How to find the Trajectory of a Particle

  • #1
PKM
49
16

Homework Statement


The velocity of a particle at a certain point is given as [itex]
\vec v=5(y\hat i - x\hat j)[/itex]. How to find the general equation of the path of the particle?

Homework Equations


Here, [itex]\frac{d\vec x}{dt}=5(y\hat i - x\hat j)[/itex].

The Attempt at a Solution


As the velocity is not given as a function of time, but that of position, how may I proceed? Plese help.
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
19,874
10,411
What you have is a differential equation for the position as a function of time. If you write out the components of your differential equation separately, what do you get?
 
  • #3
PKM
49
16
What you have is a differential equation for the position as a function of time. If you write out the components of your differential equation separately, what do you get?
Sorry, I didn't get it properly, Sir. The differential equation may be framed as [itex]\frac{d\vec x}{y\hat i-x\hat j}=5dt[/itex]. Is it what you are suggesting? Please clarify.
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
19,874
10,411
No. You cannot divide by a vector! Just write out the different components. How do ##\vec x## and ##d\vec x/dt## look in components?
 
  • #5
PKM
49
16
No. You cannot divide by a vector! Just write out the different components. How do ##\vec x## and ##d\vec x/dt## look in components?
OK, we take [itex]\vec x=x\hat i+y\hat j[/itex]. Also, [itex]x[/itex] comp. of velocity, [itex]v_x=5y[/itex], and, [itex]v_y=-5x[/itex]. What may I do next? It follows that [itex]\frac{dx}{y}=5dt[/itex]. Again, [itex]\frac{dy}{x}=-5dt[/itex]. It yields, after some manipulations, [itex]x^2+y^2=constant[/itex]. That means that the trajectory is a circle in fact, centred at the origin.
Is my process correct?
 
Last edited:
  • #6
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
23,243
14,754
OK, we take [itex]\vec x=x\hat i+y\hat j[/itex]. Also, [itex]x[/itex] comp. of velocity, [itex]v_x=5y[/itex], and, [itex]v_y=-5x[/itex]. What may I do next? It follows that [itex]\frac{dx}{y}=5dt[/itex]. How may I integrate this? Please help
To stop you going round in circles, why not differentiate the equations again.
 
  • #7
PKM
49
16
To stop you going round in circles, why not differentiate the equations again.
Which eqns do you mean? Is it [itex]d\vec x/dt=5(y\hat i-x\hat j)[/itex]?
 
  • #8
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
38,802
8,198
Which eqns do you mean? Is it [itex]d\vec x/dt=5(y\hat i-x\hat j)[/itex]?
Yes, but it is hard to work with that because two different notations are being used for vectors. Write ##\vec x## in terms of ##\hat i## and ##\hat j##.
 
  • #9
PKM
49
16
Yes, but it is hard to work with that because two different notations are being used for vectors. Write ##\vec x## in terms of ##\hat i## and ##\hat j##.
Yeah, I've hit this method. Please see post #5. Is it worth what I've done?
 
  • #10
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
38,802
8,198
Yeah, I've hit this method. Please see post #5. Is it worth what I've done?
Ah, I didn't read PeroK's remark with sufficient context.

You differentiated once and got an equation with ##\dot x## and y, and another with ##\dot y## and x.
Try differentiating each a second time. Can you now get an equation involving x and its derivatives but no mention of y?
 
  • #11
PKM
49
16
Ah, I didn't read PeroK's remark with sufficient context.

You differentiated once and got an equation with ##\dot x## and y, and another with ##\dot y## and x.
Try differentiating each a second time. Can you now get an equation involving x and its derivatives but no mention of y?
Let's see. I get [itex]\ddot x=-25\dot x[/itex], and [itex]\ddot y=-25\dot y[/itex].
Okay, these two fine differential equations should represent the trajectory, I suppose? Now, what should be the shape of the path?
Is my process in Post #5 is acceptable?
 
