug 28, 2010

### PhMichael

Aha, I think I got you but i'm still not getting the right answer =/

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

$$\frac{dV_{||}}{dt}=\frac{-\vec{A} \cdot \vec{V}}{m}$$

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 $$\vec{A}$$, no force acts on the body so it moves with a constant speed, which is:

$$V=\sqrt{2^{2}+1^{2}+(-4)^{2}}$$

and the kinetic energy is:

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

What's wrong in this solution?

6. Aug 28, 2010

### D H

Staff Emeritus
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. Aug 28, 2010

### PhMichael

I have no clue. I'm totally confused now :|

8. Aug 28, 2010

### D H

Staff Emeritus
What are these two components of the velocity vector, initially?

9. Aug 28, 2010

### PhMichael

Well, at t<0 the velocity vector was: $$\vec{V}=2 \hat{x}+ \hat {y}-4 \hat{z}$$ and at t=0 , when the force acts on the body, the time dependent velocity which is parallel to $$\vec{A}$$ is given by (after solving the 1st order ODE):

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

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

This is all I know ... I think.

Last edited: Aug 28, 2010
10. Aug 28, 2010

### PhMichael

anyone?

11. Aug 28, 2010

### vela

Staff Emeritus
In symbols, you have

$$\vec{V} = \vec{v}_\perp + \vec{v}_\parallel$$

where $$\vec{A}$$ defines the perpendicular and parallel directions. You've figured out how $\vec{v}_\parallel$ evolves with time. What about $\vec{v}_\perp$?

12. Aug 28, 2010

### PhMichael

Well, $$\vec{v}_\perp=constant$$ because no forces act in this direction, so it should be the given $$\vec{V}$$. No?

13. Aug 28, 2010

### PhMichael

wait a minute. $$\vec{v}_\perp= \vec{V} \cdot \hat{A}_{\perp}$$

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

So there are infinite vectors which are perpendicular to $$\vec{A}$$ ! How can I figure out which is the right one?

14. Aug 28, 2010

### vela

Staff Emeritus
Think of $$\vec{v}_\perp$$ as the part of $$\vec{V}$$ that's not parallel to $$\vec{A}$$. In other words,

$$\vec{v}_\perp = \vec{V}-\vec{v}_\parallel$$

15. Aug 28, 2010

### vela

Staff Emeritus
Just wanted to add that this approach would work, but you have to add the constraint that $$\hat{A}_\perp$$ lies in the plane defined by $$\vec{A}$$ and $$\vec{V}$$.

16. Aug 28, 2010

### vela

Staff Emeritus
To be more precise, its magnitude is constant. The vector, however, isn't, because its direction changes with time.

17. Aug 28, 2010

### PhMichael

well, in that case after a very long time the parallel component drops downs and we're left with,

$$\vec{v}_{\perp} = \vec{V}$$

so the perpendicular component equals the velocity given.

Something stinks here for sure ... I can feel it ... but what it that? :D

18. Aug 28, 2010

### PhMichael

$$\hat{A}=\frac{1}{\sqrt{41}} (1,-6,2)$$

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

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

so as $$t \to infinity$$

$$\vec{v}_{\perp} = (2,1,-4)$$

Is this, so far, right?

19. Aug 28, 2010

### D H

Staff Emeritus
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. Aug 28, 2010

### PhMichael

Then how am I supposed to solve it? :|