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Motion in three dimensions

  • Thread starter Matty R
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  • #1
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Hello :smile:

I was hoping someone could help me with this mechanics problem.

Homework Statement



A particle of 2kg mass moves in three dimensions under the action of a force [tex]\underline{F}=6\underline{i}-6t^2 \underline{j}+(4-2t)\underline{k}[/tex].

At [tex]t=0[/tex], it is at rest with position vector [tex]\underline{r}=\underline{j}[/tex]

Find the velocity and position vectors as functions of time.

Homework Equations



[tex]\underline{F}=m \underline{a}[/tex]

[tex]\underline{a}=\frac{d \underline{v}}{dt}[/tex]

[tex]\underline{v}=\int(\underline{a})dt[/tex]

[tex]\underline{v}=\frac{d \underline{r}}{dt}[/tex]

[tex]\underline{r}=\int(\underline{v})dt[/tex]

The Attempt at a Solution



VELOCITY:

[tex]\underline{F}=m \underline{a}[/tex]

[tex]\underline{a}=\frac{\underline{F}}{m}[/tex]

[tex]=\frac{6\underline{i}-6t^2 \underline{j}+(4-2t)\underline{k}}{2}[/tex]

[tex]=3\underline{i}-3t^2 \underline{j}+(2-t)\underline{k}[/tex]

[tex]\underline{a}=\frac{d \underline{v}}{dt}[/tex]

[tex]\underline{v}=\int(\underline{a})dt[/tex]

[tex]=\int(3-3t^2+(2-t))dt[/tex]

[tex]=(3t+c_1)\underline{i}-(t^3+c_2)\underline{j}+(2t-\frac{1}{2}t^2+c_3)\underline{k}[/tex]

I don't know what to do with the constants. I keep getting them all as 0, unless that is the right thing to do.

POSITION:

[tex]\underline{v}=\frac{d \underline{r}}{dt}[/tex]

[tex]\underline{r}=\int(\underline{v})dt[/tex]

[tex]=\int(3t-t^3+(2t-\frac{1}{2}t^2))[/tex]

[tex]=\frac{3}{2}t^2 \underline{i}-\frac{1}{4}t^4 \underline{j}+(t^2-\frac{1}{6}t^3) \underline{k}+ \underline{j}[/tex]

[tex]=\frac{3}{2}t^2 \underline{i}+\frac{3}{4}t^4 \underline{j}+(t^2-\frac{1}{6}t^3) \underline{k}[/tex]

I find mechanics very interesting, but I also find it quite confusing.

Could anyone help me please? :smile:
 

Answers and Replies

  • #2
Cyosis
Homework Helper
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Not sure what you want help with, but your method is correct. The constants vanish, because of the given boundary conditions. Example: You found that [itex]v_x=3t+c[/itex] and it is given that [itex]v_x(0)=0[/itex]. Therefore [itex]0=3*0+c[/itex].
 
  • #3
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Oh great. So both of my workings look correct?

This is what I find confusing with mechanics. Alot of the time I don't know when I'm done. :smile:
 
  • #4
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Oh great. So both of my workings look correct?

This is what I find confusing with mechanics. Alot of the time I don't know when I'm done. :smile:
The problem statement should be a pretty good clue.
Matty R said:
Find the velocity and position vectors as functions of time.
When you have found v(t) and s(t), you're done.
 
  • #5
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The problem statement should be a pretty good clue.
Usually, but I tend to overlook subtle bits. :redface:

I've just been writing up my solutions for the sheets I'll be handing in for marking, and I've hit a confusing part with my answer for the position vector.

Shouldn't I be taking the three constants into account?

[tex]r_x=\frac{3}{2}t^2+c_1[/tex]

[tex]c_1=0[/tex]


[tex]r_y=-(\frac{1}{4}t^4+c_2)[/tex]

[tex]c_2=1[/tex]


[tex]r_z=t^2-\frac{1}{6}t^3+c_3[/tex]

[tex]c_3=0[/tex]

[tex]\underline{r}=\frac{3}{2}t^2 \underline{i}-(\frac{1}{4}t^4+1) \underline{j}+(t^2-\frac{1}{6}t^3) \underline{k}[/tex]
 
  • #6
33,496
5,188
Yes, you should. When you solved for v, it would have involved some constant. At t = 0, the particle was at rest, hence its velocity was zero. That gets rid of one constant.

When you integrate the velocity to get position, you must have gotten another constant. Use the initial condition that r(0) = j to solve for that constant.
 
  • #7
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Yes, you should. When you solved for v, it would have involved some constant. At t = 0, the particle was at rest, hence its velocity was zero. That gets rid of one constant.

When you integrate the velocity to get position, you must have gotten another constant. Use the initial condition that r(0) = j to solve for that constant.
So would you say this is correct?

[tex]\underline{r}=\frac{3}{2}t^2 \underline{i}-(\frac{1}{4}t^4+1) \underline{j}+(t^2-\frac{1}{6}t^3) \underline{k}[/tex]

In the lectures, we learnt the velocity as :

[tex]\underline{v}=v_x \underline{i}+v_y \underline{j}+v_z \underline{k}[/tex]

This gave each axis it's own constant ([tex]c_1, c_2, c_3[/tex])

I did the same thing with the position vector and got the above answer.

I'd also like to ask about the second part of the question. I've been looking at it since I started this thread, and I just can't get my head around it.

Homework Statement

At what time (other than t = 0) is the velocity perpendicular to the z-axis?