  • #12
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
19,874
10,411
Let's see. I get [itex]\ddot x=-25\dot x[/itex], and [itex]\ddot y=-25\dot y[/itex].
Okay, these two fine differential equations should represent the trajectory, I suppose? Now, what should be the shape of the path?
Is my process in Post #5 is acceptable?
Please show how you got there and not just the final result you got. It is impossible to identify where you have gone wrong if you don't.
 
  • #13
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
38,802
8,198
Let's see. I get [itex]\ddot x=-25\dot x[/itex], and [itex]\ddot y=-25\dot y[/itex].
You have made a mistake somewhere. If you cannot find it, please post your working.
 
  • #15
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
38,802
8,198
your process in #5 is not correct as it leads to false conclusions
Does it? The circular motion looks right to me.
 
  • #16
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
19,874
10,411
Does it? The circular motion looks right to me.
Yes, I deleted that already.
 
  • #17
PKM
49
16
Please show how you got there and not just the final result you got. It is impossible to identify where you have gone wrong if you don't.
Okay, I better show the way I've reached the conclusion of Post #5.
I get [itex]\frac{dx}{y}=5dt[/itex] and [itex]\frac{dy}{x}=-5dt[/itex]. From these results, I concluded that [itex]\frac{dx}{y}=-\frac{dy}{x}[/itex](?), Or,[itex]xdx+ydy=0[/itex]. Hence it follows that [itex]x^2+y^2=constant[/itex].
I have a confusion with the step marked (?). Is is acceptable?
 
  • #18
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
19,874
10,411
It is correct yes. However, it does not tell you what ##x## and ##y## are as functions of ##t##. It only tells you that the motion is in a circle. Can you think of a way to parametrise ##x^2 + y^2 = R^2##?
 
  • #19
PKM
49
16
You have made a mistake somewhere. If you cannot find it, please post your working.
Oh, I had blotched it! It should have been [itex]\ddot x=-25x[/itex], and [itex]\ddot y=-25y[/itex]. It represents SHM.
 
Last edited:
  • #20
PKM
49
16
It is correct yes. However, it does not tell you what ##x## and ##y## are as functions of ##t##. It only tells you that the motion is in a circle. Can you think of a way to parametrise ##x^2 + y^2 = R^2##?
The parametric eqns of the circle may be [itex]x=R\cos \theta, y=R\sin \theta[/itex].
Then?
 
  • #21
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
19,874
10,411
Yes, then what? What is ##\theta## as a function of ##t##?
 
  • #22
PKM
49
16
Yes, then what? What is ##\theta## as a function of ##t##?
Of course, [itex]\theta =\omega t[/itex]. If the particle has constant speed, then [itex]\omega[/itex] is the angular velocity. Therefore the eqn of the x-coordinate of the particle is [itex]x=R\cos (\omega t)[/itex], or simple harmonic, isn't it?
 
  • #23
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
19,874
10,411
Of course, [itex]\theta =\omega t[/itex].
You need to show that this is the case. You cannot just assume it.

If the particle has constant speed, then [itex]\omega[/itex] is the angular velocity. Therefore the eqn of the x-coordinate of the particle is [itex]x=R\cos (\omega t)[/itex], or simple harmonic, isn't it?
What makes you think that the particle has constant speed? (It has, but you need to argue for this!)
 
  • #24
PKM
49
16
You need to show that this is the case. You cannot just assume it.


What makes you think that the particle has constant speed? (It has, but you need to argue for this!)
I find it quite simple to account for.
The magnitude of the velocity of the particle, at any point is, [itex]\sqrt {x^2+y^2}=R[/itex] (follows from the given velocity of the particle). That suggests that it is constant.
Moreover, as the speed is constant, ##\theta## must equal to ##\omega t##.

I feel this method very fine, but we may reach at the same conclusion through calculus method (as suggested by Mr haruspex). Ref. Post #19.
 
Last edited:

Suggested for: How to find the Trajectory of a Particle

Replies
7
Views
84
Replies
1
Views
258
Replies
16
Views
645
Replies
3
Views
436
Replies
76
Views
2K
  • Last Post
Replies
3
Views
355
Replies
4
Views
352
Top