Trajectory of a particle

  • A
  • Thread starter Taufik
  • Start date
  • #1
Taufik
Suppose a particle is moving in an X-Y plane. It's velocity in X direction will be dx/dt and in Y direction will be dy/dt. Suppose at a certain point it's velocity in x direction is zero and in y direction is also zero. Then, dx/dt =0 & dy/dt =0 at that point.
Now, what will the trajectory be near that point? Since dy/dx = 0/0, which is not defined, therefore the slope at that point is not defined. So, will it be something like a sharp edge or a smooth curve?
 

Answers and Replies

  • #2
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,869
1,446
Since dy/dx = 0/0
That is not right. The rule that ##\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}## only applies at points where ##\frac{dx}{dt}\neq 0##, and hence not at that point.

In fact, the particle's path may have no direction at that point. Consider a particle that moves from the origin (0,0) to (0,1), decelerating to come to a stop at (0,1), then waits a few seconds there before heading up towards (1,1) in a straight line. The path of the particle has a kink/corner in it at (0,1), so there is no 'trajectory' at that point. On the other hand, if instead of going from (0,1) to (1,1) it goes from (0,1) to (0,2) we can speak meaningfully of the 'trajectory' or 'direction of the path' at the point (0,1), which is rightwards along the X axis.

In a real physical system, there will always be a specific velocity, and hence direction, at any point in time, but not necessarily at a point in space.

Another way to see the problems of asking for direction at a point in space is to consider a particle mapping out a figure eight. What is the direction at the crossing point?
 
  • #3
Taufik
What about the path traced out by a particle on the circumference of a disc undergoing pure rolling motion? The particle at the bottommost has zero velocity in both X and Y direction. Can we directly say that we cannot find dy/dx at that point since velocity is zero in both X & Y direction, there must be something like a kink or a sharp curve instead of a smooth curve?
 
  • #4
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,869
1,446
Yes. That path is called a cycloid. See this wiki page. And it does have a kink at the point it touches the ground, so there is no value of dy/dx at those points. But the reason the path is not differentiable at that point is not that the X and Y velocities are zero. In my previous post I gave an example of a path that has those velocities zero at the point (0,1), but which is still differentiable at that point.

 
  • #5
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
2019 Award
25,719
8,905
Then, dx/dt =0 & dy/dt =0 at that point.
Then it's stopped.

Now, what will the trajectory be near that point?
It's still stopped.

The fact that it's stopped doesn't tell you how it was moving before it stopped, nor how it will move at some future time.
 
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
16,195
7,523
It's not necessarily stopped (except for a free particle, i.e., if no forces are acting). Take an isotropic harmonic oscillator as a simple example. You have
$$\ddot{\vec{x}}=-\omega^2 \vec{x}.$$
Take the initial condition ##\vec{v}(0)=0## and ##\vec{x}(0)=\vec{x}_0 \neq 0## then the solution obviously is
$$\vec{x}(t)=\vec{x}_0 \cos(\omega t).$$
Everything is smooth as a function of ##t##. As a function ##y=y(x)## there's of course a singularity since ##t=0## (by construction) is the turning point of the linear motion along the direction ##\vec{x}_0##.
 

Related Threads on Trajectory of a particle

  • Last Post
Replies
7
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
9
Views
1K
Replies
2
Views
314
Replies
23
Views
560
Replies
4
Views
11K
Replies
0
Views
1K
Replies
5
Views
626
  • Last Post
Replies
2
Views
1K
Top