How to find the Trajectory of a Particle

[PKM]

Homework Statement

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

Homework Equations

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

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.

 

[Orodruin]

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?

 

[PKM]

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 \frac{d\vec x}{y\hat i-x\hat j}=5dt. Is it what you are suggesting? Please clarify.

 

[Orodruin]

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?

 

[PKM]

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 \vec x=x\hat i+y\hat j. Also, x comp. of velocity, v_x=5y, and, v_y=-5x. What may I do next? It follows that \frac{dx}{y}=5dt. Again, \frac{dy}{x}=-5dt. It yields, after some manipulations, x^2+y^2=constant. That means that the trajectory is a circle in fact, centred at the origin.
Is my process correct?

 

[PeroK]

OK, we take \vec x=x\hat i+y\hat j. Also, x comp. of velocity, v_x=5y, and, v_y=-5x. What may I do next? It follows that \frac{dx}{y}=5dt. How may I integrate this? Please help

To stop you going round in circles, why not differentiate the equations again.

 

[PKM]

To stop you going round in circles, why not differentiate the equations again.

Which eqns do you mean? Is it d\vec x/dt=5(y\hat i-x\hat j)?

 

[haruspex]

Which eqns do you mean? Is it d\vec x/dt=5(y\hat i-x\hat j)?

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 $\hat j$.

 

[PKM]

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?

 

[haruspex]

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?

 

[PKM]

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 \ddot x=-25\dot x, and \ddot y=-25\dot y.
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?

 

[Orodruin]

Let's see. I get \ddot x=-25\dot x, and \ddot y=-25\dot y.
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.

 

[haruspex]

Let's see. I get \ddot x=-25\dot x, and \ddot y=-25\dot y.

You have made a mistake somewhere. If you cannot find it, please post your working.

 

[Orodruin]

To stop you going round in circles

 

[haruspex]

your process in #5 is not correct as it leads to false conclusions

Does it? The circular motion looks right to me.

 

[Orodruin]

Does it? The circular motion looks right to me.

Yes, I deleted that already.

 

[PKM]

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 \frac{dx}{y}=5dt and \frac{dy}{x}=-5dt. From these results, I concluded that \frac{dx}{y}=-\frac{dy}{x}(?), Or,xdx+ydy=0. Hence it follows that x^2+y^2=constant.
I have a confusion with the step marked (?). Is is acceptable?

 

[Orodruin]

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 $x^2 + y^2 = R^2$?

 

[PKM]

You have made a mistake somewhere. If you cannot find it, please post your working.

Oh, I had blotched it! It should have been \ddot x=-25x, and \ddot y=-25y. It represents SHM.

 

[PKM]

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 x=R\cos \theta, y=R\sin \theta.
Then?

 

[Orodruin]

Yes, then what? What is $\theta$ as a function of $t$?

 

[PKM]

Yes, then what? What is $\theta$ as a function of $t$?

Of course, \theta =\omega t. If the particle has constant speed, then \omega is the angular velocity. Therefore the eqn of the x-coordinate of the particle is x=R\cos (\omega t), or simple harmonic, isn't it?

 

[Orodruin]

Of course, \theta =\omega t.

You need to show that this is the case. You cannot just assume it.

If the particle has constant speed, then \omega is the angular velocity. Therefore the eqn of the x-coordinate of the particle is x=R\cos (\omega t), 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!)

 

[PKM]

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, \sqrt {x^2+y^2}=R (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.