Homework Help: Projectile motion particle problem

1. Dec 17, 2005

vaishakh

A particle starts moves in the plane xy with constant acceleration w directed along the negative y direction. The equation of the motion of the particle has the form y = ax - bx^2, where a and b are positive constants. Find the velocity of the particle at the origin of the co-ordinates.
The problem maker has completed the question without giving us from where did the particle start. So we even do not know the velocity of the particle along the y direction, thus e cannot relate it along the X-axis. So I think that the problem must be with the particle moving with constant velocity w along the negative Y-axis.
But anyway we have not been given when does the object pass through the origin. Thus we cannot make a time equation to solve the question. Anyway the slope of the XY graph as it reaches the origin is a. therefore the velocity of the particle along Y-axis is -a times the velocity of the particle along the X-axis. Therefore if the velocity of the particle along the X-axis is v, then v = -aw. Therefore the magnitude of velocity is (v^2 + w^2)^1/2 which we finally get as w(1 + a^2)^1/2. I don't know whether my work is correct in this case. Also I want to know whether the problem can be solved with respect to a, b and w when w is the constant acceleration of the particle along the negative Y- direction.

2. Dec 17, 2005

mrjeffy321

I think this problem looks a whole lot like a projectile motion problem if you think about it.
We know that the derivative of the position-time graph is the velocity with respect to time, or in this case x. So if we want to know what the velocity is when x and y = 0, then we take the derivative, then plug in the value for x that makes y zero. In this case it is obvious that is x = 0, y = 0, and in tha case, the velocity is zero too (which makes sense since the object is at the top of its parabolic path).

3. Dec 17, 2005

Astronuc

Staff Emeritus
Correct.
Or perhaps the way to say it is find the x,y when dy/dx = 0, which is when dy/dt = 0. Also, for x, y > 0, solve for x when y = 0.

There is some initial vx and vy at (x,y) = (0,0), where the problem starts at t0.

But more generally, is the case where vx is constant but vy = f(t), including for x,y ≤ 0

Last edited: Dec 17, 2005
4. Dec 18, 2005

vaishakh

Sorry, I didn't understand what you mean? the particle can still continue moving after it reached origin. The parabolic graph ax + bx^2 has never a maxima or minima to say that the particle will be at rest. Or infact the particle is given to be constantly moving along the Y-direction.

5. Dec 18, 2005

Astronuc

Staff Emeritus
As mrjeffy321 indicated the equation y = ax-bx2 is a parabolic trajectory, which happens to pass through the origin x,y = 0, 0. However, the initial conditions are not given. The trajectory is simply a path or constraint on the motion of the particle.

Certainly taking dy/dx = 0 indicates an extremum. dy/dt = 0 at that point.

The original problem statement - "A particle starts moves in the plane xy with constant acceleration w directed along the negative y direction." Where or when does the particle start moving?

There is no initial position or velocity given, and one has a general equation.

There are two cases, vx>0 (particle moving right, +x), or <0 (particle moving left).

The parabolic trajectory indicates that vx is constant if vy is vy,o-gt, where g = W/m. g is acting downward.

Last edited: Dec 18, 2005
6. Dec 19, 2005

vaishakh

But that way we don't know t, then how can we comment on the velocity? Okay now I understood the case when w is constant acceleration along -Y direction. But equating to t cannot yield any result.

7. Dec 20, 2005

vaishakh

I don't agree with the point that the velocity along x direction should be constant with respect to time. The path of motion depends on values of a and b, so the path would remain the same for differant values of w. This makes it clear that there should be some appropriate acceleration in x-direction during certian conditions.

8. Dec 20, 2005

Staff: Mentor

But the first sentence of your original post stated:
And the given equation describes a vertical parabola, consistent with that statement.

9. Dec 20, 2005

mukundpa

Differentiate the equation wrt t you will get relation between vx and vy

differentiate again this equation wrt t and put ax =0; ay = -w and x =0 you will get vx at origin. with relation between velocities get vy at origin.

MP

10. Dec 21, 2005

vaishakh

Ya, I feel it from my heart that that's true Mukundpa. Anyway I solved it using the same concept earlier.
Doc Al-acceleration in y axis being constant has nothing to say about x-axis.

11. Dec 21, 2005

Staff: Mentor

Reread the first line of your first post in this thread.

The way I read it, it doesn't say "the acceleration along the y-axis is constant"; it says "the acceleration is constant and is directed along the negative y-axis" (a much stronger statement).

In any case, you should recognize that trajectory (a vertical parabola) as that of a projectile under gravity!

12. Dec 21, 2005

Astronuc

Staff Emeritus
Just adding to what Doc Al and others have mentioned, in addition to the constant acceleration directed along the - y axis, the equation

y = ax - bx2, provides a constraint on x, with which one can show that y = 0 when dy/dx = 0.

Determine y = y(t) and one can then find x = x(t).

One should be able to show that vx must be some constant, but it could be in either positive or negative direction. However, no IC are specified.

13. Dec 22, 2005

vaishakh

How dare you argue there again? We had once accepted each other that the maxima does not lie at origin for the given equation.

14. Dec 22, 2005

Astronuc

Staff Emeritus
Sorry, I typed incorrectly
should read vy=0 i.e. dy/dt=0, at dy/dx=0. I thought that perhaps not clear.

15. Dec 22, 2005

Staff: Mentor

Yes, Astronuc made a typo. Give him a break!

Still messing with this one? Here's what I'd do:

(1) Find the apex of the motion by setting dy/dx = 0.
(2) Find the points where y = 0.
(3) Since you know the vertical (and horizontal) distance traveled from origin to apex, and the vertical acceleration, calculate the time.
(4) Use that to find the horizontal (and vertical) component of the velocity.