Classical Mechanics

  • #1
1,444
2
given a particle of mass m where r is the position vector for its trajectory
F is the force it experiences
p is its momentum
r0 represents the initial position vector
and r0 F, p are all constant vectors
show that
[tex] \vec{r}' = \frac{1}{2} \frac{\vec{F}}{m} t^2 + \frac{\vec{p}}{m} t [/tex] where
[tex] \vec{r}' = \vec{r} - \vec{r_0} [/tex]
represents teh parametric equations of a parabola if F is not parallel to p
ok so the angular momentum of the particle is
[tex] \vec{L} = \vec{r}' \times \vec{p} = (\frac{1}{2} \frac{\vec{F}}{m} t^2 + \frac{\vec{p}}{m} t) \times \vec{p} = \frac{t^2}{2m} (\vec{F} \times \vec{p}) + \frac{t}{m} (\vec{p} \times \vec{p}) [/tex]

now certainly p cross p is zero
and we end up with
[tex] \vec{L} = \frac{t^2}{2m} (\vec{F} \times \vec{p}) [/tex]
and this is not zero because F is not parallel to p
so this is a parabola because it is second order with responect to t?
this is the part that im not too sure about...
Is this right at all first of all?

Thank you for your help!
 
Last edited:

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,484
1,210
I don't understand why you're looking at angular momentum. Why don't you just integrate both sides of Newton's second law twice with respect to time?

How can force and monentum both be contant? Did you mean to say that p is the initial momentum?

Regards,
George
 
Last edited:
  • #3
1,444
2
are you saying that I should integrate [tex] \frac{dp}{dt} = F [/tex] with respect to time twice?

I dont understand how that would relate the question at hand

r0, F and p are the initial values
Im trying to prove that r' is the parametric equation of a parabola...
 
  • #4
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,484
1,210
Sorry, I miunderstood your original question. One way to do this is to choose an xyz coordinate system such that: F is along the y-axis; FxP is along the z-axis; z'_0 = 0. Then it is fairly easy to show that r' = (x', y', z') is such that z' = 0 and y' = Ax' + Bx'^2, where A and B are constants.

Regards,
George
 
  • #5
1,444
2
how or why are we making the assumption that z0 = 0?? Isn't what you're suggesting making hte problem lose its generality?
 
  • #6
Tide
Science Advisor
Homework Helper
3,089
0
All you have to do is integrate the equation

[tex]\frac {d^2 (\vec r - \vec r_0)} {dt^2} = \frac {\vec F}{m}[/tex]

when [itex]\vec F[/itex] is a constant vector. Your result should follow with little effort.
 
  • #7
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,484
1,210
stunner5000pt said:
how or why are we making the assumption that z0 = 0?? Isn't what you're suggesting making hte problem lose its generality?

The assumption z0 = 0 is not being made - you're equation implies that when t = 0, r' = 0, so z'0 = 0. Notice the prime.

It seems that Tide has interpreted your question the same way that I initially did. Let's see if I understand correctly now - you are *not* asking how to show how r' = F/(2m) t^2 + p/m t follows if F is constant.

Regards,
George
 
  • #8
Tide
Science Advisor
Homework Helper
3,089
0
George,

Yes, I did interpret the problem that way. I see now that stunner is trying to prove that the path is a parabola.

One way to do it is to show there exists a vertex and a directrix such that points on the trajectory are equidistant from the vertex and corresponding points along the directrix.
 
  • #9
1,444
2
i think i like tide's suggested method. First question i asked myself was how would you show something was a parabola - using a vertex and a directrix
 
  • #10
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,484
1,210
Let me expand on my previous comments.

First, choose an orthonormal basis [itex]\left\{ e_{1}, e_{2}, e_{3} \right\}[/itex] for space that exploits the symmetries of the problem.

[itex]\vec{r}' \cdot \left( \vec{F} \times \vec{p} \right) = 0[/itex] gives that the motion takes place in the plane spanned by [itex]\vec{F}[/itex] and [itex]\vec{p}[/itex], so choose one of the basis vectors, [itex]e_{3}[/itex] say, perpendicular to this plane. Consequently, the other 2 basis vectors are in the plane spanned by [itex]\vec{F}[/itex] and [itex]\vec{p}[/itex].

It would be nice to choose the other basis vectors to be in the directions of [itex]\vec{F}[/itex] and [itex]\vec{p}[/itex], but we aren't given that [itex]\vec{F}[/itex] and [itex]\vec{p}[/itex] are orthogonal to each other. The best we can do is choos one of the basis vectors along either [itex]\vec{F}[/itex] or [itex]\vec{p}[/itex].

Which one? Think of the physical situation - motion with a constant force. An example of this is motion under the influence of gravity (and neglecting air resistance) near the EartH's surface. A thrown ball follows a parabola, with the symmtry axis of the parabola parallel to the force of gravity. Thus, choose [itex]e_{2}[/itex] to be in the direction of [itex]-\vec{F}[/itex], and [itex]e_{1}[/itex] to be perpendicular to both [itex]e_{2}[/itex] and [itex]e_{3}[/itex]. Hence, motion is confined to the plane spanned by [itex]e_{1}[/itex] and [itex]e_{2}[/itex].

Since [itex]\vec{p}[/itex] is in the plane spanned by [itex]e_{1}[/itex] and [itex]e_{2}[/itex], we can write

[tex]\vec{p} = p \mathrm{cos} \theta e_{1} + p \mathrm{sin} \theta e_{2},[/tex]

where [itex]\theta[/itex] is the constant angle between (the initial momentum) [itex]\vec{p}[/itex] and [itex]e_{1}[/itex].

We now have

[tex]\vec{r}' = x' e_{1} + y' e_{2} + z' e_{3} = - \frac{F}{2m} t^2 e_{2} + \frac{p}{m} t \mathrm{cos} \theta e_{1} + \frac{p}{m} t \mathrm{sin} \theta e_{2}.[/tex]

Note that [itex]z' = 0[/itex]. Equate components of [itex]e_{1}[/itex] on both side, and solve for [itex]t[/itex] in terms of [itex]x'[/itex]. Use this to eliminate all t's in the expression.

What do you get?

Are you at U of T or York? My wife graduated from both.

Regards,
George
 
Last edited:
  • #11
1,444
2
so the basic idea of this problem is to show that y is proportional to x of the second order, right? Because such motion describes a parabola.

i am at york

thank you for the help
 
  • #12
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,484
1,210
stunner5000pt said:
so the basic idea of this problem is to show that y is proportional to x of the second order, right?
Almost - it turns out (I thin) that y' = Ax'^2 + Bx', where A and B are (somewhat complicated looking) constants. This is the equation of a parabola. If you complete the square (but you don't to do this), you'll find that (y' - constant) is proportional to (x' - constant)^2.

There might be a much simpler approach to this question.

i am at york

I was in (actually, just north of) Toronto for the holidays. I meant to get down to Steacie (if it was open), but, unfortunately, this didn't happen. I got quite a surprise the first time I walked into the 'new' Steacie.

I used to like to eat at Jimmy the Greek and Falafel Hut.

Regards,
George
 
  • #13
1,444
2
oh i have a question... why is it -F?? Is it a typo or is there a specific reason for it ? Convenience?
 

Related Threads on Classical Mechanics

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
6
Views
790
  • Last Post
Replies
6
Views
943
  • Last Post
Replies
2
Views
2K
Top