I don't recognise anything like this from the work I've done. I've been thinking of it as a set of axes, like when drawing a graph. At t = 0, the velocity is at it's initial position. When it moves off, it moves in a straight line along all three axes. So would it only ever be perpendicular at t = 0?

Or do you think I'm supposed to do some maths? :smile:
 
  • #8
33,496
5,188
So would you say this is correct?

[tex]\underline{r}=\frac{3}{2}t^2 \underline{i}-(\frac{1}{4}t^4+1) \underline{j}+(t^2-\frac{1}{6}t^3) \underline{k}[/tex]
Check it and see. You've done all the hard work; checking is easy. Is r(0) = j? Now take the derivative r'(t). Is r'(0) = 0 (i.e., 0i + 0j + 0k = <0, 0, 0>)? Now take the derivative one more time. Is r''(t) = some multiple of your original force vector?
In the lectures, we learnt the velocity as :

[tex]\underline{v}=v_x \underline{i}+v_y \underline{j}+v_z \underline{k}[/tex]

This gave each axis it's own constant ([tex]c_1, c_2, c_3[/tex])
At each step in going from a(t) to r(t), you introduce a constant vector, C = <c1, c2,c3>. Note: I prefer the <,,> notation for vectors rather than the i, j, and k notation, which seems clumsy to me and take longer to write.
I did the same thing with the position vector and got the above answer.

I'd also like to ask about the second part of the question. I've been looking at it since I started this thread, and I just can't get my head around it.

Homework Statement

At what time (other than t = 0) is the velocity perpendicular to the z-axis?
v(t) will be perpendicular to the z axis when the z component is zero.
I don't recognise anything like this from the work I've done. I've been thinking of it as a set of axes, like when drawing a graph. At t = 0, the velocity is at it's initial position. When it moves off, it moves in a straight line along all three axes. So would it only ever be perpendicular at t = 0?
How can it move in a straight line along all three axes?
Or do you think I'm supposed to do some maths? :smile:
See my penultimate comment.
 
  • #9
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Note: I prefer the <,,> notation for vectors rather than the i, j, and k notation, which seems clumsy to me and take longer to write.
I've seen a few different ways of writing vectors, and we've been told to use whatever we find best unless we're specifically asked. I suppose I'm just sticking with the method that I most commonly see. :smile:

How can it move in a straight line along all three axes?
Ah. I think I worded this wrong. If you take the line y = x, when you move from the origin up the positive x-axis, the line "moves along a bit and moves up a bit" with each increment, moving along the x and y axes. Thats what I meant when I said it moved allong all three axes.

I'm not too good with the terminology.


Ok then. I've gone through my workings again and got answers for the velocity and position vectors. I've also thought about what you said about my last question, and had a go. :smile:

Part a
[tex]\underline{v}=3t\underline{i}-t^3\underline{j}+(2t-\frac{1}{2}t^2)\underline{k}[/tex]

[tex]\underline{r}=\frac{3}{2}t^2 \underline{i}-(\frac{1}{4}t^4-1) \underline{j}+(t^2-\frac{1}{6}t^3) \underline{k}[/tex]

Part b
Velocity is perpendicular to the z-axis when the z-component of the velocity ([tex]v_z[/tex]) is equal to zero.

[tex]\underline{v}=v_x \underline{i}+v_y \underline{j}+v_z \underline{k}[/tex]

[tex]\underline{v}=(3t)\underline{i}-(t^3)\underline{j}+(2t-\frac{1}{2}t^2)\underline{k}[/tex]

[tex]v_z=2t-\frac{1}{2}t^2[/tex]

[tex]v_z=0[/tex]

[tex]0=-\frac{1}{2}t^2+2t+0[/tex]

[tex]t=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

[tex]t=\frac{-2 \pm 2}{-1}[/tex]

[tex]t=0 or 4[/tex]

The velocity will be perpendicular to the z-axis at t = 4s, along with t = 0s.

Any better? :smile:
 
  • #10
33,496
5,188
As I said in post #8, you can check it for yourself.
Check it and see. You've done all the hard work; checking is easy. Is r(0) = j? Now take the derivative r'(t). Is r'(0) = 0 (i.e., 0i + 0j + 0k = <0, 0, 0>)? Now take the derivative one more time. Is r''(t) = some multiple of your original force vector?
 
  • #11
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0
As I said in post #8, you can check it for yourself.
Yep. Sorry.

[tex]\underline{r}=\frac{3}{2}t^2 \underline{i}-(\frac{1}{4}t^4-1) \underline{j}+(t^2-\frac{1}{6}t^3) \underline{k}[/tex]

[tex]\underline{r}(0)=(\frac{3}{2}.0^2)-(\frac{1}{4}.0^4-1)+(0^2-(\frac{1}{6}.0^3))[/tex]



[tex]\underline{r'}=3t \underline{i}-t^3 \underline{j}+(2t-\frac{1}{2}t^2) \underline{k}[/tex]

[tex]\underline{r'}= \underline{v}[/tex]

[tex]\underline{r'}(0)= (3.0)-0^3+((2.0)-(\frac{1}{2}.0^2))[/tex]

[tex]\underline{r'}= 0[/tex]



[tex]\underline{r''}=3 \underline{i}-3t^2 \underline{j}+(2-t) \underline{k}[/tex]

[tex]\underline{r''}= \underline{a}[/tex]

So it must be correct. :smile:
 
Last edited:

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